canadian maths

The Canadian Mathematical Societyin collaboration with

The CENTRE for EDUCATIONin MATHEMATICS and COMPUTING

Time: 2 1

2 hours

Calculators are NOT permitted.

Do not open this booklet until instructed to do so.There are two parts to the paper.

PART AThis part of the paper consists of 8 questions, each worth 5 marks. You can earn full value for eachquestion by entering the correct answer in the space provided. Any work you do in obtaining an answerwill be considered for part marks if you do not have the correct answer, provided that it is done in thespace allocated to that question in your answer booklet.

PART BThis part of the paper consists of 4 questions, each worth 10 marks. Finished solutions must be writtenin the appropriate location in the answer booklet. Rough work should be done separately. If you requireextra pages for your finished solutions, foolscap will be provided by your supervising teacher. Any extrapapers should be placed inside your answer booklet.Marks are awarded for completeness, clarity, and style of presentation. A correct solution poorlypresented will not earn full marks.

NOTE: At the completion of the contest, insert the information sheet inside the answer booklet.

© 2000 Canadian Mathematical Society

The

Canadian OpenMathematics Challenge

Wednesday, November 29, 2000

Canadian Open Mathematics Challenge

NOTE: 1. Please read the instructions on the front cover of this booklet.2. Write solutions in the answer booklet provided.3. It is expected that all calculations and answers will be expressed as exact numbers such

as 4 2 7p, + , etc.4. Calculators are not allowed.

PART A

1. An operation “D ” is defined by a ba

bD =1 – , b π 0.

What is the value of 1 2 3 4D D D( ) ( )?

2. The sequence 9, 18, 27, 36, 45, 54, … consists of successive multiples of 9. This sequence is thenaltered by multiplying every other term by –1, starting with the first term, to produce the new

sequence – , , – , , – , ,...9 18 27 36 45 54 . If the sum of the first n terms of this new sequence is 180,determine n.

3. The symbol n! is used to represent the product n n n– –1 2 3 2 1( )( ) ( )( )( )L .

For example, 4 4 3 2 1!= ( )( )( ). Determine n such that n!= ( )( )( )( )( )( )2 3 5 7 11 1315 6 3 2 .

4. The symbol xÎ û means the greatest integer less than or equal to x. For example,

5 7 5.Î û = , pÎ û = 3 and 4 4Î û = .

Calculate the value of the sum

1 2 3 4 48 49 50Î û + Î û + Î û + Î û + + Î û + Î û + Î ûL .

5. How many five-digit positive integers have the property that the product of their digits is 2000?

6. Solve the equation 4 16 22 6sin sinx xÊ

Ë�¯ = , for 0 2£ £x p .

7. The sequence of numbers …, a a a a a a a– – –, , , , , ,3 2 1 0 1 2 3, … is defined by a n a nn n– –+( ) = +( )1 322 ,

for all integers n. Calculate a0.

8. In the diagram, D ABC is equilateral and the radius of itsinscribed circle is 1. A larger circle is drawn through the verticesof the rectangle ABDE . What is the diameter of the larger circle?

E C D

A B

PART B

1. Triangle ABC has vertices A 0 0,( ), B 9 0,( ) and C 0 6,( ). The points P and Q lie on side AB such

that AP PQ QB= = . Similarly, the points R and S lie on side AC so that AR RS SC= = .The vertex C is joined to each of the points P and Q. In the same way, B is joined to R and S.

(a) Determine the equation of the line through the points R and B.

(b) Determine the equation of the line through the points P and C.

(c) The line segments PC and RB intersect at X, and the line segments QC and SB intersect at Y.

Prove that the points A, X and Y lie on the same straight line.

2. In D ABC , the points D, E and F are on sides BC , CA and AB,

respectively, such that � = �AFE BFD , � = �BDF CDE , and

� = �CED AEF .

(a) Prove that � = �BDF BAC .(b) If AB = 5, BC = 8 and CA = 7, determine the length of BD.

3. (a) Alphonse and Beryl are playing a game, starting with thegeometric shape shown in Figure 1. Alphonse begins thegame by cutting the original shape into two pieces alongone of the lines. He then passes the piece containing theblack triangle to Beryl, and discards the other piece.Beryl repeats these steps with the piece she receives; that is to say, she cuts along the length ofa line, passes the piece containing the black triangle back to Alphonse, and discards the otherpiece. This process continues, with the winner being the player who, at the beginning of his orher turn, receives only the black triangle. Show, with justification, that there is always awinning strategy for Beryl.

(b) Alphonse and Beryl now play a game with the same rulesas in (a), except this time they use the shape in Figure 2and Beryl goes first. As in (a), cuts may only be madealong the whole length of a line in the figure. Is there astrategy that Beryl can use to be guaranteed that she willwin? (Provide justification for your answer.)

4. A sequence t t t tn1 2 3, , , ..., of n terms is defined as follows:

t1 1= , t2 4= , and t t tk k k= +– –1 2 for k n= 3 4, , ..., .

Let T be the set of all terms in this sequence; that is, T t t t tn= { }1 2 3, , , ..., .

(a) How many positive integers can be expressed as the sum of exactly two distinct elements ofthe set T ?

(b) How many positive integers can be expressed as the sum of exactly three distinct elements ofthe set T ?

Figure 1

Figure 2

A

E

CDB

F



Time: 212

hours




PART BThis part of the paper consists of 4 questions, each worth 10 marks. Finished solutions must be writtenin the appropriate location in the answer booklet. Rough work should be done separately. If you requireextra pages for your finished solutions, paper will be provided by your supervising teacher. Any extrapapers should be placed inside your answer booklet.Marks are awarded for completeness, clarity, and style of presentation. A correct solution poorlypresented will not earn full marks.



TheCanadian Open

Mathematics Challenge




as 4 2 7p, + , etc.4. Calculators are not allowed.

1. An operation “ —” is defined by a b a b— = +2 3 .

What is the value of 2 0 0 1—( )— —( )?

2. In the given diagram, what is the value of x?

3. A regular hexagon is a six-sided figure which has all of its angles equal and all of its side lengths equal.If P and Q are points on a regular hexagon which has a side length of 1, what is the maximum possiblelength of the line segment PQ?

4. Solve for x:

2 2 4 642x x( ) = + .

5. Triangle PQR is right-angled at Q and has side lengths PQ = 14and QR = 48. If M is the midpoint of PR, determine the cosine of

– MQP .

6. The sequence of numbers t t t1 2 3, , , ... is defined by and tt

tnn

n+ =

+111

–, for every positive integer n.

Determine the numerical value of t999 .

7. If a can be any positive integer and

2x a ya y xx y z

+ =+ =+ =

determine the maximum possible value for x + y + z.

D

E G

F

C

B

A3x∞

4x∞

5x∞

6x∞

2x∞

P

Q R

M

t1 2=

8. The graph of the function y g x= ( ) is shown.

Determine the number of solutions of the equation

g x( ) =– 1 12

.

PART B

1. The triangular region T has its vertices determined by the intersections of the three lines x + 2y = 12,x = 2 and y = 1.(a) Determine the coordinates of the vertices of T, and show this region on the grid provided.(b) The line x + y = 8 divides the triangular region T into a quadrilateral Q and a triangle R.

Determine the coordinates of the vertices of the quadrilateral Q.(c) Determine the area of the quadrilateral Q.

2. (a) Alphonse and Beryl are playing a game, starting with a pack of 7 cards. Alphonse begins bydiscarding at least one but not more than half of the cards in the pack. He then passes the remainingcards in the pack to Beryl. Beryl continues the game by discarding at least one but not more than halfof the remaining cards in the pack. The game continues in this way with the pack being passed backand forth between the two players. The loser is the player who, at the beginning of his or her turn,receives only one card. Show, with justification, that there is always a winning strategy for Beryl.

(b) Alphonse and Beryl now play a game with the same rules as in (a), except this time they start witha pack of 52 cards, and Alphonse goes first again. As in (a), a player on his or her turn must discardat least one and not more than half of the remaining cards from the pack. Is there a strategy thatAlphonse can use to be guaranteed that he will win? (Provide justification for your answer.)

3. (a) If f x x x c( ) = + +2 6 , where c is an integer, prove that f f0 1( ) + ( )– is odd.

(b) Let g x x px qx r( ) = + + +3 2 , where p, q and r are integers. Prove that if g 0( ) and g –1( ) are both odd,

then the equation g x( ) = 0 cannot have three integer roots.

4. Triangle ABC is isosceles with AB = AC = 5 and BC = 6. Point D

lies on AC and P is the point on BD so that – = ∞APC 90 . If

– = –ABP BCP , determine the ratio AD:DC.

y

x0

2

4

– 4 – 2 2 4

– 2

– 4

A

B C

D

P



Time: 212

hours







TheCanadian Open





as 4 2 7π, + , etc.4. Calculators are not allowed.

PART A

1. In triangle PQR, F is the point on QR so that PF is perpendicularto QR. If PR = 13, RF = 5, and FQ = 9, what is the perimeterof ∆PQR?

2. If x y+ = 4 and xy = −12, what is the value of x xy y2 25+ + ?

3. A regular pentagon is a five-sided figure which has all of itsangles equal and all of its side lengths equal. In the diagram,TREND is a regular pentagon, PEA is an equilateral triangle, andOPEN is a square. Determine the size of ∠ EAR .

4. In a sequence of numbers, the sum of the first n terms is equal to 5 62n n+ . What is the sum of the

3rd, 4th and 5th terms in the original sequence?

5. If m and n are non-negative integers with m n< , we define m n∇ to be the sum of the integers fromm to n, including m and n. For example, 5 8 5 6 7 8 26∇ = + + + = .

For every positive integer a, the numerical value of 2 1 2 1

1 1

a a

a a

−( )∇ +( )[ ]−( )∇ +( )[ ] is the same. Determine this

value.

6. Two mirrors meet at an angle of 30o at the point V. A beam oflight, from a source S, travels parallel to one mirror and strikesthe other mirror at point A, as shown. After a number ofreflections, the beam comes back to S. If SA and AV are both1 metre in length, determine the total distance travelled by thebeam.

7. N is a five-digit positive integer. A six-digit integer P is constructed by placing a 1 at the right-handend of N. A second six-digit integer Q is constructed by placing a 1 at the left-hand end of N. If P isthree times Q, determine the value of N.

P

Q F R9 5

13

AS

V30°

A

P

O ND

R

TE

8. Suppose that M is an integer with the property that if x is randomly chosen from the set

1 2 3 999 1000, , , , ,K{ } , the probability that x is a divisor of M is 1100 . If M ≤ 1000, determine the

maximum possible value of M.

PART B

1. Square ABCD has vertices A 0 0,( ) , B 0 8,( ), C 8 8,( ) , and D 8 0,( ). The points P 0 5,( ) and Q 0 3,( ) are onside AB , and the point F 8 1,( ) is on side CD.(a) What is the equation of the line through Q parallel to the line through P and F?(b) If the line from part (a) intersects AD at the point G, what is the equation of the line through F

and G?(c) The centre of the square is the point H 4 4,( ) . Determine the equation of the line through H

perpendicular to FG.(d) A circle is drawn with centre H that is tangent to the four sides of the square. Does this circle

intersect the line through F and G? Justify your answer. (A sketch is not sufficient justification.)

2. (a) Let A and B be digits (that is, A and B are integers between 0 and 9 inclusive). If the product ofthe three-digit integers 2 5A and 13B is divisible by 36, determine with justification the fourpossible ordered pairs A B,( ) .

(b) An integer n is said to be a multiple of 7 if n k= 7 for some integer k.(i) If a and b are integers and 10 7a b m+ = for some integer m, prove that a b− 2 is a multiple

of 7.(ii) If c and d are integers and 5 4c d+ is a multiple of 7, prove that 4c d− is also a multiple

of 7.

3. There are some marbles in a bowl. Alphonse, Beryl and Colleen each take turns removing one or twomarbles from the bowl, with Alphonse going first, then Beryl, then Colleen, then Alphonse again, andso on. The player who takes the last marble from the bowl is the loser, and the other two players arethe winners.(a) If the game starts with 5 marbles in the bowl, can Beryl and Colleen work together and force

Alphonse to lose?(b) The game is played again, this time starting with N marbles in the bowl. For what values of N

can Beryl and Colleen work together and force Alphonse to lose?

4. Triangle DEF is acute. Circle C1 is drawn with DF as its

diameter and circle C2 is drawn with DE as its diameter.

Points Y and Z are on DF and DE respectively so that EY and

FZ are altitudes of ∆DEF . EY intersects C1 at P, and FZ

intersects C2 at Q. EY extended intersects C1 at R, and FZ

extended intersects C2 at S. Prove that P, Q, R, and S are

concyclic points.

D

E F

C2

C1



Time: 212

hours



PART AThis part of the paper consists of 8 questions, each worth 5. You can earn full value for each questionby entering the correct answer in the space provided. Any work you do in obtaining an answer will beconsidered for part marks if you do not have the correct answer, provided that it is done in the spaceallocated to that question in your answer booklet.

PART BThis part of the paper consists of 4 questions, each worth 10. Finished solutions must be written in theappropriate location in the answer booklet. Rough work should be done separately. If you require extrapages for your finished solutions, paper will be provided by your supervising teacher. Any extra papersshould be placed inside your answer booklet.Marks are awarded for completeness, clarity, and style of presentation. A correct solution poorlypresented will not earn full marks.



TheCanadian Open





as 4 2 7π, + , etc.4. Calculators are not allowed.

PART A

1. Jeff, Gareth and Ina all share the same birthday. Gareth is one year older than Jeff, and Ina is two yearsolder than Gareth. This year the sum of their ages is 118. How old is Gareth?

2. The point 4 2,−( ) is reflected in the x-axis. The resultingpoint is then reflected in the line with equation y x= . Whatare the coordinates of the final point?

3. A circle of radius 1 is centred at the origin. Two particles start

moving at the same time from the point 1 0 ,( ) and move aroundthe circle in opposite directions. One of the particles movescounterclockwise with constant speed v and the other moves

clockwise with constant speed 3v. After leaving 1 0 ,( ), the twoparticles meet first at point P, and continue until they meet nextat point Q. Determine the coordinates of the point Q.

4. Two different numbers are chosen at random from the set 0 1 2 3 4, , , ,{ }. What is the probability that

their sum is greater than their product?

5. In the diagram, square ABCD has a side length of 6. Circulararcs of radius 6 are drawn with centres B and D. What is the areaof the shaded region?

6. The symbol a means the greatest integer less than or equal to a.

For example, 5 7 5 4 4 4 2 5. , . = = − = − and .

Determine all values of x for which 3 4

5x x

+

= .

y

x

(4, – 2)

y = x

y

x(1, 0)

D C

A B

7. Each of the points P 4 1 ,( ), Q 7 8 ,−( ) and R 10 1, ( ) is the midpoint of a radius of the circle C. Determinethe length of the radius of circle C.

8. Determine the number of triples k l m, , ( ) of positive integers such that

k l m

k l m

+ + =

+ + =

97

45

56

67

82

eg

a b

df

h

i

c

k

PART B

1. In the diagram shown, whole numbers are to beplaced in the ten circles so that the sum of thenumbers in the circles along any of the ten straightlines is 15. For example, a g k+ + = 15 ande i+ = 15.(a) If k = 2 and e = 5, fill in the whole numbers

that go in all of the circles in the diagram.(b) Suppose that k = 2 and the value of e is

unknown.(i) Find a formula for each of b and c in terms

of e. A clearly labelled diagram is sufficientexplanation.

(ii) Show that e must be equal to 5.

(c) Suppose now that k x= , where x is unknown. Prove that e must still be equal to 5.

2. A barn has a foundation in the shape of a trapezoid, with three sidesof length 6 m, and one side of length 12 m, as shown.(a) Determine each of the interior angles in the trapezoid.(b) Chuck the Llama is attached by a chain to a point on the outside

wall of the barn. Chuck is smarter than the average llama, and sorealizes that he can always reach the area between the barn andwhere the chain is fully extended.(i) If Chuck is attached at the point A with a chain of length 8 m, what is the area outside the barn

that Chuck can reach?(ii) If Chuck is attached at some point P along the wall between A and B with a chain of length

15 m, determine the location of P which restricts Chuck to the minimum area.

3. (a) In the diagram, the two circles C1 and C2 have acommon chord AB. Point P is chosen on C1 so that it isoutside C2. Lines PA and PB are extended to cut C2 atX and Y, respectively. If AB = 6, PA = 5, PB = 7 andAX = 16, determine the length of XY.

(b) Two circles C3 and C4 have a common chord GH. Point Q is chosen on C3 so that it is outsideC4. Lines QG and QH are extended to cut C4 at V and W, respectively. Show that, no matterwhere Q is chosen, the length of VW is constant.

D A

C B

6 6

6

12

X

Y

B

AP

C2

C1

over ...

2003C

anadian Open

Mathem

aticsC

hallenge(E

nglish)

4. The polynomial equation x x x3 26 5 1 0− + − = has three real roots a, b and c.

(a) Determine the value of a b c5 5 5+ + .

(b) If a b c< < , show that c2004 is closer to its nearest integer than c2003 is to its nearest integer.



presents the

Canadian OpenMathematics Challenge


Time: 2 12 hours c©2004 Canadian Mathematical Society


Do not open this booklet until instructed to do so.There are two parts to this paper.

PART AThis part of the paper consists of 8 questions, each worth 5 marks. You can earn full value for eachquestion by entering the correct answer in the space provided. Any work you do in obtaining an answerwill be considered for part marks if you do not have the correct answer,provided that it is done in thespace allocatedto that question in your answer booklet.


NOTE: At the completion of the contest, insert the information sheet inside the answerbooklet.


NOTE: 1. Please read the instructions on the front cover of this booklet.2. Write solutions in the answer booklet provided.3. It is expected that all calculations and answers will be expressed as

exact numbers such as4π, 2 +√

7, etc.4. Calculators arenot allowed.

PART A

1. If x + 2y = 84 = 2x + y, what is the value ofx + y?

2. LetS be the set of all three-digit positive integers whose digits are 3, 5 and 7, with no digitrepeated in the same integer. Calculate the remainder when the sum of all of the integers inS is divided by 9.

3. In the diagram, pointE has coordinates(0, 2), andB lies onthe positivex-axis so thatBE =

√7. Also, pointC lies on

the positivex-axis so thatBC = OB. If point D lies in thefirst quadrant such that∠CBD = 30◦ and∠BCD = 90◦,what is the length ofED?

Ox

y

B

E

D

C

4. A functionf(x) has the following properties:

i) f(1) = 1

ii) f(2x) = 4f(x) + 6

iii) f(x + 2) = f(x) + 12x + 12

Calculatef(6).

5. The Rice Tent Company sells tents in two different sizes, large and small. Last year, theCompany sold 200 tents, of which one quarter were large. The sale of the large tents pro-duced one third of the company’s income. What was the ratio of the price of a large tent tothe price of a small tent?

6. In the diagram, a square of side length 2 has semicirclesdrawn on each side. An “elastic band” is stretched tightlyaround the figure. What is the length of the elastic band inthis position?

7. Leta andb be real numbers, witha > 1 andb > 0.If ab = ab and

a

b= a3b, determine the value ofa.

8. A rectangular sheet of paper,ABCD, hasAD = 1 andAB = r, where1 < r < 2. Thepaper is folded along a line throughA so that the edgeAD falls onto the edgeAB. Withoutunfolding, the paper is folded again along a line throughB so that the edgeCB also lieson AB. The result is a triangular piece of paper. A region of this triangle is four sheetsthick. In terms ofr, what is the area of this region?

PART B

1. The pointsA(−8, 6) andB(−6,−8) lie on the circlex2 + y2 = 100.

(a) Determine the equation of the line throughA andB.

(b) Determine the equation of the perpendicular bisector ofAB.

(c) The perpendicular bisector ofAB cuts the circle at two points,P in the first quadrantandQ in the third quadrant. Determine the coordinates ofP andQ.

(d) What is the length ofPQ? Justify your answer.

2. (a) Determine the two values ofx such thatx2 − 4x− 12 = 0.

(b) Determine theonevalue ofx such thatx−√

4x + 12 = 0. Justify your answer.

(c) Determine all real values ofc such that

x2 − 4x− c−√

8x2 − 32x− 8c = 0

has precisely two distinct real solutions forx.

3. A map shows all Beryl’s Llamaburgers restaurant locations in North America. On this map,a line segment is drawn from each restaurant to the restaurant that is closest to it. Everyrestaurant has a unique closest neighbour. (Note that ifA andB are two of the restaurants,thenA may be the closest toB withoutB being closest toA.)

(a) Prove that no three line segments on the map can form a triangle.

(b) Prove that no restaurant can be connected to more than five other restaurants.

4. In asumac sequence, t1, t2, t3, . . ., tm, each term is an integer greater than or equal to 0.Also, each term, starting with the third, is the difference of the preceding two terms (thatis, tn+2 = tn − tn+1 for n ≥ 1). The sequence terminates attm if tm−1 − tm < 0. Forexample, 120, 71, 49, 22, 27 is a sumac sequence of length 5.

(a) Find the positive integerB so that the sumac sequence 150,B, . . . has the maximumpossible number of terms.

(b) Let m be a positive integer withm ≥ 5. Determine the number of sumac sequencesof lengthm with tm ≤ 2000 and with no term divisible by 5.

2004C

anadianO

penM

athematics

Challenge

(English)



presents the

Canadian Open

Mathematics ChallengeWednesday, November 23, 2005




PART AThis part of the paper consists of 8 questions, each worth 5 marks. You can earn full valuefor each question by entering the correct answer in the space provided. If you do not have thecorrect answer, any work you do in obtaining an answer will be considered for part marks,provided that it is done in the space allocated to that question in your answer booklet.

PART BThis part of the paper consists of 4 questions, each worth 10 marks. Finished solutions mustbe written in the appropriate location in the answer booklet. Rough work should be doneseparately. If you require extra pages for your finished solutions, paper will be provided byyour supervising teacher. Any extra papers should be placed inside your answer booklet. Besure to write your name and school name on any inserted pages.Marks are awarded for completeness, clarity, and style of presentation. A correct solutionpoorly presented will not earn full marks.

NOTE: At the completion of the contest, insert the information sheet inside theanswer booklet.



exact numbers such as 4π, 2 +√

7, etc.4. Calculators are not allowed.

PART A

1. Determine the value of 102 − 92 + 82 − 72 + 62 − 52 + 42 − 32 + 22 − 12.

2. A bug in the xy-plane starts at the point (1, 9). It moves first to the point (2, 10) andthen to the point (3, 11), and so on. It continues to move in this way until it reachesa point whose y-coordinate is twice its x-coordinate. What are the coordinates ofthis point?

3. If ax3 + bx2 + cx + d = (x2 + x− 2)(x− 4)− (x + 2)(x2 − 5x + 4) for all values of x,what is the value of a + b + c + d?

4. A fractionp

qis in lowest terms if p and q have no common factor larger than 1.

How many of the 71 fractions172

,272

, . . . ,7072

,7172

are in lowest terms?

5. An office building has 50 storeys, 25 of which are painted black and the other 25 ofwhich are painted gold. If the number of gold storeys in the top half of the buildingis added to the number of black storeys in the bottom half of the building, the sumis 28. How many gold storeys are there in the top half of the building?

6. In the grid shown, each row has a value assignedto it and each column has a value assigned to it.The number in each cell is the sum of its row andcolumn values. For example, the “8” is the sumof the value assigned to the 3rd row and the valueassigned to the 4th column. Determine the valuesof x and y.

3 0 5 6 −2−2 −5 0 1 y

5 2 x 8 00 −3 2 3 −5−4 −7 −2 −1 −9

7. In the diagram, the semi-circle has centre O and diameter AB. A ray of light leavespoint P in a direction perpendicular to AB. It bounces off the semi-circle at point Din such a way that ∠PDO = ∠EDO. (In other words, the angle of incidence equalsthe angle of reflection at D.) The ray DE then bounces off the circle in a similarway at E before finally hitting the semicircle again at B. Determine ∠DOP .

A B

E

D

P O

8. The number 18 is not the sum of any 2 consecutive positive integers, but is the sumof consecutive positive integers in at least 2 different ways, since 5 + 6 + 7 = 18 and3+4+5+6 = 18. Determine a positive integer less than 400 that is not the sum ofany 11 consecutive positive integers, but is the sum of consecutive positive integersin at least 11 different ways.

PART B

1. A line with slope −3 intersects the positive x-axis at A and the positive y-axis at B.A second line intersects the x-axis at C(7, 0) and the y-axis at D. The lines intersectat E(3, 4).

(a) Find the slope of the line through C and E.

(b) Find the equation of the line through Cand E, and the coordinates of the point D.

(c) Find the equation of the line through Aand B, and the coordinates of the point B.

(d) Determine the area of the shaded region.

x

D

B

O A C (7, 0)

E (3, 4)

y

2. (a) Determine all possible ordered pairs (a, b) such that

a− b = 12a2 + ab− 3b2 = 22

(b) Determine all possible ordered triples (x, y, z) such that

x2 − yz + xy + zx = 82y2 − zx + xy + yz = −18z2 − xy + zx + yz = 18

3. Four tiles identical to the one shown, with a > b > 0,are arranged without overlap to form a square with asquare hole in the middle.

a

a

b

b

(a) If the outer square has area (a + b)2, show that the area of the inner squareis (a− b)2.

(b) Determine the smallest integer value of N for which there are prime numbersa and b such that the ratio of the area of the inner square to the area of theouter square is 1 : N .

(c) Determine, with justification, all positive integers N for which there are oddintegers a > b > 0 such that the ratio of the area of the inner square to thearea of the outer square is 1 : N .

2005C

anadianO

penM

athematics

Challenge

(English)4. Triangle ABC has its base on line segment PN and vertex A on line PM . Circles

with centres O and Q, having radii r1 and r2, respectively, are tangent to the triangleABC externally and to each of PM and PN .

P N

M

A

B C

E

D

GF

KL

QO

(a) Prove that the line through K and L cuts the perimeter of triangle ABC intotwo equal pieces.

(b) Let T be the point of contact of BC with the circle inscribed in triangle ABC.Prove that (TC)(r1) + (TB)(r2) is equal to the area of triangle ABC.



presents the

Canadian Open

Mathematics ChallengeWednesday, November 22, 2006

Supported by:




PART AThis part of the paper consists of 8 questions, each worth 5 marks. You can earn full valuefor each question by entering the correct answer in the space provided. If you do not havethe correct answer, any work you do in obtaining an answer will be considered for part marks,provided that it is done in the space allocated to that question in your answer booklet.

PART BThis part of the paper consists of 4 questions, each worth 10 marks. Finished solutions mustbe written in the appropriate location in the answer booklet. Rough work should be doneseparately. If you require extra pages for your finished solutions, paper will be provided byyour supervising teacher. Any extra papers should be placed inside your answer booklet. Besure to write your name and school name on any inserted pages.Marks are awarded for completeness, clarity, and style of presentation. A correct solution poorlypresented will not earn full marks.

NOTES:At the completion of the contest, insert the information sheet inside the answerbooklet.The names of top scoring competitors will be published on the Web sites of theCMS and CEMC.




7, etc.4. Calculators are not allowed.

PART A

1. What is the value of(1 + 1

2

) (1 + 1

3

) (1 + 1

4

) (1 + 1

5

)?

2. If f(2x + 1) = (x− 12)(x + 13), what is the value of f(31)?

3. In4ABC, M is the midpoint of BC, as shown. If ∠ABM = 15◦

and ∠AMC = 30◦, what is the size of ∠BCA?

MB

A

C

4. Determine all solutions (x, y) to the system of equations

4x

+5y2

= 12

3x

+7y2

= 22

5. In 4ABC, BC = 4, AB = x, AC = x + 2, and cos(∠BAC) =x + 82x + 4

.

Determine all possible values of x.

6. Determine the number of integers n that satisfy all three of the conditions below:• each digit of n is either 1 or 0,

• n is divisible by 6, and

• 0 < n < 107.

7. Suppose n and D are integers with n positive and 0 ≤ D ≤ 9.Determine n if

n

810= 0.9D5 = 0.9D59D59D5 . . . .

8. What is the probability that 2 or more successive heads will occur at least once in10 tosses of a fair coin?

PART B

1. Piotr places numbers on a 3 by 3 grid using the following rule, called “Piotr’sPrinciple”:

For any three adjacent numbers in a horizontal, vertical or diagonalline, the middle number is always the average (mean) of its twoneighbours.

(a) Using Piotr’s principle, determine the missing numbers in thegrid to the right. (You should fill in the missing numbers inthe grid in your answer booklet.)

3 198

(b) Determine, with justification, the total of the nine numberswhen the grid to the right is completed using Piotr’sPrinciple.

x

5 23

(c) Determine, with justification, the values of x and y when thegrid to the right is completed using Piotr’s Principle.

x 79 y

20

2. In the diagram, the circle x2 + y2 = 25 intersectsthe x-axis at points A and B. The line x = 11intersects the x-axis at point C. Point P movesalong the line x = 11 above the x-axis and APintersects the circle at Q.

(a) Determine the coordinates of P when4AQB has maximum area. Justify youranswer.

(b) Determine the coordinates of P when Q isthe midpoint of AP . Justify your answer.

(c) Determine the coordinates of P when thearea of 4AQB is 1

4 of the area of 4APC.Justify your answer.

y

xA B

PQ

C

x = 11

3. (a) In the diagram, trapezoid ABCD has parallel sidesAB and DC of lengths 10 and 20, respectively. Also,the length of AD is 6 and the length of BC is 8.Determine the area of trapezoid ABCD.

A B

CD

(b) In the diagram, PQRS is a rectangle andT is the midpoint of RS. The inscribedcircles of 4PTS and 4RTQ each haveradius 3. The inscribed circle of 4QPThas radius 4. Determine the dimensions ofrectangle PQRS.

R ST

PQ

2006C

anadianO

penM

athematics

Challenge

(English)4. (a) Determine, with justification, the fraction

p

q, where p and q are positive integers

and q < 100, that is closest to, but not equal to, 37 .

(b) The baseball sum of two rational numbersa

band

c

dis defined to be

a + c

b + d.

(A rational number is a fraction whose numerator and denominator are bothintegers and whose denominator is not equal to 0.) Starting with the rationalnumbers 0

1 and 11 as Stage 0, the baseball sum of each consecutive pair of

rational numbers in a stage is inserted between the pair to arrive at the nextstage. The first few stages of this process are shown below:

STAGE 0: 01

11

STAGE 1: 01

12

11

STAGE 2: 01

13

12

23

11

STAGE 3: 01

14

13

25

12

35

23

34

11

Prove that

(i) no rational number will be inserted more than once,(ii) no inserted fraction is reducible, and(iii) every rational number between 0 and 1 will be inserted in the pattern at

some stage.



presents the

Sun Life Financial CanadianOpen Mathematics Challenge



Calculators are NOT permitted.Do not open this booklet until instructed to do so.There are two parts to this paper.

PART AThis part of the paper consists of 8 questions, each worth 5 marks. You can earn full valuefor each question by entering the correct answer(s) in the space provided. If your answer isincorrect, any work that you do will be considered for part marks, provided that it is donein the space allocated to that question in your answer booklet.

PART BThis part of the paper consists of 4 questions, each worth 10 marks. Finished solutions mustbe written in the appropriate location in the answer booklet. Rough work should be doneseparately. If you require extra pages for your finished solutions, paper will be provided byyour supervising teacher. Any extra papers should be placed inside your answer booklet. Besure to write your name and school name on any inserted pages.Marks are awarded for completeness, clarity, and style of presentation. A correct solution poorlypresented will not earn full marks.

NOTES:At the completion of the contest, insert the information sheet inside the answerbooklet.The names of top scoring competitors will be published on the Web sites of theCMS and CEMC.

Sun Life Financial Canadian Open Mathematics Challenge



7, etc., rather than as 12.566 . . .or 4.646 . . ..

4. Calculators are not allowed.

PART A

1. If a = 15 and b = −9, what is the value of a2 + 2ab + b2?

2. A circular wind power generator turns at a rate of 30 complete revolutions perminute. Through how many degrees does it turn in one second?

3. In the diagram, ABCD is a rectangle with A on theline y = x + 10, B on the line y = −2x + 10, andC and D on the x-axis. If AD = 4, what is the areaof rectangle ABCD?

y

x

A B

CD

4. In June, the ratio of boys to girls in a school was 3 : 2. In September, there were80 fewer boys and 20 fewer girls in the school and the ratio of boys to girls was 7 : 5.What was the total number of students at the school in June?

5. The numbers 1, 2, 3, . . . , 9 are placed in a square array. The sum of the three rows,the sum of the three columns, and the sum of the two diagonals are added togetherto form a “grand sum”, S.For example, if the numbers are placed as shown, the grand sum is

1 2 34 5 67 8 9

S = row sums + column sums + diagonal sums= 45 + 45 + 30= 120 .

What is the maximum possible value of the grand sum S?

6. In the diagram, O is the centre of the circle, AN is tangent tothe circle at A, P lies on the circle, and PN is perpendicularto AN . If AN = 15 and PN = 9, determine the radius ofthe circle. P

NA

O

7. Determine all ordered triples of real numbers, (x, y, z), that satisfy the system ofequations

xy = z2

x + y + z = 7x2 + y2 + z2 = 133 .

8. In the diagram, there are 28 line segments of length 1 arranged as shown to form9 squares. There are various routes from A to B travelling along the segments sothat no segment is travelled more than once. Of these possible routes, determine

• the length of route that occurs the most often, and

• the number of different routes of this length.

A

B

PART B

1. An arithmetic sequence a, a + d, a + 2d, . . . is a sequence in which successive termshave a common difference d. For example, 2, 5, 8, . . . is an arithmetic sequence withcommon difference d = 3 because 5− 2 = 8− 5 = 3.

(a) If x− 1, 2x + 2 and 7x + 1 are the first three terms of an arithmetic sequence,determine the value of x.

(b) For the value of x from (a), what is the middle term of the arithmetic sequencex− 1, 2x + 2, 7x + 1, . . . , 72?

A geometric sequence a, ar, ar2, . . . is a sequence in which successive terms have acommon ratio r. For example, the sequence 2, 10, 50, . . . is a geometric sequence with

common ratio r = 5 because102

=5010

= 5.

(c) If y − 1, 2y + 2 and 7y + 1 are the first three terms of a geometric sequence,determine all possible values of y.

(d) For each of the values of y from (c), determine the 6th term of the geometricsequence y − 1, 2y + 2, 7y + 1, . . . .

2. In the diagram, ∠ABC = ∠BCD = 90◦. Also, AB = 9, BC = 24 and CD = 18.The diagonals AC and BD of quadrilateral ABCD meet at E.

(a) Determine the area of the quadrilateral ABCD.

(b) Show that the ratio DE : EB = 2 : 1.

(c) Determine the area of triangle DEC.

(d) Determine the area of triangle DAE.

C

B A

D

E

9

24

18

2007Sun

Life

Fin

ancia

lC

anadia

nO

pen

Math

ematics

Challen

ge

(English

)3. Alphonse and Beryl are back! They are playing a two person game with the following

rules:

• Initially there is a pile of N stones, with N ≥ 2.• The players alternate turns, with Alphonse going first. On his first

turn, Alphonse must remove at least 1 and at most N −1 stones fromthe pile.

• If a player removes k stones on their turn, then the other player mustremove at least 1 and at most 2k − 1 stones on their next turn.

• The player who removes the last stone wins the game.

(a) Determine who should win the game when N = 7, and explain the winningstrategy.

(b) Determine who should win the game when N = 8, and explain the winningstrategy.

(c) Determine all values of N for which Beryl has a winning strategy. Explain thisstrategy.

4. A cat is located at C, 60 metres directly west of a mouse located at M . The mouseis trying to escape by running at 7 m/s in a direction 30◦ east of north. The cat, anexpert in geometry, runs at 13 m/s in a suitable straight line path that will interceptthe mouse as quickly as possible.

(a) If t is the length of time, in seconds, that ittakes the cat to catch the mouse, determinethe value of t.

(b) Suppose that the mouse instead chooses adifferent direction to try to escape. Show thatno matter which direction it runs, all points ofinterception lie on a circle.

(c) Suppose that the mouse is intercepted afterrunning a distance of d1 metres in a particulardirection. If the mouse would have beenintercepted after it had run a distance of d2

metres in the opposite direction, show thatd1 + d2 ≥ 14

√30.

N

EW

S

60 mC M

307 m/s

13 m/s


Comments on the Paper

Part A

1. An operation “∆ ” is defined by a ∆b =1–ab

, b ≠ 0.

What is the value of 1∆ 2( )∆ 3∆ 4( )?

Solution

By the definition of “∆ ”

1∆ 2 =1– 1

2= 1

2

3∆ 4 =1– 34

= 14

and so 1∆ 2( )∆ 3∆ 4( ) = 12

∆

14

=1–

1214

=1– 2 = –1

ANSWER: –1 2. The sequence 9, 18, 27, 36, 45, 54, … consists of successive multiples of 9. This sequence is then altered by multiplying

every other term by –1, starting with the first term, to produce the new sequence –9, 18, – 27, 36, – 45, 54,... . If the sum of the first n terms of this new sequence is 180, determine n.

Solution The terms in the sequence are paired, by combining each odd-numbered term with the next term (that is, we combine terms 1 and 2, 3 and 4, 5 and 6, etc). The sum of each of these pairs is 9. So we need 20 of these pairs to reach a sum of 180. Thus we need 2 ×20 or 40 terms. ANSWER: 40

3. The symbol n! is used to represent the product n n –1( ) n – 2( ) … 3( ) 2( ) 1( ) . For example,4!= 4 3( ) 2( )1( ) . Determine n such that n!= 215( )36( )53( )72( )11( ) 13( ) .

Solution Since n! has a prime factor of 13, n must be at least 13. Since n! has no prime factor of 17, n must be less than 17. These two facts are true because if m ≤ n , then m divides n!. Since n! has 53 as a factor, then n ≥ 15 , since we need n! to have 3 factors which are multiples of 5. We must thus determine if n =15 or n =16. So we look at the number of factors of 2 in 16!. 16! gets 1 factor of 2 each from 2, 6, 10, 14 2 factors of 2 each from 4, 12 3 factors of 2 from 8 4 factors of 2 from 16 We have a total of 15 twos which then corresponds to n =16. ANSWER: 16 4. The symbol x means the greatest integer less than or equal to x. For example, 5.7 = 5, π =3 and 4 = 4 .

Calculate the value of the sum 1 + 2 + 3 + 4 + … + 48 + 49 + 50 .

Solution We note that for k a positive integer and k 2 ≤ n < k +1( )2 , then k ≤ n < k +1 and so n = k .

Thus for 1 ≤ n ≤ 3, n =1

4 ≤ n ≤ 8, n = 2

9 ≤ n ≤ 15, n = 3

etc. So the sum equals 1+1+1( )+ 2 +2 +2 + 2 +2( )+ (3 + … + 3) + … + (6 + … + 6) + (7 + 7)

= 3 1( )+5 2( )+ 7 3( )+9 4( )+11 5( )+13 6( )+2 7( )= 3 +10 +21+ 36+ 55+ 78+14= 217

ANSWER: 217 5. How many five-digit positive integers have the property that the product of their digits is 2000? Solution Let a five-digit number have the form a b c d e where 0 ≤ a, b, c, d, e ≤ 9, a ≠ 0 . Since the product of the digits is 2000, we must have the product abcde =2000 = 2453 .

Since the product of the digits is 2000, then 3 of the digits have to be 5. The remaining 2 digits must have a product of

16 or 24 . Thus the two remaining digits must be 4 and 4, or 2 and 8. Possibility 1

Case 1 Using the numbers 5, 5, 5, 4, 4 there are 5!

3!2!=10 possible numbers.

Case 2 Using the numbers 5, 5, 5, 2, 8 there are 5!3!

= 20 possible numbers.

There are 30 possible such numbers. OR Possibility 2

We choose 3 of the 5 positions for the “5s” in 53

ways; there are 3 possibilities for the remaining two digits (including

order): 2, 8; 4, 4; 8, 2.

So there are 3×53

=3 ×10 =30 possible such 5 digit numbers.

ANSWER: 30 6. Solve the equation 4 16sin 2 x( )=26 sin x , for 0 ≤ x ≤ 2π.

Solution We write all factors as powers of 2. Thus

4 16sin2 x( )= 26 sin x

22 24 sin2 x( )= 26 sin x

24sin 2 x +2 = 26 sin x (*)

Equating exponents (which we can do by taking base 2 logarithms),

4 sin2 x + 2 = 6 sin x

2 sin2 x –3 sin x +1= 02 sin x –1( ) sin x –1( )= 0

Therefore, sin x = 12

or sin x =1.

Since 0 ≤ x ≤ 2π, x =π6

,5π6

or π2

.

ANSWER:π6

,5π6

, π2

7. The sequence of numbers …, a–3, a– 2, a–1, a0, a1, a2, a3 , … is defined by an – n +1( )a2–n = n +3( )2 , for all integers n.

Calculate a0 .

Solution Using the given general equation, we note that there are only two choices of n which yield equations containing a2 ,

n = 0 or 2. i.e. a0 – a2 = 9 from n = 0 a2 – 3a0 = 25 from n =2 Adding the first equation to the second, we obtain –2a0 = 34, so a0 = –17.

ANSWER: –17 8. In the diagram, ∆ ABC is equilateral and the radius of its

inscribed circle is 1. A larger circle is drawn through the vertices of the rectangle ABDE .

What is the diameter of the larger circle?

E C D

A B Solution First, we calculate the side length of the equilateral triangle ABC . Let O be the centre of the smaller circle and P the point of tangency of

the circle to the side AB . Join OP and OB . Then ∠ OPB = 90° by tangency and ∠ OBP =30°

by symmetry since ∠ CBA = 60° .

C

A P B

O 30°

Since OP =1 and ∆ BOP is 30°-60°-90°, then OB =2 and BP = 3 . Thus AB = 2 3 . Also by symmetry, CO = OB = 2 , so CP = 3. Since ABDE is a rectangle and CP ⊥ AB , then AE =3 . We now look at the rectangle ABDE and its circumcircle. Since ABDE is a rectangle, ∠ EAB = 90° .

So BE is a diameter. By Pythagoras,

BE2 = EA2 + AB2

=32 + 2 3( )2

= 21

The diameter is 21 .

E D

A B

3

2 3

ANSWER: 21

Part B

1. Triangle ABC has vertices A 0, 0( ), B 9, 0( ) and C 0, 6( ). The points P and Q lie on side AB such that AP = PQ = QB .

Similarly, the points R and S lie on side AC so that AR = RS = SC . The vertex C is joined to each of the points P and Q. In the same way, B is joined to R and S.

(a) Determine the equation of the line through the points R and B. (b) Determine the equation of the line through the points P and C. (c) The line segments PC and RB intersect at X, and the line segments QC and SB intersect at Y. Prove that the

points A, X and Y lie on the same straight line. Solution Since A 0, 0( ), B 9, 0( ) and AP = PQ = QB , then P has coordinates 3, 0( ) and Q has coordinates 6, 0( ). Similarly, R is the point (0, 2) and S is the point (0, 4). (a) Since R is (0, 2) and B is (9, 0), then the slope of RB is m =

2 – 00 – 9

= –29

and so the equation of the line is

y – 2 = – 2

9x – 0( )

y = – 29

x + 2

(b) Since P is (3, 0) and C is (0, 6), then the slope of PC is m =

0 – 63 – 0

= – 2 and so the equation of the line is

y – 0 = –2 x – 3( )

y = –2x +6

(c) First, we determine the coordinates of X. Equating the lines from (a) and (b), we have

– 29

x +2 = – 2x +6

169

x = 4

x = 94

C(0, 6)

S(0, 4)

R(0, 2)

A(0, 1) P(3, 0) Q(6, 0) B(9, 0)

X

Y

and substituting into y = –2x +6 = –2 9

4

+6 = 32

, so X is the point 94

, 32

.

We calculate the equations of the lines QC and SB as in (a) and (b).

For QC , slope =0 – 66 – 0

= –1 and so y – 0 = –1 x – 6( ) or y = –x +6.

ForSB , slope = 0 – 49 – 0

= –49

and so y – 0 = –49

x – 9( ) or y = – 49

x + 4 .

So the point Y, which lies at the intersection QC and SB , we obtain by equating these lines

–x + 6 = – 49

x +4

2 = 59

x

x = 185

and so y = –x +6 = – 185

+6 = 125

and thus Y is the point 185

, 125

.

Now the line through A 0, 0( ) and X 94

, 32

has slope m =

32

– 094

– 0= 2

3 and so is y = 2

3x .

The point Y lies on this line, as 125

= 23

185

. [This could be done with L.S./R.S. format using equation of line.]

Therefore A, X, Y lie on the same line. 2. In ∆ ABC , the points D, E and F are on sides BC , CA and AB ,

respectively, such that ∠ AFE = ∠ BFD , ∠ BDF = ∠ CDE , and ∠ CED = ∠ AEF .

(a) Prove that ∠ BDF = ∠ BAC .

(b) If AB = 5, BC = 8 and CA = 7, determine the length of BD .

F

A

B D C

E

Solution (a) Let ∠ AFE = ∠ BFD = x

∠ BDF = ∠ CDE = y∠ CED = ∠ AEF = z

Thus ∠ FAE =180° – x – z

∠ FBD =180° – x – y∠ ECD =180° – y – z

and these 3 angles add to 180° , so

F

A

B D C

Ex z

x

y y

z

540° – 2 x + y + z( )=180°

x + y + z =180°

Since ∠ FAE + ∠ AFE + ∠ AEF =180° (from ∆ AEF )

∠ FAE + x + z = x + y + z

∠ FAE = y

Therefore ∠ BDF = ∠ BAC .

(b) Similarly to part (a), ∠ ECD = ∠ BFD = x , ∠ FBD = ∠ CED = z .

By equal angles, ∆ ABC ~ ∆ DBF ~ ∆ DEC ~ ∆ AEF and so BDBF

=BABC

=58

, CDCE

=CACB

=78

, AEAF

=ABAC

=57

.

Therefore, let BD = 5k , BF =8k , CD = 7l , CE =8l , AE = 5m , AF = 7m for some k, l, m. Then 5k +7l =8 (1) 7m +8k = 5 (2) 5m + 8l = 7 (3) Determining 7 × (3) – 5 × (1) to eliminate m, we get 56l – 40k = 49 – 25 = 24

7l – 5k = 3 (4)

F

A

B D C

Ex z

x

y y

z

7m 5my

z x

8k

5k 7l

8l

Calculating (1) – (4), we get 10k = 5 or BD = 5k = 5

2.

3. (a) Alphonse and Beryl are playing a game, starting with the

geometric shape shown in Figure 1. Alphonse begins the game by cutting the original shape into two pieces along oneof the lines. He then passes the piece containing the black triangle to Beryl, and discards the other piece.

Figure 1

Beryl repeats these steps with the piece she receives; that is to say she cuts along the length of a line, passes the piece containing the black triangle back to Alphonse, and discards the other piece. This process continues, with the winner being the player who, at the beginning of his or her turn receives only the black triangle. Show, with justification, that there is always a winning strategy for Beryl.

Solution We first consider Alphonse’s possible moves to begin the game. We can assume, without loss of generality, that he cuts

on the left side of the black triangle.

Case 1

Alphonse removes two white triangles, leaving .

In this case, Beryl removes only one white triangle, and passes the shape back to Alphonse, forcing him to

remove the last white triangle and lose. Case 2

Alphonse removes one white triangle only, leaving .

Beryl removes both of the white triangles on the right, leaving Alphonse in the same position as in Case 1 for his second turn.

Therefore Beryl can always win, regardless of Alphonse’s strategy.

(b) Alphonse and Beryl now play a game with the same rules

as in (a), except this time they use the shape in Figure 2 and Beryl goes first. As in (a), cuts may only be made along the whole length of a line in the figure. Is there a strategy that Beryl can use to be guaranteed that she will win? (Provide justification for your answer.)

Figure 2 Solution We show that, again, Beryl always has a winning strategy. The strategy is to reduce the shape in Figure 2 to the shape in Figure 1, and to have Alphonse make the first cut at this

stage. Beryl also knows that if she is forced into a position of being the first to cut when Figure 2 is reduced to Figure 1, then Alphonse can force her to lose.

We number the lines on the diagram for convenience. (1)

(9)

(2)

(3)

(8)

(7)

(4) (5) (6) We can assume without loss of generality (because of symmetry) that Beryl cuts along (1), (2) or (3) to begin. If she cuts (2) or (3), then Alphonse cuts the other of these two and leaves Beryl with Figure 1, where she will lose. Therefore Beryl cuts (1) to begin.

If Alphonse now cuts (2) or (3), Beryl cuts the other of these two and passes the shape in Figure 1 back to Alphonse, and

so he loses. If Alphonse cuts (8) or (9), Beryl cuts the opposite and passes the shape in Figure 1 to Alphonse, and so he loses.

(Similarly, if he cuts (5) or (6)). So assume that Alphonse cuts (4) or (7), say (4) by symmetry.

If Beryl now cuts any of (2), (3), (5), (6), (8), or (9), then Alphonse can force Beryl to lose, in the same way as she could

have forced him to lose, as above. So Beryl cuts (7).

Now Alphonse is forced to cut one of (2), (3), (5), (6), (8), or (9), and so Beryl makes the appropriate cut, passing the

shape in Figure 1 back to Alphonse, and so he must lose.

Therefore Beryl always can have a winning strategy.

4. A sequence t1, t2, t3, ..., tn of n terms is defined as follows: t1 =1, t2 = 4 , and tk = tk –1 + tk–2 for k =3, 4, ..., n . Let T be the set of all terms in this sequence; that is, T = t1, t2 , t3, ..., tn{ } .

(a) How many positive integers can be expressed as the sum of exactly two distinct elements of the set T ? Summary Part (a) 4 ✓ Part (b) 6 ✓ Solution tk > 0 for all k, 1 ≤ k ≤ n . Also tk < tk+1 for all k ≤ n –1 since tk +1 = tk + tk–1 .

Hence the sequence is monotone increasing.

The sum of any pair of terms is an integer and there are n2

pairs.

Can any two pairs produce the same integer? Consider ta + tb and tc + td . Clearly if tb = td then ta = tc and vice versa, implying the same pair. Hence none of the four terms is equal, so we can assume one term, say td to be the largest. Then td = td –1 + td –2 ≥ ta + tb , since the maximum values of ta and tb are td –1 and td –2 and they cannot be alike.

But since tc >0 , tc + td > ta + td and there are no two pairs that add to the same integer, so there are exactly n2

integers

possible.

(b) How many positive integers can be expressed as the sum of exactly three distinct elements of the set T ? Solution Consider ta + tb + tc and td + te + t f . If any of the first three equals any of the second three we are left with pair

sums of the remaining ones being equal, which is impossible from part (a). Hence there are six unlike terms, and again we can assume one, say t f , to be the greatest.

It is clearly possible for equality by setting ta and tb equal to t f –1 and t f –2 and then td and te equal to tc –1 and

tc –2 . In how many ways can this be done for given f. Clearly,6 ≤ f ≤ n , and since 2 < c < f – 2, for any given f there

are f – 5 choices for c and the number of ways possible is f – 5( )f =6

n

∑ =1+2 + 3+ + n – 5( )=n – 4

2

.

There are a maximum of n3

–

n – 42

possible integers.

Of these, are any two of like sum? In ta + tb + tc , the maximal values are t f +1, t f –3 , and t f –4 , since if one is t f –1 and one t f –2 we revert to the

discussed state. Hence

ta + tb + tc ≤ t f –1 + t f –3 + t f –4

= t f –1 + t f –2

= t f .

But td + te + t f > t f .

Hence there are no other triples for which equality of sums exist, and the number of possible integers is n3

–

n – 42

.

6

Comments on the Paper Commentaires sur l'épreuve

2 0 2 3 4 1 5

0 1 0 3 0 3 3

2 0

2 1

∇ = + = + =

∇ = + = + =

2 0 0 1 5 3

5 3

25 27

52

2 3

∇( )∇ ∇( ) = ∇

= += +=

PART A1. Solution

By the definition,

and so

CommentsThis question was quite well done. Most students correctlyinterpreted the given operation to do the required calcula-tions.Average: 3.6

2. SolutionFrom the diagram,

∠ = −ACB x180 7o oand

∠ = −FEG x180 8o o .Therefore, ∠ = −DCE x180 7o o

D

E G

F

C

B

A3x°

4x°

5x°

6x°2x°

180° – 7x° 180° – 8x°

A B

F C

E D

and ∠ = −DEC x180 8o o, so from ∆CDE ,

CommentsExtremely well done! Almost all of the competitors hada good handle on dealing with angles in triangles.Average: 4.1

5 180 7 180 8 180

360 10 180

10 180

18

x x x

x

x

x

o o o o o o

o o o

o o

+ − + − =

− =

==

(*)

PARTIE A1. Solution

D’après la définition, on a :

Donc :

CommentairesLes élèves ont bien réussi cette question. La plupart desélèves ont correctement interprété les opérations donnéesnécessaires aux calculs requis.Moyenne: 3,6

2 0 2 3 4 1 5

0 1 0 3 0 3 3

2 0

2 1

∇ = + = + =

∇ = + = + =2 0 0 1 5 3

5 3

25 27

52

2 3

∇( )∇ ∇( ) = ∇

= += +=

D

E G

F

C

B

A3x°

4x°

5x°

6x°2x°

180° – 7x° 180° – 8x°

2. SolutionD’après le diagramme,

∠ = −ACB x180 7o o et

∠ = −FEG x180 8o o .Donc ∠ = −DCE x180 7o o

et ∠ = −DEC x180 8o o.Dans le triangle CDE, on a donc :

CommentairesChapeau! La quasi-totalité des concurrents ont bien maîtrisécette question qui nécessitait l’application des connaissancesdes angles aux triangles.Moyenne: 4,1

5 180 7 180 8 180

360 10 180

10 180

18

x x x

x

x

x

o o o o o o

o o o

o o

+ − + − =

− =

==

(*)

3. Solution 1Soit ABCDEF un hexagone régulier dontles côtés ont une longueur de 1. Chacunde ses angles mesure 120o. Les segmentsFC, EB et DA sont des bissectrices desangles. L’hexagone est donc décomposéen 6 triangles équilatéraux.La distance maximale possible entre deux points surl’hexagone est celle entre deux sommets opposés. Puisquechaque triangle équilatéral a des côtés de longueur 1, lalongueur maximale possible du segment est égale à 2.

Version abrégée de la solution 1Comme l’indique le diagramme, l’hexagone régulier peutêtre divisé en 6 triangles équilatéraux ayant des côtés delongueur 1. La distance maximale possible entre deuxpoints sur l’hexagone est celle entre deux sommets opposés.Elle est égale à 2.

A B

F C

E D

3. Solution 1Let ABCDEF be a regular hexagon witha side length of 1. Each angle is 120o.Thus, if we join FC, EB, DA, each of theinterior angles is bisected, and so eachpart is 60o. Thus the hexagon is decom-posed into 6 equilateral triangles, as shown.The maximum distance between any two points on thehexagon is the distance between two opposite vertices.Since each of the triangles is equilateral with a side lengthof 1, the diagonal distance is 2, ie. the maximum possiblelength of PQ is 2.

Brief version of Solution 1A regular hexagon with side length 1 can be decomposedinto 6 equilateral triangles with a side length of 1, as shown.The maximum distance between any two points is betweenopposite vertices, and this distance is 2.

7


BA

F CR S60° 60°

Solution 2The maximum distance is betweentwo opposite vertices, say F and Cby symmetry.Drop perpendiculars from A and Bto meet FC at R and S respectively.Since AB = 1 and AB is parallel to RS, then RS = 1.By symmetry, FR CS= . But

FR AF= = ( ) =cos60 1 12

12

o . Therefore, CF = 2, and so

the maximum possible distance is 2.

CommentsThe key problems here were to interpret the question andto then figure out that the longest distance between any twopoints is the distance between opposite vertices. Theeasiest way to calculate this length was to break thehexagon up into 6 equilateral triangles each of side lengthone. If you didn’t notice this, have a look at this idea.Average: 3.6

4. Solution

or2 2 4 64

2 4 4 64

4 64

3

2x x

x x

x

x

( ) = +

( ) = +

==

2 2 4 64

2 2 2 64

2 64

2 6

3

2

2 2

2

x x

x x

x

x

x

( ) = +

( ) = +

===

Solving the equation,

CommentsThis question was reasonably well done. Students who arecomfortable dealing with exponents had a great deal ofsuccess on this question. Many students continue to havedifficulty with exponents.Average: 3.8

5. Solution 1Join M to Q.Through M, draw a line parallelto QR meeting PQ at F.

Therefore, PF FQ= = 7 and MF = 24 .By Pythagoras, MQ = 25, and so

cos cos∠( ) = ∠( )=

=

MQP MQF

FQ

MQ

725

P

Q R

MF

Solution 2Par symétrie, la distance maximaleentre deux points sur l’hexagone estcelle entre deux sommets opposés,disons F et C. Aux points A et B, onabaisse des perpendiculaires AR et AS au segment FC.Puisque AB = 1 et que AB est parallèle à RS, alors RS = 1.

Par symétrie, FR CS= . Or FR AF= = ( ) =cos60 1 12

12

o .

Donc CF = 2 et la longueur maximale possible est doncégale à 2.

CommentairesUne grande partie des difficultés de cette question résidaitdans son interprétation. Les élèves devaient par la suitedéduire que la distance entre deux sommets opposésreprésentait la distance maximale entre deux points. La façonla plus rapide de la calculer consistait à diviser l’hexagone ensix triangles équilatéraux dont chacune des arêtes équivalaità un. Nous invitons les élèves qui n’ont pas envisagé cetteapproche de résolution de l’examiner de plus près.Moyenne: 3,6

BA

F CR S60° 60°

4. SolutionOn a :

CommentairesLes élèves ont généralement bien réussi cette question, toutparticulièrement ceux qui maîtrisaient bien le conceptd’exposant (qui d’ailleurs pose encore des difficultés pourbon nombre d’élèves).Moyenne: 3,8

ou2 2 4 64

2 4 4 64

4 64

3

2x x

x x

x

x

( ) = +

( ) = +

==

2 2 4 64

2 2 2 64

2 64

2 6

3

2

2 2

2

x x

x x

x

x

x

( ) = +

( ) = +

===

P

Q R

MF

5. Solution 1On joint M et Q.Au point M, on trace un segmentparallèle au côté QR. Ce segmentcoupe le côté PQ en F.Donc PF FQ= = 7 et MF = 24 .D’après le théorème de Pythagore dans le triangle MFQ,MQ = 25, d’où :

cos cos∠( ) = ∠( )=

=

MQP MQF

FQ

MQ

725

8


Solution 2Join M to Q. By Pythagoras,

PR = + =14 48 502 2 .Since M is the midpoint of the

hypotenuse, then MQ MP MR= = . (This is because the

circle circumscribed around ∆PQR has PR as diameter

(since ∠ =PQR 90o) and so M is the centre and thus MP,

MQ and MR are radii.) Therefore, ∠ = ∠MQP MPQ and

so cos cos∠( ) = ∠( ) = =MQP MPQ 1450

725 .

CommentsThis is a nice question because it can be done with aEuclidean approach or a more analytic approach. Somestudents had difficulty when they encountered a triangle thatwas not right-angled in which they had to calculate a cosine.Average: 2.7

6. SolutionWe calculate the first few terms in the series

t

tt

t

tt

t

tt

t

tt

t

1

21

1

32

2

1313

43

3

1212

54

4

2

11

2 12 1

13

11

1

112

11

1

13

11

3 13 1

2

=

= −+

= −+

=

= −+

=−+

= −

= −+

=− −− +

= −

= −+

= − −− +

=

P

Q R

M14

Since a term in the sequence depends only on the previousone, then the sequence will cycle with a period of 4. Thus,

t t t t1 5 997 1001 2= = = = =L . Therefore, t99813= and t999

12= − .

CommentsThis question was very well done! Most students quicklydetermined after a few calculations that the sequence wascyclic, and on this basis determined the value of the requiredterm. The nicest solution by a student was to repeatedlyapply the definition:

tt

t

t t

t t tnn

n

tttt

n n

n n n

n

n

n

n

+

−+−+

− −

− − −=

−+

=−

+=

−( ) − +( )−( ) + +( ) = −

−

−

−

−

1

1111

1 1

1 1 1

11

1

1

1 1

1 11

1

1

1

1

In a similar way,

and so

ttnn

−−

= −13

1

t tnt

nn

+ −= −−

=−

1 1 31

3

Solution 2On joint M et Q.D’après le théorème dePythagore dans le triangle

PQR, PR = + =14 48 502 2 .Puisque M est le milieu de l’hypoténuse,

MQ MP MR= = . (Puisque ∠ =PQR 90o, PQ est lediamètre du cercle circonscrit au triangle et M en est lecentre. Donc MP, MQ et MR sont des rayons.)Donc ∠ = ∠MQP MPQ , d’où .

CommentairesIl s’agissait d’une question intéressante en raison de sesdeux méthodes de résolution : l’approche euclidienne ouune approche analytique plus générale. Quelques élèvesn’ont pu contourner la difficulté que posait la présence d’untriangle sans angle droit, ce qui nécessitait le calcul ducosinus.Moyenne: 2,7

P

Q R

M14

cos cos∠( ) = ∠( ) = =MQP MPQ 1450

725

t

tt

t

tt

t

tt

t

tt

t

1

21

1

32

2

1313

43

3

1212

54

4

2

11

2 12 1

13

11

1

112

11

1

13

11

3 13 1

2

=

= −+

= −+

=

= −+

=−+

= −

= −+

=− −− +

= −

= −+

= − −− +

=

tt

t

t t

t t tnn

n

tttt

n n

n n n

n

n

n

n

+

−+−+

− −

− − −=

−+

=−

+=

−( ) − +( )−( ) + +( ) = −

−

−

−

−

1

1111

1 1

1 1 1

11

1

1

1 1

1 11

1

1

1

1

6. SolutionLes premiers termes de la suite sont :

Puisque chaque terme de la suite dépend seulement duterme précédent, les valeurs se reproduisent à tous les 4termes. Les valeurs ont donc un cycle de longueur 4.Donc t t t t1 5 997 1001 2= = = = =L . Donc t998

13= et

t99912= − .

CommentairesCette question a été très bien réussie! La plupart des élèvesont conclu, moyennant quelques calculs, qu’ils se trouvaienten présence d’une séquence cyclique : ils ont par la suite étéen mesure de trouver la valeur du terme requis. La solutionla plus astucieuse consistait à appliquer la définition defaçon répétitive :

9


y

x0

2

4

– 4 – 2 2 4

– 2

y = g x( )

y

x0

2

4

– 4 – 2 2 4

– 2

y = g x( ) – 1

y

x0

2

4

– 4 – 2 2 4

– 2

y = g x( ) –1

y =12

De la même manière,

On peut conclure que

Par conséquent, on peut affirmer que la séquence se répéteraà tous les quatre termes. Il suffisait d’y penser!Moyenne: 2,7

ttnn

−−

= −13

1

t tnt

nn

+ −= −−

=−

1 1 31

3

7. SolutionOn exprime d’abord x, y et z en fonction de a.D’après la deuxième équation, y x a= − (*) .On reporte y = x – a dans la première équation :

On reporte x = –2a dans l’équation (*) pour obtenir y a= −3 .

On reporte y a= −3 dans la troisième équation pour obtenir

z x y a= + = −5 .

Donc x y z a+ + = −10 .Puisque a peut prendre la valeur de n’importe quel entierstrictement positif, la valeur maximale possible del’expression x y z+ + est –10. On l’obtient lorsque a = 1.

CommentairesNous avons été agréablement surpris du succès des élèves àrésoudre cette question qui se voulait extrêmement difficile;de nombreux élèves ont facilement résolu ce systèmed’équations où il fallait trouver la valeur de x, y et de z enfonction de a.Moyenne: 2,8

2

2

x a x a

x a

+ = −= −

8. Solution 1 (Graphique) y

x0

2

4

– 4 – 2 2 4

– 2

y = g x( )

y

x0

2

4

– 4 – 2 2 4

– 2

y = g x( ) – 1

y

x0

2

4

– 4 – 2 2 4

– 2

y = g x( ) –1

y =12

Thus, the sequence repeats every four terms. What a nicesolution!Average: 2.7

7. SolutionWe treat a as a constant and solve for x, y, z in terms ofa.

From the second equation, y x a= − (*) Therefore, from the first equation,

2

2

x a x a

x a

+ = −= −

Substituting into (*), y a= −3 .

Substituting into the 3rd equation, z x y a= + = −5 .

So x y z a+ + = −10 .Since a is a positive integer, the maximum value forx y z+ + is –10 (which occurs when a = 1).

CommentsThis question was done very well. We anticipated that thiswould be quite a difficult question, and so we wereextremely happy with the results. Many students handledthis system of equations with relative ease by figuring outthat they had to solve for x, y and z in terms of a.Average: 2.8

8. Solution 1 (Graphical)

so the number of solutions

of g x( ) − =1 12 is 8,

from the third graph.

Selon le troisième graphique,

l’équation g x( ) − =1 12

admet 8 solutions.

10


y

x

x = 2

(2, 1) (7, 1)(10, 1)

y = 1

x + 2y = 12

(2, 5)(4, 4)

QR

Solution 2 (Algébrique)

D’après l’équation g x( ) − =1 12 , on a :

D’après le graphique donné :

a) g x( ) = 32 admet 1 solution,

b) g x( ) = − 32 admet 1 solution,

c) g x( ) = 12 admet 3 solutions,

d) g x( ) = − 12 admet 3 solutions.

Donc g x( ) − =1 12 admet 8 solutions.

CommentairesCette question a permis aux élèves de mettre à l’épreuveleurs connaissances de la valeur absolue dans le contexted’une représentation graphique. Certains élèves ont choisiune approche graphique pour déterminer le graphiquerequis à partir de la transposition du premier graphique. Enfait, plusieurs élèves ont eu recours à cette méthode derésolution pour trouver les valeurs possibles de g x( ) oug x( ) ; ils ont ensuite déterminer le nombre de solutions

appropriées à partir des informations du graphique. Lesélèves qui ont tenté de déterminer l’équation de la courbeont en général éprouvé des difficultés au niveau des calculs.Moyenne: 2,0

g x g x

g x g x

g x g x

( ) ( )

( ) ( )

( ) ( )

− = − = −

= =

= ± = ±

1 112

12

32

12

32

12

or

or

or

ou

ou

ou

y

x

x = 2

(2, 1)(10, 1)

y = 1

x + 2y = 12

(2, 5)

Partie B

1. Solutiona) Les droites définies

par x = 2 et y = 1 se

coupent au point 2 1,( ) .Les droites définiespar x = 2 et x y+ =2 12se coupent au point

2 5,( ), car en reportant

x = 2 dans x y+ =2 12, on obtient 2 + 2y = 12, d’où y

= 5. Les droites définies par y = 1 et x y+ =2 12 se

coupent au point 10 1,( ) , car en reportant y = 1 dans

x y+ =2 12, on obtient x + 2 = 12, d’où x = 10.b) Les droites d’équations

x y+ = 8 et se coupent

au point 2 6,( ). Cepoint est au-dessus dupoint d’intersection desdroites d’équationsx = 2 et x y+ =2 12.

y

x

x = 2

(2, 1) (7, 1)(10, 1)

y = 1

x + 2y = 12

(2, 5)(4, 4)

QR

y

x

x = 2

(2, 1)(10, 1)

y = 1

x + 2y = 12

(2, 5)

Solution 2 (Algebraic)

From the original equation g x( ) − =1 12 , using the

definition of absolute value we obtain,

g x g x

g x g x

g x g x

( ) ( )

( ) ( )

( ) ( )

− = − = −

= =

= ± = ±

1 112

12

32

12

32

12

or

or

or

From the original graph,

(a) g x( ) = 32 has 1 solution,

(b) g x( ) = − 32 has 1 solution,

(c) g x( ) = 12 has 3 solutions,

(d) g x( ) = − 12 has 3 solutions.

Therefore, g x( ) − =1 12 has 8 solutions.

CommentsThis question was a good test of the concept of absolutevalue from a graphical perspective. Some students used agraphical approach to convert the original graph to thedesired one. Quite a few students used the graphical ap-proach to determine the potential values for either g x( ) org x( ) and then read the appropriate number of solutions off

the graph. Students who tried to determine the actualequation of the curve tended to get bogged down in theircalculations.Average: 2.0

Part B

1. Solution(a) The lines x = 2 and

y = 1 intersect at 2 1,( ) .The lines x = 2 and

x y+ =2 12 intersect

at 2 5,( ), since

x y y= ⇒ + = ⇒ =2 2 2 12 5.

The lines y = 1 and x y+ =2 12 intersect at 10 1,( ) ,

since y x x= ⇒ + = ⇒ =1 2 12 10.

(b) x y+ = 8 intersects

x = 2 at 2 6,( ), whichis above the point ofintersection of x = 2

and x y+ =2 12.

x y+ = 8 intersects

y = 1 at 7 1,( ).To find the intersection point of x y+ = 8 and

x y+ =2 12, subtract the first equation from the sec-

ond to obtain y = 4 , so x = 4.

11


Therefore, the vertices of Q are 2 1,( ) , 2 5,( ), 4 4,( ),

7 1,( ).(c) Area of Area of Area of Q T R= −

= ( )( ) − ( )( )= −

=

12

12

92

232

8 4 3 3

16

CommentsThis question was exceptionally well done. Studentseither approached this strictly graphically or with a com-bination of graphical and analytic approaches. In eithercase, they tended to do very well. Part (c) was quite welldone. Students managed to determine one of the manyways to calculate the area of Q, either by subtracting thearea of R from the area of T, by breaking Q up into twotriangles, or by breaking Q up into one rectangle and twotriangles.Average: 8.4

2. (a) Solution 1We define a “losing position” to be a number ofcards, such that if a player receives this number ofcards at the beginning of his or her turn, he or she isguaranteed to lose assuming best play by both play-ers. A “winning position” is defined similarly.

Clearly, by the rules of the game, 1 is a losingposition.For a player to receive 1 card at the beginning of aturn, the previous player must start with 2 cards.(This is true since a player can never remove morethan half of the deck, so the number of cards at thebeginning of the previous turn can never be morethan double that of the current turn.) Therefore, 2 isa winning position, since a player starting with 2cards can only remove 1, and so passes 1 card to theother player, who loses.

Is 3 a winning position or a losing position?Given a pack of 3 cards, the rules of the game say thata player can only remove 1 card, and so pass a packof 2 cards (a winning position) to the other player.Therefore, 3 is a losing position.

We can then see that 4, 5 and 6 are all winningpositions, as given 4, 5 or 6 cards, a player canremove 1, 2 or 3 cards respectively to pass the otherplayer 3 cards, a losing position.

Les droites d’équations x y+ = 8 ett y = 1 se coupent au

point 7 1,( ).Pour déterminer le point d’intersection des droitesdéfinies par x y+ = 8 et x y+ =2 12 , on soustrait lapremière équation de la deuxième, membre par membre,pour obtenir y = 4 , d’où x = 4.

Les coordonnées du quadrilatère Q sont 2 1,( ) , 2 5,( ),

4 4,( ) et 7 1,( ).c) Aire de Q = Aire de T – Aire de R

CommentairesLes élèves ont exceptionnellement bien réussi cette questionpeu importe la méthode de résolution utilisée. Des élèves ontainsi choisi la méthode de résolution graphique tandis qued’autres ont opté pour un agencement des méthodes graphiqueet analytique. Ils ont tout particulièrement bien réussi la partie(c). Ils ont trouvé l’une des nombreuses façons de déterminerl’aire de Q, soit en soustrayant l’aire de R de celle de T, ou endivisant Q en deux triangles ou encore en un rectangle et deuxtriangles.Moyenne: 8,4

= ( )( ) − ( )( )= −

=

12

12

92

232

8 4 3 3

16

2. a) Solution 1On définit une « position perdante » comme étant unnombre de cartes qu’un joueur ou une joueuse reçoit audébut de son tour et qui lui assure une défaite si les deuxadversaires jouent à leur meilleur. On définit une « positiongagnante » de façon semblable.

Selon les règles du jeu, 1 est une position perdante.Pour qu’un joueur reçoive 1 carte au début de son tour, lejoueur précédent doit commencer avec 2 cartes. (Un joueurne peut retirer plus de la moitié des cartes, donc le nombrede cartes qu’il reçoit ne peut être supérieur au double dunombre de cartes remises au joueur suivant.)Donc 2 est une position gagnante, puisqu’un joueur quireçoit 2 cartes peut seulement en enlever une et il remet 1carte à l’adversaire qui perd.

Est-ce que 3 est une position gagnante ou perdante?Si on reçoit 3 cartes, on peut seulement retirer une carte dujeu et on remet 2 cartes à l’adversaire qui reçoit une positiongagnante. Donc 3 est une position perdante.

On peut constater que 4, 5 et 6 sont des positions gagnantes,car on peut retirer respectivement 1, 2 ou 3 cartes etremettre 3 cartes à l’adversaire qui reçoit alors une positionperdante. On constate que 7 est une position perdante, caron doit retirer 1, 2 ou 3 cartes et on remet respectivement6, 5 ou 4 cartes à l’adversaire qui reçoit à chaque fois uneposition gagnante.

12


Therefore, 7 is a losing position, since a player remov-ing 7 cards must remove 1, 2 or 3 cards, thus giving theother player 6, 5 or 4 cards respectively, any of whichis a winning position. So if Alphonse starts with 7cards, Beryl can always win.

Summary of Beryl’s Strategy• She will receive 4, 5 or 6 cards from Alphonse.• Remove 1, 2 or 3 cards in order to pass 3 cards back

to Alphonse.• Alphonse is forced to remove 1 only, and pass back

2 to Beryl.• Beryl removes 1 and passes 1 back, so Alphonse

loses.

Solution 2 (Sufficient for full marks)Alphonse starts with 7 cards, and so can remove 1, 2 or3 cards, passing 6, 5 or 4 cards to Beryl.Beryl should remove 3, 2 or 1 cards respectively,leaving 3 cards only, and pass these 3 cards back toAlphonse.Alphonse now is forced to remove 1 card only, andpass 2 back to Beryl.Beryl removes 1 card (her only option) and passes 1back to Alphonse, who thus loses.Therefore, Beryl is guaranteed to win.

(b) Solution 1We must determine if 52 is a winning position or alosing position. By a similar argument to above, since7 is a losing position, 8 through 14 are all winningpositions, since they can all be reduced to 7 in one turn.Therefore, 15 is a losing position, since given 15 cards,a player is forced to reduce to some number between8 and 14, since no more than 7 cards can be removed.

Similarly, 16 through 30 are winning positions, 31 isa losing position, and 32 through 62 are winningpositions.Therefore, 52 is a winning position, so Alphonse hasa winning strategy.

Summary of Alphonse’s strategy• Alphonse removes 21 cards from original 52, and

pass 31 cards to Beryl.• If Beryl removes b1 cards with 1 151≤ ≤b ,

Alphonse removes 16 1− b cards to reduce thepack to 15 cards. [Notice that this is always a

legal move, since 2 16 32 2 311 1 1−( ) = − ≤ −b b b ,

so 16 1− b is never more than half of the pack.]

Si Alain reçoit 7 cartes, Brigitte peut donc toujoursgagner.

Résumé de la stratégie de Brigitte• Elle recevra 4, 5 ou 6 cartes d’Alain.• Elle retirera 1, 2 ou 3 cartes de manière à remettre 3

cartes à Alain.• Alain est forcé à retirer 1 carte et à remettre 2 cartes à

Brigitte.• Brigitte retire 1 carte et remet 1 carte à Alain qui perd.

Solution 2 (suffisante pour recevoir le maximum de points)Alain reçoit 7 cartes et il peut retirer 1, 2 ou 3 cartes pourremettre 6, 5 ou 4 cartes à Brigitte.Brigitte doit retirer 3, 2 ou 1 carte de manière à remettre 3cartes à Alain.Alain est forcé à retirer 1 carte et à remettre 2 cartes àBrigitte.Brigitte retire 1 carte, (elle n’a pas d’autre choix) et elleremet 1 carte à Alain qui perd.Brigitte a donc une stratégie gagnante.

b) Solution 1On doit déterminer si 52 est une position gagnante ouperdante.Comme dans la partie précédente, on peut démontrer que 8,9, 10, …, 14 sont des positions gagnantes, puisqu’on peutretirer suffisamment de cartes pour remettre 7 cartes àl’adversaire qui reçoit alors une position perdante.Donc 15 est une position perdante, puisqu’en recevant 15cartes, on doit remettre de 8 à 14 cartes à l’adversaire quireçoit alors des positions gagnantes.

De la même manière, les nombres de 16 à 30 sont despositions gagnantes, 31 est une position perdante et lesnombres de 32 à 62 sont des positions gagnantes.

Donc 52 est une position gagnante. Alain peut donc utiliserune stratégie gagnante.

Résumé de la stratégie d’Alain• Alain retire 21 cartes du jeu de 52 cartes et remet 31

cartes à Brigitte.• Si Brigitte retire b1 cartes, 1 151≤ ≤b , Alain retire alors

16 1− b cartes et remet 15 cartes à Brigitte. [On remarqueque c’est toujours permis, car

2 16 32 2 311 1 1−( ) = − ≤ −b b b et 16 1− b n’est jamaissupérieur à la moitié du nombre de cartes reçues.]

• Si Brigitte retire b2 cartes, 1 72≤ ≤b , retire alors 8 2− bcartes et remet 7 cartes à Brigitte. [Ce nombre esttoujours permis selon un argument semblable.]

• Alain adopte maintenant la même stratégie que Brigittedans la partie a).

13


• If Beryl removes b2 cards with 1 72≤ ≤b ,Alphonse removes 8 2− b to reduce the pack to 7cards. [This move is always legal by a similarargument.]

• Beryl now has 7 cards, so Alphonse should adoptBeryl’s strategy from (a).

Solution 2Alphonse removes 21 cards from original 52, andpasses 31 cards to Beryl.

If Beryl removes b1 cards with 1 151≤ ≤b , Alphonse

removes 16 1− b cards to reduce the pack to 15 cards.[This is always a legal move, since

2 16 32 2 311 1 1−( ) = − ≤ −b b b , so 16 1− b is never

more than half of the pack.]

If Beryl removes b2 cards with 1 72≤ ≤b , Alphonse

removes 8 2− b to reduce the pack to 7 cards. [Thismove is always legal by a similar argument.]

Beryl now has 7 cards, so Alphonse should adoptBeryl’s strategy from (a), so Alphonse has a winningstrategy.

CommentsThe "Alphonse and Beryl" questions continue to be ahighlight of the COMC. Part (a) met with a good degreeof success. Most students quickly realized that the posi-tion of 3 cards was the important one on which to focus.Part (b) did not meet with as much success – many studentsthought that the strategy had something to do with parity(ie. even or odd numbers of cards) or with a "matching"strategy (if Beryl takes 5 cards, Alphonse should take 5cards). Competitors should have a look at the solution andthen try playing the game with an unsuspecting friend!Average: 4.0

3. (a) SolutionCalculating,

Now 2c is always even for c an integer, so 2 5c −is always odd.

(b) Solution 1

Assume that g x( ) = 0 has three integer roots a, b, c,ie.

g x x a x b x c x px qx r( ) = −( ) −( ) −( ) = + + +3 2.

Now g abc( )0 = − from the above and is odd, soeach of a, b and c must be odd for their product to beodd.

f f c c

c

( ) ( )0 1 1 6

2 5

+ − = + − +( )= −

Solution 2Alain retire 21 cartes du jeu de 52 cartes et remet 31 cartes àBrigitte.

Si Brigitte retire b1 cartes, 1 151≤ ≤b , Alain retire alors

16 1− b cartes et remet 15 cartes à Brigitte. [On remarque que

c’est toujours permis, car 2 16 32 2 311 1 1−( ) = − ≤ −b b b et

16 1− b n’est jamais supérieur à la moitié du nombre de cartesreçues.]

Si Brigitte retire b2 cartes, 1 72≤ ≤b , retire alors 8 2− bcartes et remet 7 cartes à Brigitte. [Ce nombre est toujourspermis selon un argument semblable.]

Alain adopte maintenant la même stratégie que Brigitte dansla partie a). Il a donc une stratégie gagnante.

CommentairesLes questions pertinentes à « Alphonse et Beryl » demeurentsans contredit un des points marquants du concours du Défiouvert canadien de mathématiques (DOCM). Les élèves ontbien réussi la partie (a). La plupart ont rapidement comprisque la position des trois cartes constituait l’aspect essentieldu problème. Les élèves ont cependant éprouvé des difficultésà la partie (b) où bon nombre ont supposé que la méthode derésolution était liée au nombre égal de cartes paires etimpaires ou à la correspondance du nombre de cartes (siBeryl prend cinq cartes Alphonse devrait faire de même). Lesconcurrents devraient lire la solution puis inviter un camaradeà se laisser prendre au jeu!Moyenne: 4,0

3. a) Solution

Puisque c est un entier, 2c est toujours pair et 2 5c − estdonc toujours impair.

b) Solution 1Supposons que l’équation g x( ) = 0 admet trois racinesentières, a, b et c.

Donc g x x a x b x c x px qx r( ) = −( ) −( ) −( ) = + + +3 2 .Donc g abc( )0 = − et puisque g(0) est impair, alorschacune des racines doit être impaire pour que leurproduit soit impair.

On a g a b c−( ) = − −( ) − −( ) − −( )1 1 1 1 .Puisque a, b et c sont impairs, alors − −1 a, –1 – b et

–1 – c sont pairs et g −( )1 est donc pair, ce qui est unecontradiction.Donc l’équation g x( ) = 0 ne peut admettre trois racinesentières.

f f c c

c

( ) ( )0 1 1 6

2 5

+ − = + − +( )= −

14


Therefore, g a b c−( ) = − −( ) − −( ) − −( )1 1 1 1 .

Since a is odd, − −1 a is even, and so g −( )1 is even, a

contradiction. (In fact, g −( )1 is divisible by 8.)

Thus, g x( ) = 0 cannot have three integer roots.

Solution 2

Assume that g x( ) = 0 has three integer roots a, b, c,

ie. g x x a x b x c x px qx r( ) = −( ) −( ) −( ) = + + +3 2

CommentsThe purpose of part (a) was to have a relatively straightfor-ward "proof" question, which required students to write alogical argument. In general, students did exceptionallywell on (a). Part (b) was a fair bit more difficult. Manystudents recognized that r had to be odd and then that oneof p and q is even and the other odd, but were then stuck. Afew students ingeniously pointed out that not only can thiscubic equation not have 3 integer roots, it cannot even have1 integer root, as they showed with the following proof:Suppose that q and r are both odd and p even.

Let a be an integer root of x px qx r3 2 0+ + + = , ie.

a pa qa r3 2 0+ + + = .

If a is even, then a3, pa2, and qa are all even, and r is odd,

so a pa qa r3 2+ + + is odd and so cannot be 0.

If a is odd, then a3, qa, and r are all odd, and is even (since

p is even), so a pa qa r3 2+ + + is odd and so cannot be 0.Therefore, there cannot be any integer roots.(A similar argument works for p odd and q even.)Congratulations to those students who had this brilliantinsight!Average: 4.0

Solution 2Supposons que l’équation g x( ) = 0 admet troisracines entières, a, b et c.Donc

g x x a x b x c x px qx r( ) = −( ) −( ) −( ) = + + +3 2 .On développe pour obtenir :

x a b c x ab ac bc x abc

x px qx r

3 2

3 2

− + +( ) + + +( ) −

= + + +Puisque g(0) et g(–1) sont impairs, et que g r( )0 =et g p q r−( ) = − + − +1 1 alors r et –1 + p – q + rsont impairs.Donc p q− est impair.Donc p ou q doit être pair (ils ne peuvent êtreimpairs tous les deux, car leur différence seraitpaire).

Puisque r est impair et que r abc= − , alors a, b et csont tous impairs. Donc les nombres

sont impairs tous les deux, ce qui est unecontradiction car on vient de conclure que p ou qdoit être pair.Donc l’équation g x( ) = 0 ne peut admettre troisracines entières.

CommentairesLes élèves ont généralement très bien réussi la partie (a)où il fallait résoudre une question au moyen d’unraisonnement logique, mais ont éprouvé davantage dedifficultés avec la partie (b). De nombreux élèves ontcompris que r était impair et que de p et q l’un devaitêtre pair et l’autre impair sans toutefois aller plus loindans la résolution du problème. Quelques élèvesingénieux ont indiqué que l’équation cubique ne pouvaitavoir trois racines entières ni même une comme ledémontre le raisonnement suivant :

Supposons que q et r sont tous deux pairs et que p estimpair.Soit a est une racine entière de l’équation

x px qx r3 2 0+ + + = , c’est-à-dire

a pa qa r3 2 0+ + + = .

Si a est impair, alors a3, pa2 et qa sont tous pairs, et

r est impair, a pa qa r3 2+ + + a pour solution unnombre impair autre que 0.

Si a est impair, alors a3, qa et r sont tous impairs et pa2

est impair (puisque p est impair, alors a pa qa r3 2+ + +a pour solution un nombre pair autre que 0).

p a b c

q ab ac bc

= − + +( )= + +

or, expanding,

x a b c x ab ac bc x abc

x px qx r

3 2

3 2

− + +( ) + + +( ) −

= + + +Now g r( )0 = so r is odd, and

g p q r−( ) = − + − +1 1 , which is odd.Combining these, since r is odd, then p q− is odd too.Therefore, one of p and q is even (they cannot both beodd, since odd – odd = even).Since r is odd and r abc= − , then each of a, b and cis odd. This implies that

p a b c

q ab ac bc

= − + +( )= + +

are both odd, a contradiction (since we have shownabove that one of p and q must be even.)

Therefore, g x( ) = 0 cannot have three integer roots.

15

x

ABC

x

sin sin

cos

cos

sin(**)

α θ α

θ α

α

=+ +( )

=+( )

=∠( )

=

=

5

90

5

5

5

253

35

o


4. Solution 1 (Trigonometry)

Let ∠ = ∠ =BCP ABP αand ∠ =ACP θ .Then ∠ =PBC θ since∆ABC is isosceles. Also,

∠ = −PAC 90o θ (from∆APC), ∠ = +ADP 2θ α(exterior angle), and ∠ = + +APB 90o θ α(exterior angle).

Let AP x= . Then from ∆ APC ,

sin sinθ θ= = ⇒ =x

AC

xx

55 (*)

By the sine law in ∆ABP,

A

B C

D

P

90 – θ

2+ α

θ

θαα

θ

90 + θ + αx

A

B C

D

P

90 – θ

2+ α

θ

θαα

θ

90 + θ + αx

Par conséquent, il n’existe pas de racine entière.(Le raisonnement semblable voulant que p soit impair et qpair est également valable.)Félicitations aux élèves qui sont arrivés à cette brillantedéduction!Moyenne: 4,0

4. Solution 1 (par trigonométrie et géométrie analytique)Soit ∠ = ∠ =BCP ABP α et ∠ =ACP θ .Donc ∠ =PBC θ , puisque le triangleABC est isocèle.Dans le triangle APC,

∠ = −PAC 90o θ . ∠ = +ADP 2θ α(angle extérieur du triangle BCD)

∠ = + +APB 90o θ α (angle extérieur du triangle ADP)

Soit AP x= . Dans le triangle APC, on a sin θ = =x

AC

x

5,

d’où x = 5 sin θ (*).

D’après la loi des sinus dans le triangle ABP :

[On remarque que cos ∠( ) =ABC 35 , car la hauteur au point A

est aussi la médiatrice de BC. On a aussi sin ∠( ) =ABC 45 .]

On utilise (*) et (**) pour obtenir :

Pour déterminer le rapport de AD à DC, on utilise un repèrecartésien.Soit (0, 0) les coordonnées du point B et (6, 0) les coordonnéesdu point C. Les coordonnées du point A sont (3, 4), puisque

la hauteur de A à BC a une longueur de 4. Puisque tanθ = 23 ,

x

ABC

x

sin sin

cos

cos

sin(**)

α θ α

θ α

α

=+ +( )

=+( )

=∠( )

=

=

5

90

5

5

5

253

35

o

[Note that cos ∠( ) =ABC 35 since drawing a perpen-

dicular from A bisects BC. Also, sin ∠( ) =ABC 45 .]

Therefore combining (*) and (**)

5 253

3 5

3 5

3 5 5

3 4 3

6 423

sinsinsin sin

sin sin

sin sin cos cos sin

sin cos sin

sin cos

tan

θαθ αθ θθ θ θθ θ θθ θθ

=

== ∠ −( )= ∠( ) − ∠( )= −=

=

ABC

ABC ABC

To determine the ratio of AD to DC, we use coordinates.

Let B have coordinates 0 0,( ) and C have coordinates

6 0,( ). Thus, A has coordinates 3 4,( ), since the altitudefrom A to BC has length 4.

Since tanθ = 23 , the line from B to D has equation

y x= 23 . Also, the line from A to C has equation

y x= − −( )43 6 . To find D, we find the intersection of

these two lines:

5 253

3 5

3 5

3 5 5

3 4 3

6 423

sinsinsin sin

sin sin

sin sin cos cos sin

sin cos sin

sin cos

tan

θαθ αθ θθ θ θθ θ θθ θθ

=

== ∠ −( )= ∠( ) − ∠( )= −=

=

ABC

ABC ABC

16

23

43 8

2 8

4

x x

x

x

= − +==

Therefore, D has coordinates 4 83,( ) . Since the x-coordi-

nate of D is 13 of the way between those of A and C, then

AD DC: := 1 2.

Solution 2 (Similar triangles)Draw a perpendicular from A tomeet BC at M. Then sinceAB AC= , BM MC= = 3 and soAM = 4.Let ∠ = ∠ =BCP ABP α and∠ =ACP θ . Then ∠ =PBC θsince ∆ABC is isosceles.Draw circle with AC has diameter. This circle passes

through both P and M, since ∠ = ∠ =APC AMC 90o.Join P to M. Then ∠ =PAM α since ∠ = ∠PAM PCM

(subtended by the same chord). Also, ∠ =AMP θ forsimilar reasons. Therefore, ∆MPA is similar to ∆BPC .Thus,

PA

PC

MA

BC

PA

PC= = ⇒ = =4

623

tanθ

So now we must compute the length of DC. Consider∆BDC . By the sine law,

DC BC

BDC

DCDCB

DCB

DCB DCB

DCB DCB

sin sin

sin

sin

sinsin

sinsin cos cos sin

cos cot sin

θθ

θ

θθ

θθ θ

θ

=∠( )

=− − ∠( )

=+ ∠( )

=∠( ) + ∠( )

=∠( ) + ∠( )

=+ ( )

=

6

180

6

6

6

6

103

35

32

45

o

which yields also that AD = − =5 103

53 and so

AD DC: := 1 2.


θα

αθ

α

θ

A

B M C

D

P

3 3

5

θα

αθ

α

θ

A

B M C

D

P

3 3

5

DC BC

BDC

DCDCB

DCB

DCB DCB

DCB DCB

sin sin

sin

sin

sinsin

sinsin cos cos sin

cos cot sin

θθ

θ

θθ

θθ θ

θ

=∠( )

=− − ∠( )

=+ ∠( )

=∠( ) + ∠( )

=∠( ) + ∠( )

=+ ( )

=

6

180

6

6

6

6

103

35

32

45

o

la droite qui passe par B et D a pour équation y x= 23 . De

plus, la droite qui passe par A et C a pour équation

y x= − −( )43 6 . Le point D est le point d’intersection de ces

deux droites. Donc :

Les coordonnées de D sont donc 4 83,( ) .

On compare les déplacements horizontaux de A à D et deD à C :

Donc AD DC: := 1 2.

Solution 2 (par triangles semblables)On abaisse une perpendiculaire AM deA à BC. Puisque AB AC= , alorsBM MC= = 3, d’où AM = 4.Soit ∠ = ∠ =BCP ABP α et∠ =ACP θ . Donc ∠ =PBC θ ,puisque le triangle ABC est isocèle.On trace un cercle ayant AC pour diamètre. Ce cercle

passe aux points P et M, puisque ∠ = ∠ =APC AMC 90o.On joint P et M. Donc ∠ =PAM α , puisque∠ = ∠PAM PCM (ces angles interceptent le même arc).De même, ∠ =AMP θ . Les triangles MPA et BPC sontdonc semblables.

Donc PA

PC

MA

BC= = 4

6, d’où tan θ = =PA

PC

23

.

On cherche maintenant à déterminer DC. Selon la loi dessinus dans le triangle BDC :

Donc AD = − =5 103

53 , d’où AD DC: := 1 2.

23

43 8

2 8

4

x x

x

x

= − +==

4 36 4

12

––

=

17

A

B

K

M C

D

OX

P


Solution 3 (Geometry)Draw a line from A perpen-dicular to BC at M.Let O be the midpoint of AC.Draw circle with centre Oand radius OC.Then this circle passes through A (since AO OC= ), P

and M (since ∠ = ∠ =APC AMC 90o).Join M to O and extend this line segment to meet thecircle at K.

Since CO CA= 12 and CM CB= 1

2 , then MK is parallel

to BA.Extend BD to meet MK at ′K .Then ∠ ′ = ∠ ′MK B ABK because of parallel lines.But ∠ ′ = ∠ = ∠ ⇒ ∠ ′ = ∠ABK ABP BCP MK P MCP.Therefore, ′K lies on the circle; that is, ′K coincides withK.

Next, join A to K and K to C.Then AKCM is a rectangle, since KM and AC are diametersof the circle, so the quadrilateral has four right angles.Therefore, AK MC BM= = .Then AKMB is a parallelogram, which implies that AM andBK bisect each other (meeting at X).

Consider now ∆AKM . Then KX and AO are medians, andso intersect in the ratio 2 1: , ie. AD DO: := 2 1. SinceAO OC: := 1 1, then AD DC: := 1 2.

CommentsThis question was extremely difficult, but it was gratifyingto see many students at least making an effort to start thequestion. The first step was to determine some of theangles in the diagram, which was relatively straightfor-ward. The second step of recognizing the presence of acircle and then finding similar triangles required a greatdeal of insight. Of the three solutions presented, thesecond is probably the nicest. The first solution requiresless insight, but is more difficult computationally.Average: 0.3

A

B

K

M C

D

OX

P

Solution 3 (par géométrie)On abaisse une perpendiculaire AM de Aà BC. Soit O le milieu de AC. On trace uncercle de centre O et de rayon OC. Cecercle passe au point A, puisqueAO OC= , et aux points P et M,puisque ∠ = ∠ =APC AMC 90o.On joint M et O et on prolonge le segment obtenu jusqu’aucercle au point K.Puisque CO CA= 1

2 et CM CB= 12 , alors MK est parallèle à

BA. On prolonge BD jusqu’à MK au point ′K .Donc ∠ ′ = ∠ ′MK B ABK à cause des segments parallèles.Or ∠ ′ = ∠ = ∠ABK ABP BCP, d’où ∠ ′ = ∠MK P MCP.Donc ′K est situé sur le cercle, ce qui indique que ′K et Kcoïncident.On joint A et K, ainsi que K et C.Puisque KM et AC sont des diamètres du cercle, le quadrilatèreAKCM est un rectangle.Donc AK MC BM= = .Donc AKMB est un parallélogramme. Donc AM et BK secoupent en leur milieu au point X.On considère maintenant le triangle AKM. KX et AO sont desmédianes de ce triangle. Elles se coupent donc dans le rapport2:1. Donc AD DO: := 2 1. Puisque AO OC: := 1 1, alorsAD DC: := 1 2.

CommentairesNous nous réjouissons des tentatives de bon nombre d’élèvesà résoudre cette question d’un haut niveau de difficulté. Dansun premier temps, il fallait déterminer la valeur de quelquesangles du diagramme, une étape somme toute simple. Dansun deuxième temps, il fallait discerner la présence d’uncercle, puis trouver des triangles semblables par une approchedéductive complexe. Parmi les trois solutions données par lesélèves, la seconde comportait sans doute les éléments les plusintéressants tandis que la première nécessitait un raisonnementmoins complexe mais davantage de calculs difficiles.Moyenne: 0,3


The CENTRE for EDUCATION in MATHEMATICS and COMPUTING

TheCanadian Open



Solutions


2002 COMC Solutions 2

Part A

1. By Pythagoras in ∆PFR, PF 2 2 213 5 144= − = , orPF = 12.By Pythagoras in ∆PFQ, PQ2 2 29 12 225= + = , orPQ = 15.Therefore, the side lengths of ∆PQR are 13, 14 and 15,i.e. the perimeter is 42.

P

Q F R9 5

1312

2. Solution 1

x xy y x xy y xy

x y xy

2 2 2 2

2

2

5 2 3

3

4 3 12

20

+ + = + + +

= +( ) +

= + −( )= −

Solution 2Examining the two given equations, we see that x = 6 and y = −2 is a solution.

Therefore, x xy y2 2 2 25 6 5 6 2 2 36 60 4 20+ + = + ( ) −( ) + −( ) = − + = − .

Solution 3We solve the first equation for x and substitute into the second equation.From the first equation, x y= −4 .

Substituting into the second equation, 4 12−( ) = −y y or 0 4 122= − −y y .

Factoring, 0 6 2= −( ) +( )y y , i.e. y = 6 or y = −2. The corresponding values of x are

x = −2 and x = 6, which give the same answer as in Solution 2, i.e. x xy y2 25 20+ + = − .

3. To determine ∠EAR , we look at the angles around thepoint E.

We know that ∠ + ∠ + ∠ + ∠ =AER REN NEP PEA 360o.Since ∠PEA is an angle in an equilateral triangle,

∠ =PEA 60o.

Since ∠NEP is an angle in a square, ∠ =NEP 90o.

A

P

O ND

R

TE

102°108°90°

60°

Since ∠REN is an angle in a regular pentagon, ∠ = ( ) =REN 15 540 108o o .


Therefore,

∠ = − ∠ − ∠ − ∠

= − − −

=

AER REN NEP PEA360

360 108 90 60

102

o

o o o o

o

Now since PEA is an equilateral triangle, OPEN is a square, and TREND is a regularpentagon, then their side lengths must all be the same, since OPEN and TREND share aside, and since OPEN and PEA share a side. In particular, AE ER= .Therefore, ∆ARE is an isosceles triangle, and so

∠ = − ∠( ) = −( ) =ARE AER12

12180 180 102 39o o o o.

4. Solution 1The sum of the 3rd, 4th and 5th terms of the sequence is equal to the sum of the first fiveterms of the sequence minus the sum of the first two terms of the sequence.

Thus, the sum is 5 5 6 5 5 2 6 2 155 32 1232 2( ) + ( )[ ] − ( ) + ( )[ ] = − = .

Solution 2We determine the first 5 terms in the sequence and then add up the 3rd, 4th and 5th terms.From the formula given, the sum of the first 1 terms is 11.This tells us that the first term is 11.From the formula given, the sum of the first 2 terms is 32. Since the first term is 11, thenthe second term is 21.Next, the sum of the first 3 terms is 63, and so the third term is 31, since the first two termsare 11 and 21. (We could use the fact that the sum of the first two terms is 32, instead.)Next, the sum of the first 4 terms is 104, and so the fourth term is 41.Lastly, the sum of the first 5 terms is 155, and so the fifth term is 51.Therefore, the sum of the 3rd, 4th and 5th terms is 31 41 51 123+ + = .

Solution 3Since the sum of the first n terms has a quadratic formula, then the terms in the sequencehave a common difference, i.e. The sequence is an arithmetic sequence.Therefore, the sum of the 3rd, 4th and 5th terms is equal to three times the 4th term.The 4th term is the sum of the first four terms minus the sum of the first three terms, i.e.104 63 41− = .Thus, the sum of the 3rd, 4th and 5th terms is 3 41 123( ) = .


5. Solution 1Since the value of this expression is the same for every positive integer a, then we can findthe value by substituting in a = 1.Thus,

2 1 2 1

1 11 30 2

1 2 30 1 2

63

2a a

a a

−( )∇ +( )[ ]−( )∇ +( )[ ] =

∇[ ]∇[ ] =

+ ++ +

= =

Therefore, the value required is 2.

Solution 2If a is a positive integer, the only integer between 2 1a − and 2 1a + is 2a . Similarly, theonly integer between a −1 and a +1 is a.Thus,

2 1 2 1

1 12 1 2 2 1

1 163

2a a

a a

a a a

a a a

a

a

−( )∇ +( )[ ]−( )∇ +( )[ ] =

−( ) + + +( )−( ) + + +( ) = =

Therefore, the value required is 2.

6. Label the two ends of the mirrors U and W, as shown.

Since the initial beam is parallel to the mirror WV, then ∠ =UAS 30o. Since the angle ofincidence equals the angle of reflection, then the reflected beam of light also makes an

angle of 30o with the mirror UV.

Let B be the point on the mirror WV that the light next strikes.

Since ∠ = ∠ =VAB AVB 30o, then the angle of incidence, ∠ABW , is equal to 60o, becauseit is an external angle of ∆ABV . (∠ABW could also have been calculated using the facts

that ∠ =SAB 120o and SA is parallel to WV.)

Therefore, the angle of reflection is also 60o.Let C be the point on the mirror UV where the light next

strikes. Since ∠ =CVB 30o and ∠ =VBC 60o, then

∠ =BCV 90o. This tells us that the light is reflectedstraight back along its path from C back to S.Therefore, the required distance is 2 SA AB BC+ +( ).

AS

30°B

W

U

60°60°

30° 12

12

130°

V

C

Since we are given that SA AV= = 1, then since ∆ABV is isosceles with BC an altitude,then AC CV= = 1

2 , and so BC AC= ( ) =13

12 3

and AB AC= ( ) =23

13

.

Therefore, the required distance is

2 2 1 2 2 313

12 3

33

SA AB BC+ +( ) = + +( ) = + = +

Thus, the total distance travelled by the beam is 2 3+( ) m, or about 3.73 m.


7. Solution 1Since P is formed by adding a 1 at the end of N, then P N= +10 1.Since Q is formed by adding a 1 in front of the 5 digits of N, then Q N= +100000 .Since P Q= 3 ,

10 1 3 100000

10 1 300000 3

7 299999

42857

N N

N N

N

N

+ = +( )+ = +

==

Therefore, N is 42857.

Solution 2Suppose N has digits abcde. Then since P Q= 3 , we have abcde abcde1 3 1= ( ) .

Since the units digit on the left side is 1, then the units digit on the right is also 1, whichmeans that e = 7.Thus, abcd abcd71 3 1 7= ( ) . Since the tens digit on the left side is 7 and we get a “carry” of

2 from multiplying the last digit on the right side by 3, then 3× d has a units digit of 5, i.e.d = 5.Thus, abc abc571 3 1 57= ( ). Since the hundreds digit on the left side is 5 and we get a carry

of 1 from multiplying the last two digits on the right side by 3, then the units digit of 3× cmust be a 4, i.e. c = 8.Thus, ab ab8571 3 1 857= ( ). In a similar fashion, we see that b = 2 and a = 4 .

Therefore, N = 42857.

8. We are not told that M must be a positive integer, but it makes sense to look for a positiveinteger M that satisfies these conditions, since we want the maximum possible value of M.Since there are 1000 numbers in the set 1 2 3 999 1000, , , , ,K{ } and the probability that an x

chosen randomly from this set is a divisor of M is 1100 , then M must have 10 divisors

between 1 and 1000.Since we are told that M ≤ 1000, then M must have exactly 10 positive divisors.

Therefore, M must be of the form p9 where p is a prime number, or p q4 where p and q are

both primes.(Recall that to find the number of positive divisors of M, we find the prime factorization ofM and then take each of the exponents, add 1, and find the product of these numbers. For

example, if M = =48 2 34 , then the number of positive divisors is 4 1 1 1 10+( ) +( ) = .)

Now, we want to determine the maximum M in each of these two forms.

Case 1 M p= 9

Since M ≤ 1000, then we must have p = 2, i.e. M = 512.

(If p = 3, then p9 19683= is too large.)


Case 2 M p q= 4

Since M ≤ 1000 and 5 6254 = , then we must have p = 2 or p = 3.If p = 2, then the largest q can be so that q is prime and M ≤ 1000 is 61, i.e.M = ( )( ) =16 61 976.If p = 3, then the largest q can be so that q is prime and M ≤ 1000 is 11, i.e.M = ( )( ) =81 11 891.

Therefore, the maximum possible value of M is 976.

Part B

1. (a) The slope of the line through P and F is5 10 8

12

−−

= −

and so the slope of the desired line is also − 12 .

Since the point Q lies on the y-axis, then the y-intercept of the line is 3.Therefore, the line is y x= − +1

2 3.

(b) Since AD lies along the x-axis, then G is thepoint where the line from (a) crosses the x-axis.To find the coordinates of G, we set y = 0 in

the line from (a) to get 0 312= − +x or x = 6.

Therefore, the desired line passes through thepoints G 6 0,( ) and F 8 1,( ). Thus its slope is1 08 6

12

−−

= , and so its equation is

y x− = −( )0 612 or y x= −1

2 3.

y

xD(8, 0)G(6, 0)A(0, 0)

Q(0, 3)

P(0, 5)

B(0, 8)C(8, 8)

F(8, 1)

(c) Since FG has slope 12 , then a line perpendicular to FG has slope −2, the negative

reciprocal of 12 .

Since the desired line passes through H 4 4,( ) , it has equation y x− = − −( )4 2 4 ory x= − +2 12.


(d) The circle has centre H 4 4,( ) , and it is tangent to

all four sides of the square, and so its radius mustbe 4, since the distance from the centre to each ofthe four sides is 4.Does this circle intersect the line y x= −1

2 3, i.e.

the line through F and G?

y

xD(8, 0)G(6, 0)A(0, 0)

Q(0, 3)

P(0, 5)

B(0, 8)C(8, 8)

F(8, 1)

H(4, 4)

We must find the shortest distance between the centre of the circle and the line, i.e. theperpendicular distance. We already have the equation of a line through H that isperpendicular to the line through F and G, the line y x= − +2 12. Where do these lines

intersect? Setting y-coordinates equal,12

52

3 2 12

15

6

x x

x

x

− = − +

==

i.e. the lines intersect at the point G 6 0,( )! Therefore, the shortest distance from H

to the line through F and G is the distance from H to G, which is

6 4 4 0 202 2−( ) + −( ) =i.e. is greater than than 4 16= , the radius of the circle.Therefore, the circle does not intersect the line.

2. (a) For the product 2 5 13A B( )( ) to be divisible by 36, we need it to be divisible by both 4

and 9. Since 2A5 is odd, it does not contain a factor of 2.Therefore, 13B must be divisible by 4.For a positive integer to be divisible by 4, the number formed by its last two digits mustbe divisible by 4, i.e. 3B is divisible by 4, i.e. B = 2 or B = 6.

Case 1 B = 2In this case, 132 is divisible by 3, but not by 9. Therefore, for the original product to bedivisible by 9, we need 2A5 to be divisible by 3.For a positive integer to be divisible by 3, the sum of its digits is divisible by 3, i.e.2 5 7+ + = +A A is divisible by 3.Therefore, A = 2 5 or or 8.

Case 2 B = 6In this case, 136 contains no factors of 3, so for the original product to be divisible by 9,we need 2A5 to be divisible by 9.


For a positive integer to be divisible by 9, the sum of its digits is divisible by 9, i.e.2 5 7+ + = +A A is divisible by 9. Therefore, A = 2.

Therefore, the four possible ordered pairs are A B, , , , , , , ,( ) = ( ) ( ) ( ) ( )2 2 8 2 5 2 2 6 .

(b) (i) If 10 7a b m+ = , then b m a= −7 10 . Thus,a b a m a a m a m− = − −( ) = − = −( )2 2 7 10 21 14 7 3 2

Since 3 2a m− is an integer, then by definition, a b− 2 is divisible by 7.

(ii) Solution 1If 5 4c d+ is divisible by 7, then 5 4 7c d k+ = for some integer k.Therefore, d k c= −( )1

4 7 5 .

So 4 4 7 5 21 77 3

414

14c d c k c c k

c k− = − −( ) = −( ) =

−( ).

Since 4c d− is an integer, then 7 3c k−( ) must be divisible by 4. But 4 has no

common factors with 7, so 4 must divide into 3c k− , i.e. 3

4c k−

is an integer.

Therefore, 4 73

4c d

c k− =

−

, i.e. 4c d− is divisible by 7.

Solution 2We note that 4 14 7 2 5 4c d c d c d− = +( ) − +( ) .

Since both terms on the right side are divisible by 7, then 4c d− is divisible by 7.

Solution 3Multiplying the expression 4c d− by 5 does not affect its divisibility by 7.Thus, we can consider whether or not 20 5c d− is divisible by 7, and this will beequivalent to considering 4c d− .Since we are told that 5 4 7c d t+ = for some integer t, then we know that4 5 4 20 16 28c d c d t+( ) = + = or 20 28 16c t d= − .

If we now consider 20 5c d− , we see20 5 28 16 5

28 21

7 4 3

c d t d d

t d

t d

− = −( ) −= −= −( )

Since 20 5c d− is divisible by 7 by definition, then 4c d− is divisible by 7.


3. (a) We consider the possible cases. On his first turn, Alphonse can take either 1 marble or2 marbles.If Alphonse takes 1 marble, Beryl can take 2 marbles and then Colleen 1 marble, toleave Alphonse with 1 marble left in the bowl. Therefore, Alphonse loses. (Note thatBeryl and Colleen can agree on their strategy before the game starts.)If Alphonse takes 2 marbles, Beryl can take 1 marble and then Colleen 1 marble, toleave Alphonse again with 1 marble left in the bowl. Therefore, Alphonse loses.In either case, Beryl and Colleen can work together and force Alphonse to lose.

(b) Solution 1On their two consecutive turns, Beryl and Colleen remove in total 2, 3 or 4 marbles.On his turn, Alphonse removes either 1 marble or 2 marbles. Therefore, by workingtogether, Beryl and Colleen can ensure that the total number of marbles removed onany three consecutive turns beginning with Alphonse’s turn is 4 or 5. (Totals of 3 and 6cannot be guaranteed because of Alphonse’s choice.)

Therefore, if N is a number of marbles in which Alphonse can be forced to lose, then soare N + 4 and N + 5, because Beryl and Colleen can force Alphonse to choose from Nmarbles on his second turn.

From (a), we know that 5 is a losing position for Alphonse. Also, 1 is a losing positionfor Alphonse. (Since 1 is a losing position, then 5 and 6 are both losing positions,based on our earlier comment.)

Since 5 and 6 are losing positions, then we can determine that 9, 10 and 11 are alsolosing positions, as are 13, 14, 15, and 16. If we add 4 to each of these repeatedly, wesee that N is a losing position for every N ≥ 13.

What about the remaining possibilities, i.e. 2, 3, 4, 7, 8, and 12?For N = 2 or N = 3, if Alphonse chooses 1 marble, then either Beryl or Colleen isforced to take the last marble, so these are not losing positions for Alphonse, i.e. theyare winning positions.For N = 4, if Alphonse chooses 2 marbles, then either Beryl or Colleen is forced totake the last marble, so this is also not a losing position for Alphonse.

Next, we notice that if Alphonse chooses 1 marble, then the total number of marbleschosen by the three players will be 3, 4 or 5, and if Alphonse chooses 2 marbles, thenthe total number chosen will be 4, 5 or 6.So if N = 7, then Alphonse can choose 1 marble and ensure that he receives 2, 3 or 4marbles on his next turn. So 7 is a winning position for Alphonse.


If N = 8, then Alphonse can choose 2 marbles and ensure that he receives 2, 3 or 4marbles on his next turn. So 8 is also a winning position for Alphonse.

Lastly, we consider N = 12.If Alphonse chooses 1 marble, Beryl and Colleen can choose 1 each and return 9marbles to Alphonse. As we have shown, this is a losing position for Alphonse.If Alphonse chooses 2 marbles, Beryl and Colleen can choose 2 each and return 6marbles to Alphonse. This is a losing position for Alphonse.

Therefore, the values of N for which Beryl and Colleen can force Alphonse to lose are1, 5, 6, and all N for which N ≥ 9.

Solution 2First, we notice that if Alphonse chooses 1 marble, then the total number of marbleschosen by the three players will be 3, 4 or 5, and if Alphonse chooses 2 marbles, thenthe total number chosen will be 4, 5 or 6.

We define a “losing position” to be a number of marbles in the bowl so that if Alphonsestarts with this number, he can be forced to lose.

From (a), we know that 5 is a losing position for Alphonse. Also, 1 is a losing positionfor Alphonse.For N = 2 or N = 3, if Alphonse chooses 1 marble, then either Beryl or Colleen isforced to take the last marble, so these are not losing positions (ie. they are winningpositions) for Alphonse.For N = 4, if Alphonse chooses 2 marbles, then either Beryl or Colleen is forced totake the last marble, so this is a winning position for Alphonse.

How can we ensure that a starting position N ≥ 6 is not a losing position?N will not be a losing position if either none of N − 3, N − 4 or N − 5 are losingpositions, or none of N − 4, N − 5 or N − 6 are losing positions. (If either groupconsists of three non-losing positions, then Alphonse can ensure that he gets a positionfrom the appropriate set at the beginning of his next turn by choosing either 1 or 2marbles respectively.)

Also, N will be a losing position as long as at least one of N − 3, N − 4, N − 5 and atleast one of N − 4, N − 5, N − 6 are losing positions. (If there is a losing position ineach group of 3, then no matter whether Alphonse chooses 1 or 2 marbles, then Beryland Colleen will be able to force Alphonse into one of these previously known losingpositions.)

Using these two criteria for checking whether a position is a losing position or not alosing position, we can see


i) N = 6 is a losing position, since N − =5 1 is a losing position in both groups ofthree

ii) N = 7 is not a losing position, since N − 3, N − 4, N − 5 (namely, 4, 3, 2) are notlosing positions.

iii) N = 8 is not a losing position, since N − 4, N − 5, N − 6 (namely, 4, 3, 2) are notlosing positions.

iv) N = 9 is a losing position, since N − =4 5 is a losing position in both groups ofthree

v) N = 10 is a losing position, since N − =5 5 is a losing position in both groups ofthree

vi) N = 11 is a losing position, since N − =5 6 is a losing position in both groups ofthree

vii) N = 12 is a losing position, since N − =3 9 and N − =6 6 are both losingpositions

And so we have obtained 4 consecutive losing positions, which guarantees us that anyN ≥ 13 will also be a losing position, since N − 4 will be a losing position in bothgroups of 3.


But among the first eight possibilities, there are now no more sets of three consecutivenon-losing positions. This tells us that every position for N ≥ 9 is a losing position,since we cannot find three consecutive non-losing positions as described above.


4. Solution 1Join E to P, Y and R, and join F to Q, Z and S.Let O be the point of intersection of EY and FZ.Since EY and FZ are altitudes in ∆DEF , then thethird altitude, DX say, passes through O.

If we look at altitude DX, we see that ∠ =DXE 90o .Since circle C2 has DE as its diameter, then point Xmust lie on circle C2, since a right angle is subtended

by the diameter at point X.

D

E F

R

X

S

Z

P

O Y

Q

Similarly, point X lies on circle C1.Therefore, DX is a chord of both circle C1 and circle C2.We can now use the “Chord-Chord Theorem” in each of circle C1 and C2, to say


SO OQ DO OX C

RO OP DO OX C

⋅ = ⋅⋅ = ⋅

(from circle

(from circle 2

1

)

)

From this we can conclude that SO OQ RO OP⋅ = ⋅ .

Why does this allow us to conclude that P, Q, R, and S lie on the same circle?

From the equation, we obtain SO

OP

RO

OQ= , which tells us that ∆SOP is similar to ∆ROQ ,

and so ∠ = ∠ = ∠ = ∠PSQ PSO ORQ PRQ .

Since the chord PQ subtends the equal angles ∠PSQ and ∠PRQ (in an undrawn circle),then the points P, Q, R, and S are concyclic.

Solution 2In order to show that the four points lie on a circle, we will show that the points areequidistant from a fifth point, which will thus be the centre of the circle on which the fourpoints lie.Consider first the points Q and S. Any point equidistant from Q and S lies on theperpendicular bisector of the line joining these points. Since Q and S both lie on circle C2,DE is a diameter of C2, and QS is perpendicular to DE (since they lie on an altitude of the

triangle), then DE is the perpendicular bisector of QS.Similarly, DF is the perpendicular bisector of PR.Therefore, any point that is equidistant from all four of the given points must lie on bothDE and DF. Thus, the only possible candidate is point D. (And we already know thatDS DQ= and DP DR= from our discussion of perpendicular bisectors.)

Thus, if we can show that DS DR= , then we will have shown what we need to show.

Method 1Let SZ c= , DZ a= and EZ b= .

Then DS DZ SZ c a2 2 2 2 2= + = + (Pythagoras).

Now if we extract ∆DSE , we see that ∠ =DSE 90o,since DE is a diameter of circle C2. Therefore,

∆DSZ is similar to ∆SEZ , or DZ

SZ

SZ

EZ= or c ab2 = .

Thus, DS a ab a a b DZ DE2 2= + = +( ) = ⋅ .

Similarly, DR DY DF2 = ⋅ , looking at ∆DRF .

D

E F

R

SZ

P

O Y

Q

cb

a

Now consider the points E, Z, Y, and F. Since ∠ = ∠ =EZF EYF 90o, then EF must be thediameter of the circle containing points Y and Z (and points E and F).Therefore, DE and DF are secants of the circle which intersect the circle at Z and Y,respectively. By the “Secant-Secant Theorem”, DZ DE DY DF⋅ = ⋅ .


From above, we can conclude that DS DR2 2= , or DS DR= , and thusDP DQ DR DS= = = .

Method 2

As above, we can obtain that DS a ab a a b DZ DE2 2= + = +( ) = ⋅ .

Since ∠ =DZF 90o , then DZ DF ZDF DF EDF= ∠( ) = ∠( )cos cos , and so

DS DZ DE DE DF EDF2 = ⋅ = ⋅ ∠( )cos .

Repeating the process on the other side of the triangle gives us that

DR DY DF DF DE EDF2 = ⋅ = ⋅ ∠( )cos , or DR DS2 2= .

Therefore, DP DQ DR DS= = = .Therefore, we can conclude that the points P, Q, R, and S are concyclic.


The CENTRE for Educationin MATHEMATICS and COMPUTING

TheCanadian Open



Solutions


2004 Canadian Open Mathematics Challenge Solutions 2

Part A

1. Let Gareth’s present age, in years, be G.Then Jeff’s age is G − 1, and Ina’s age is G + 2.Since the sum of their three ages is 118, then

G −1( ) + G + G + 2( ) =118

G = 39Therefore, Gareth’s age is 39.

2. When the point 4,−2( ) is reflected in the x-axis, its image is 4,2( ) .When the point 4,2( ) is reflected in the line y = x , its image is 2,4( ) .Therefore, the coordinates of the final point are 2,4( ) .

3. The particle which moves clockwise is moving three times as fast as the particle movingcounterclockwise. Therefore, the particle moving clockwise moves three times as far as theparticle moving counterclockwise in the same amount of time.

This tells us that in the time that the clockwise particletravels 3

4 of the way around the circle, the counterclockwise

particle will travel 14 of the way around the circle, and so the

two particles will meet at P 0,1( ) .

y

x(1, 0)

P(0, 1)

Using the same reasoning, the particles will meet at Q −1, 0( )when they meet the second time.

y

xQ(–1, 0)

P


4. Solution 1In choosing a pair of numbers from the five given numbers, there are 10 different ways of choosingthese numbers. These pairs are 0,1( ), 0,2( ), 0,3( ), 0,4( ), 1,2( ), 1,3( ), 1,4( ), 2,3( ), 2,4( ), 3,4( ){ } . The only

pairs in which the sum is greater than the product are those containing a 0 or a 1. Since there are 7of these, the required probability is 7

10 .

Solution 2The most straightforward way to approach this problem is to make a chart:

Numbers chosen Sum Product0,1 1 00,2 2 00,3 3 00,4 4 01,2 3 21,3 4 31,4 5 42,3 5 62,4 6 83,4 7 12

So there are 10 possible ways that two different numbers can be chosen, and for 7 of thesepossibilities, the sum of the two numbers is greater than the product.Therefore, the probability is 7

10 .

5. Join A to C.This line divides the shaded region into two identical pieces.

D C

A B

Consider the shaded region above AC.This piece of the region is formed by taking the sector DAC ofthe circle, centre D and radius 6, and then removing ∆ADC .

D C

A

6

6


Since ∠ADC = 90o , then the sector is one quarter of the whole circle, and has area14 π r2 = 1

4 π 62( ) = 9π .

Also, ∆ADC is right-angled with base DC of length 6 and height DA of length 6, and so has area12 bh = 1

2 6( ) 6( ) =18.

Therefore, the area of the region above the line is 9π −18 , and so the area of the entire shadedregion is 2 9π − 18( ) = 18π −36 square units.

6. If x < 0 , then 3

x< 0 , so

3

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ ≤

3

x< 0. Similarly,

4

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ < 0 , so we cannot possibly have

3

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ +

4

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 5 . Therefore, x > 0 .

When x > 0 , we have 3

x<

4

x, so

3

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ ≤

4

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ .

Since each of 3

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ and

4

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ is an integer, then there are three possibilities:

i) 3

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 0 and

4

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 5

ii) 3

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ =1 and

4

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 4

iii) 3

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 2 and

4

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 3

If 3

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 0 , then 0 ≤

3

x<1 or x > 3. If

4

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 5, then 5 ≤

4

x< 6 or

2

3< x ≤

4

5. These intervals do

not overlap, so there are no solutions in this case.

If 3

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ =1, then 1 ≤

3

x< 2 or

3

2< x ≤ 3. If

4

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 4 , then

4

5< x ≤1. These intervals do not

overlap, so there are no solutions in this case.

If 3

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 2 , then 2 ≤

3

x< 3 or 1 < x ≤

3

2. If

4

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 3, then 1 < x ≤

4

3. In this case, the intervals do

overlap. When we combine these intervals, we see that if 1 < x ≤4

3, then

3

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ +

4

x⎢ ⎣ ⎢

⎥ ⎦ ⎥ = 5 .

Therefore, the range of values is 1 < x ≤4

3.


7. Solution 1Let the radius of circle C be r.Since P, Q and R are given as midpoints of the radius of thelarge circle, they themselves lie on a circle with the same centreas the given circle, but with half its radius.

P(4, 1)

Q(7, – 8)

1

2r

1

2r

1

2r

1

2r

1

2r

1

2r

R(10, 1)

Method 1 – Perpendicular bisectorsTo find the centre of the circle passing through P, Q and R, we must find the intersection of theperpendicular bisectors of the sides of the triangle formed by the three points.

Consider first side PR. Since PR is a line segment parallel tothe x-axis, its perpendicular bisector has equation x = 7.

R(10, 1)(7, 1)

112

,72

⎛⎝

⎞⎠

Q(7, – 8)

P(4, 1)

Consider next side PQ. Since P has coordinates 4,1( ) and Q has coordinates 7,−8( ) , then PQ

has slope –3 and has midpoint 112 ,− 7

2( ) . Therefore, the perpendicular bisector of PQ has slope13 and has equation y + 7

2 = 13 x − 11

2( ) .

Therefore, at the intersection of these two perpendicular bisectors, y + 72 = 1

3 7 − 112( ) or y = −3 .

Thus the centre of the circle is the point 7,−3( ) , and since 7,−8( ) lies on the circle, the radius of

the small circle is 5. Therefore, the radius of circle C is 10.

Method 2 – GeometricAs in Method 1, we proceed by trying to find the centre ofthe circle. Also, we again know that the centre is theintersection of the perpendicular bisectors of the sides of the∆PQR . One perpendicular bisector is very easy to find –that of PR, which has equation x = 7 as we found above.This tells us that the centre lies on the line x = 7. Thus, thecentre of the circle can be represented by O 7,b( ) .

(7, b)1

2r

1

2r

Q(7, – 8)

P(4, 1) (7, 1) R(10, 1)


Since radii of a circle are equal,

OP2 = OQ2

7 − 4( )2 + b −1( )2 = 7 − 7( )2 + b + 8( )2

9 + b2 − 2b +1= b2 +16b + 64

b = −3

The radius of the circle P, Q and R is 32 + −4( )2= 5 and the radius of the larger circle is 10.

8. The first thing that we must notice in this problem is because we are looking for positive integers k,l and m such that

4k5

+5l6

+6m7

= 82

then k must be divisible by 5, l must be divisible by 6, and m must be divisible by 7.So we make the substitution k = 5K , l = 6L and m = 7M , where K, L and M are positive integers.Therefore, we obtain, by substitution

5K + 6L + 7M = 97

4K + 5L + 6M = 82Subtracting the second equation from the first, we get

K + L + M =15

4K + 5L + 6M = 82Subtracting six times the first equation from the second equation, we get

K + L + M = 15

−2K − L = −8or

K + L + M = 15

2K + L = 8Since K, L and M are all positive integers, we obtain from the second equation the followingpossibilities for K and L which give us M from the first equation, and thus k, l and m:

K L M k l m

1 6 8 5 36 562 4 9 10 24 633 2 10 15 12 70

Therefore, there are three triples k,l,m( ) of positive integers which are solutions to the system of

equations.


Part B

1. (a) We will proceed systematically to fill in the circles based on the initial knowledge thatk = 2 and e = 5:

i = 10 (5, i)[Throughout the solution to this problem, we have used notation such as (5, 7, c) to indicatethat there is a straight line joining the circles containing 5, 7 and c. Thus, 5 + 7 + c = 15 orc = 3.]b = 8 (2, 5, b)d = 7 (8, d)c = 3 (5, 7, c)a = 4 (a, 8, 3)g = 9 (4, g, 2)f = 6 (4, f, 5)

h =1 (9, h, 5)eg

a b

df

h

i

c

k 2

10

56

19

48

3

7

We can verify that the numbers along each of the ten straight lines add to 15.

(b) (i) Solution 1We start with knowing that k = 2 and e is unknown.Theni = 15 − e (e, i)c =15 − 2 − 15 − e( ) = e − 2 (2, 15 − e , c)

b =15 − 2 − e = 13− e (2, e, b)Therefore, b =13 − e and c = e − 2.

eg

a b

df

h

i

c

k 2

15 – e

13 – e e – 2


Solution 2We start with knowing that k = 2 and e isunknown.Theni = 15 − e (e, i)c =15 − 2 − 15 − e( ) = e − 2 (2, 15 − e , c)d =15 − e − e − 2( ) = 17 − 2e (e, d, e − 2)b =15 − 17 − 2e( ) = 2e − 2 (b, 17 − 2e )

Therefore, b = 2e − 2 and c = e − 2.

eg

a b

df

h

i

c

k 2

15 – e

2e – 2 e – 2

17 – 2e

(ii) SolutionWe know from (i) that k = 2, b =13 − e,and c = e − 2.Therefore, d =15 − b =15 − 13− e( ) = 2 + e .

But we also know thate + d + c = 15

e + 2 + e( ) + e − 2( ) = 15

3e = 15

e = 5Therefore, e = 5.

eg

a b

df

h

i

c

k 2

15 – e

13 – e e – 2

2 + e

(c) SolutionWe can model our approach from (b).Starting with e being unknown and k = x , weexpress some of the circles in terms of x and e:i = 15 − e (e, i)c =15 − x − 15 − e( ) = e − x (x, 15 − e , c)

b =15 − x − e (x, e, b)d =15 − 15 − x − e( ) = x + e (15 − x − e , d)

But we know thate + d + c = 15

e + x + e( ) + e − x( ) = 15

3e = 15

e = 5Therefore, e must still be equal to 5.

eg

a b

df

h

i

c

k x

15 – e

15 – x – e e – x

x + e


2. (a) Solution 1Drop perpendiculars from D and A to BC, meeting BC at E and F respectively.Since DA is parallel to CB, then DE and AF are also perpendicular to DA.Since DAFE is a rectangle, then EF = 6.Since DC = AB , DE = AF and ∆DEC and ∆AFB are right-angled, then they are congruent triangles, and so CE = BF ,and so both of these lengths must be equal to 3.

D A

C B

6 6

6

6 3F3 E

By the Pythagorean Theorem, DE = DC 2 − CE 2 = 62 − 32 = 27 = 3 3 .Therefore, the sides of ∆DEC are in the ratio 1 to 3 to 2, so DEC is a 30-60-90 triangle, with

∠DCE = 60o and ∠CDE = 30o .

Therefore, since we have congruent triangles, ∠DCB = ∠ABC = ∠DCE = 60o and

∠CDA = ∠DAB = 90o + ∠CDE =120o .

Solution 2Join D to the midpoint M of CB.Then CM = MB = 6 .Since DM and MB are parallel and of equal length, then ABand DM will also be parallel and equal length.Thus, DM = 6 , and so ∆DCM is equilateral.

D A

C B

6 6

6

6M6Therefore, ∠DCB = ∠DCM = 60o . By symmetry, ∠ABC = ∠DCB = 60o .

Since DA and CB are parallel, ∠CDA = ∠DAB =120o .

(b) (i) If Chuck was attached to a point P and there were no obstructions, he would be able toreach a circle of radius 8 m. (If Chuck stays at the end of his chain, he could trace out acircle of radius 8 m, but Chuck can move everywhere inside this circle, since his chain doesnot have to be tight.) However, here we have an obstruction – the trapezoidal barn.

Since the interior angle of the barn at point A is 120o , then the exterior angle of the barn is

240o . So Chuck can certainly reach the area which is a 240o sector of radius 8 m, centredat A. (If Chuck extends the chain as far as possible in a straight line in the direction of Dfrom A, can then walk in a clockwise direction, keeping the chain at its full length until the

chain lies along AB. He will have moved through 240o , and the region is the sector of acircle.)


However, when the chain is fully extended in thedirection of D, Chuck will be 2 m past point D.He will thus be free to move towards side DC ofthe barn. If he does this and keeps the chain tight,he will trace out part of a circle of radius 2 mcentred at D. (Point D now serves as a “pivot”point for the chain.) Since the exterior angle of

the barn at point D is 240o , then the angle

between AD extended and DC is 60o . Therefore,

Chuck can reach a 60o sector of a circle of radius2 m, centred at D.

D A

CB

2

6

6

260° 120°

120° 22

240°

60°

When the chain is fully extended in the direction of B, Chuck will be 2 m past point B. Hewill thus be free to move towards side BC of the barn. If he does this and keeps the chaintight, he will trace out part of a circle of radius 2 m centred at B. (Point B now serves as a

“pivot” point for the chain.) Since the exterior angle of the barn at point B is 300o (the

interior angle at B is 60o ), then the angle between AB extended and BC is 120o . Therefore,

Chuck can reach a 120o sector of a circle of radius 2 m, centred at B.

The area of a sector of angle θo of a circle of radius r is

θ360

πr2 .

Therefore, the total area that Chuck can reach is240360

π( )82 +60360

π( )22 +120360

π( )22 =23

64π( ) +16

4π( ) +13

4π( ) =128π

3+ 2π =

134π3

square metres.

(ii) Let x be the distance along AB from A to P.Since the total perimeter of the barn is 30 m and Chuck is attached with a 15 m chain, thenChuck can reach the same point on the barn whether he wraps the chain around the barn ina clockwise direction or a counterclockwise direction. This point will move, however, as Pmoves. For example, if Chuck was attached at point A (ie. if x = 0), then he could reach apoint 3 m along CB from C towards B wrapping in either the clockwise or counterclockwisedirection. If Chuck was attached at point B (ie. if x = 6), he could reach the midpoint ofCD in either direction. As point P moves from A towards B, this furthest point on the barnthat Chuck can reach will slide along BC towards C and then up CD towards D. If P is atthe midpoint of AB (ie. if x = 3), the furthest point along the barn that he can reach will bepoint C. So in our analysis, we must be careful as to whether 0 ≤ x ≤ 3 or 3 ≤ x ≤ 6.

Regardless of the value of x, Chuck can certainly reach a 180o sector of a circle of radius 15centred at P.


We start in the counterclockwisedirection.Also, regardless of the value of x, Chuck

can reach a 60o sector of a circle of radius15 − x centred at A (using A as the newpivot point for the chain).Still regardless of the value of x, Chuck

can reach a 60o sector of a circle of radius9 − x = 15 − x( ) − 6 centred at D (using D

as the new pivot point for the chain).If 3 ≤ x ≤ 6, then 9 − x ≤ 6, so Chuckcannot reach past point C.

D A

C B

6 6 – x

660°

120°

9 – x 60°x

9 + x

15 – x

3 – x 120°

9 + x

If 0 ≤ x ≤ 3, then 9 − x ≥ 6, so Chuck can reach past point C, and so can reach a 120o sectorof a circle of radius 3− x = 9 − x( ) − 6 centred at C (using C as the new pivot point for the

chain).Next, we consider the clockwise direction.

Regardless of the value of x, Chuck can reach a 120o sector of a circle of radius9 + x = 15 − 6 − x( ) (the distance from B to P is 6 − x ) centred at B (using B as the new

pivot point for the chain).If 0 ≤ x ≤ 3, then 9 + x ≤ 12, so Chuck cannot reach past point C.

If 3 ≤ x ≤ 6, then 9 + x ≥ 12, so Chuck can reach past point C, and so can reach a 120o

sector of a circle of radius x − 3 = 9 + x( ) −12 centred at C (using C as the new pivot point

for the chain).

We now calculate the total area that Chuck can reach.If 0 ≤ x ≤ 3, then the area that Chuck can reach is180

360π152 +

60

360π 15 − x( )2 +

60

360π 9 − x( )2 +

120

360π 3 − x( )2 +

120

360π 9 + x( )2

=1

2π 225( ) +

1

6π 225 − 30x + x2( ) +

1

6π 81−18x + x2( ) +

1

3π 9 − 6x + x 2( ) +

1

3π 81+18x + x2( )

=16

π 675 + 225 − 30x + x 2 + 81−18x + x 2 +18 −12x + 2x 2 +162 + 36x + 2x 2( )=

1

6π 1171− 24x + 6x2( )

= πx2 − 4πx +387

2π


If 3 ≤ x ≤ 6, then the area that Chuck can reach is

180

360π152 +

60

360π 15 − x( )2 +

60

360π 9 − x( )2 +

120

360π 9 + x( )2 +

120

360π x − 3( )2

= πx2 − 4πx +387

2π

(Notice that the one term between these two initial expressions that seems to be different isactually the same!)

Therefore, no matter what the value of x is, the area that Chuck can reach is

πx 2 − 4πx +387

2π . This is a parabola opening upwards, so the vertex of the parabola gives

us the minimum of the parabola. This vertex is at x = −−4π2 π( )

= 2 . Since x = 2 is between

the endpoints of the allowable interval (0 and 6), then this will give the minimum.

Therefore, the location of P which minimizes the area that Chuck can reach is 2 m along thewall from A towards B.

3. (a) Solution 1Let ∠PAB = θ .

Then ∠XAB =180o −θ , and so ∠XYB = θ since XYBA isa cyclic quadrilateral, and so opposite angles sum to

180o .Therefore, ∆PAB is similar to ∆PYX (common angle atP, equal angle θ ).

So, XYBA

=PXPB

or XY =BA ⋅ PX

PB=

6 5 +16( )7

=18 .

X

Y

B

AP

C2

C1

5

716

180° – θ6

θ

θ

Solution 2By the cosine law in ∆APB,

AB2 = PA 2 + PB 2 − 2 PA( ) PB( )cos ∠APB( )36 = 25 + 49 − 2 5( ) 7( )cos ∠APB( )

cos ∠APB( ) = 3870 = 19

35


Now, PX and PY are both secants of circle C2 , so by the

Secant-Secant Theorem,PA ⋅ PX = PB ⋅ PY

5 5 +16( ) = 7 7 + BY( )105 = 7 7 + BY( )15 = 7 + BY

BY = 8

X

Y

B

AP

5

716

6

8

Now in ∆PXY we know the lengths of sides PX (length 21), PY (length 15) and the cosine of∠XPY = ∠APB , so we can use the cosine law to calculate the length of XY.

XY 2 = PX 2 + PY2 − 2 PX( ) PY( )cos ∠XPY( )

XY 2 = 441+ 225 − 2 21( ) 15( ) 1935( )

XY 2 = 441+ 225 − 2 3( ) 3( ) 19( )

XY 2 = 441+ 225 − 342

XY 2 = 324

XY =18Therefore, the length of XY is 18.

(b) Solution 1Since the circle C4 is fixed, the length VW will be fixed if the angle it subtends on the circle is

fixed, ie. if the angle ∠VHW does not depend on the position of Q.Now

∠VHW = 180o − ∠VHQ

= 180o − 180o − ∠GVH − ∠GQH( )= ∠GVH + ∠GQH

V

W

H

GQ

C4

C3

But since chord GH is a chord of fixed length in both circles (ie. it doesn’t change as Q moves),then the angles that it subtends in both circles are constant. In particular, ∠GVH and ∠GQHare both fixed (that is, they do not depend on the position of Q).Since both of these angles are fixed, then ∠VHW = ∠GVH + ∠GQH is also fixed.Therefore, the length of VW is fixed.


Solution 2We start by noting that chord GH has a constant length,ie. does not depend on the the position of Q. Thus, GH isa constant chord in both C3 and C4 .In C3 , let ∠GQH = α .In C4 , let ∠GVH = ∠GWH = β .

These angles are constant since GH is of constant length.

V

W

H

GQ

C4

C3

Therefore, ∠VHQ = ∠QGW = 180o − α + β( ) , and so ∠VGW = ∠WHV = α + β .

Since these last two angles are constant for all position of Q, then VW is a chord of constantlength.

Solution 3We proceed by considering two different positions for the point Q, which we call Q1 and Q2 .These points will create two different positions for the line segment VW, which we call V1W1

and V2W2 .To show that the length of VW is constant, we must show that V1W1 and V2W2 have the same

length. We do know that the points G and H are fixed, so the length of GH does not vary.

H

G

V1 W1

Q1

H

GV2

W2

Q2

As in part (a) Solution 1, ∆Q1HG and ∆Q1V1W1 are similar triangles, as are ∆Q2HG and∆Q2V2W2 . (The two positions for Q play the same role as P, points G and H play the same role

as A and B, and the points V and W play the same role as X and Y.)

Therefore, by similar triangles, V1W1

HG=

Q1V1

Q1H or V1W1 = HG ⋅

Q1V1

Q1H.

Also, by similar triangles, V2W2

HG=

Q2V2

Q2H or V2W2 = HG ⋅

Q2V2

Q2 H.

Since the length of HG (or GH) does not change, to show that V1W1 and V2W2 have the same

length, we must show that Q1V1

Q1H=

Q2V2

Q2H.

Join H to each of V1 and V2 .

Now GH is a chord of fixed length in both circles, so the angle that it subtends at any point onthe circumference of each circle is the same.


Therefore, ∠GQ1H = ∠GQ2H and ∠GV1H = ∠GV2 H .But this tells us that ∆Q1HV1 and ∆Q2HV2 are similar.

Thus, Q1V1

Q2V2=

Q1H

Q2H or

Q1V1

Q1H=

Q2V2

Q2H.

Therefore, V1W1 = HG ⋅Q1V1

Q1H= HG ⋅

Q2V2

Q2H= V2W2 , ie. the length of VW is constant.

4. (a) Solution 1Since a, b and c are the roots of the equation x3 − 6x 2 + 5x −1 = 0 , then using the properties ofthe coefficients of a cubic equation,

a + b + c = 6

ab + ac + bc = 5

abc =1Since we know that each of a, b and c is a roots of the equation, then

a3 − 6a2 + 5a −1 = 0

b3 − 6b2 + 5b −1 = 0

c3 − 6c2 + 5c −1 = 0or after rearranging

a3 = 6a2 − 5a +1

b3 = 6b2 − 5b +1 (*)

c3 = 6c2 − 5c +1Adding these three equations, we obtain

a3 + b3 + c3 = 6a2 + 6b2 + 6c2 − 5a − 5b − 5c + 3

= 6 a2 + b2 + c2( ) − 5 a + b + c( ) + 3

We already know that a + b + c = 6, so if we could determine the value of a2 + b2 + c2 , thenwe would know the value of a3 + b3 + c3 .But

a + b + c( )2 = a2 + b2 + c2 + 2ab + 2bc + 2ac

62 = a2 + b2 + c2 + 2 ab + bc + ac( )

a2 + b2 + c2 = 36 − 2 5( )

a2 + b2 + c2 = 26and so

a3 + b3 + c3 = 6 a2 + b2 + c2( ) − 5 a + b + c( ) + 3 = 6 26( ) − 5 6( ) +3 = 129

If we know take the equations in (*) and multiply both sides in the first, second and thirdequations by a, b and c, respectively, we get


a4 = 6a3 − 5a2 + a

b4 = 6b3 − 5b2 + b (**)

c4 = 6c3 − 5c2 + cwhich we then can add to obtain

a4 + b4 + c4 = 6 a3 + b3 + c3( ) − 5 a2 + b2 + c2( ) + a + b + c( )

= 6 129( ) − 5 26( ) + 6

= 650Repeating the process one more time by multiplying the first, second and third equations in(**) by a, b and c, respectively, and adding, we obtain

a5 + b5 + c5 = 6 a4 + b4 + c4( ) − 5 a3 + b3 + c3( ) + a2 + b2 + c2( )= 6 650( ) − 5 129( ) + 26

= 3281Therefore, the value of a5 + b5 + c5 is 3281.

Solution 2Since a, b and c are the roots of the equation x3 − 6x 2 + 5x −1 = 0 , then using the properties ofthe coefficients of a cubic equation,

s = a + b + c = 6

t = ab + ac + bc = 5

p = abc =1

We will attempt to express a5 + b5 + c5 in terms of s, t and p, which will thus allow us tocalculate the value of a5 + b5 + c5 .

First,a + b + c( )2 = a2 + b2 + c2 + 2ab + 2bc + 2ac

a2 + b2 + c2 = s2 − 2tNext,

a2 + b2 + c2( ) a + b + c( ) = a3 + b3 + c3 + a2b + a2c + b2a + b2c + c2a + c2b

a3 + b3 + c3 = s s2 − 2t( ) − a2b + a2c + b2a + b2c + c2a + c2b[ ]= s s2 − 2t( ) − ab + ac + bc( ) a + b + c( ) − 3abc[ ]

= s s2 − 2t( ) − ts −3p[ ]

= s3 −3st + 3p

We can now attempt to express a5 + b5 + c5 as


a5 + b5 + c5

= a2 + b2 + c2( ) a3 + b3 + c3( ) − a2b3 + a2c3 + b2a3 + b2c3 + c2a3 + c2b3[ ]= a2 + b2 + c2( ) a3 + b3 + c3( ) − a2b2 + a2c2 + b2c2( ) a + b + c( ) − a2b2c + a2bc2 + ab2c2( )[ ]= s2 − 2t( ) s3 − 3st + 3p( ) − ab + ac + bc( )2 − 2 a2bc + ab2c + abc2( )[ ] a + b + c( ) − abc ab + ac + bc( )[ ]= s2 − 2t( ) s3 − 3st + 3p( ) − t2 − 2abc a + b + c( )[ ]s − pt[ ]= s2 − 2t( ) s3 − 3st + 3p( ) − t2 − 2ps[ ]s − pt[ ]= 62 − 2 5( )( ) 63 − 3 6( ) 5( ) +3 1( )( ) − 52 − 2 1( ) 6( )[ ] 6( ) −1 5( )[ ]= 26( ) 129( ) − 13[ ] 6( ) − 5[ ]= 3354 − 73[ ]= 3281

Therefore, the value of a5 + b5 + c5 is 3281.

(b) We will proceed by dividing our proof into several steps.Step 1: Estimate the values of a, b and c

Step 2: Show that an + bn + cn is an integer for every positive integer nStep 3: Final conclusion

Step 1: Estimate the values of a, b and c

Define f x( ) = x 3 − 6x 2 + 5x −1.

If x < 0, then x 3 < 0, −6x2 < 0, 5x < 0, and −1< 0, so f x( ) = x 3 − 6x 2 + 5x −1< 0.This tells us that f x( ) = 0 cannot have any negative roots. It is also clear that 0 is not aroot of f x( ) = 0 , so each of a, b and c is positive.

If we calculate a few values of f x( ) = x 3 − 6x 2 + 5x −1, we obtain f 0( ) = −1, f 1( ) = −1,f 2( ) = −7 , f 3( ) = −13 , f 4( ) = −13 , f 5( ) = −1, and f 6( ) = 29 .

Therefore, one of the roots is between 5 and 6.However, we know from part (a) that a + b + c = 6, so since all three roots are positive, thenwe must have 5 < c < 6 and 0 < a,b <1. (Since one root is bigger than 5, each root ispositive, and the sum of the three roots is 6, then neither a nor b is bigger than 1.)Since 5 < c < 6 and a + b + c = 6, then 0 < a + b <1.We also know from part (a) that abc = 1. Since 5 < c < 6, then 1

6 < ab < 15 and since each

of a and b is less than 1, then each of a and b must be bigger than 16 .

Since a and b are each bigger than 16 and 0 < a + b <1, then 1

6 < a,b < 56 .

(We could have proceeded less formally by doing some quick calculations to see thatf 0.1( ) = −0.559 , f 0.2( ) = −0.232 , f 0.3( ) = −0.013, f 0.4( ) = 0.104 , f 0.5( ) = 0.125 ,


f 0.6( ) = 0.056 , f 0.7( ) = −0.097 , and so a must be between 0.3 and 0.4, and b must be

between 0.6 and 0.7.)

Step 2: Show that an + bn + cn is an integer for every positive integer n

In part (a), we saw that an + bn + cn is definitely an integer for n equal to 1, 2 and 3.If we return to the set of equations (*) in Solution 1 of part (a) and multiply the three

equations by an −3, bn− 3 and c n− 3 , respectively, we obtainan = 6an− 1 − 5an− 2 + an −3

bn = 6bn −1 − 5bn− 2 + bn− 3 (***)

cn = 6cn −1 − 5c n− 2 + c n− 3

and adding, we get

an + bn + cn = 6 an− 1 + bn− 1 + c n− 1( ) − 5 an− 2 + bn− 2 + c n− 2( ) + an −3 + bn −3 + cn −3( ) (****)

for every n greater than or equal to 4.

If we set n equal to 4, then since ak + bk + ck is an integer for k equal to 1, 2 and 3, then by

(****), a4 + b4 + c4 is also an integer.

If we set n equal to 5, then since ak + bk + ck is an integer for k equal to 2, 3 and 4, then by

(****), a5 + b5 + c5 is also an integer.

It is now clear than we can continue this process inductively, since if ak + bk + ck is an

integer for k equal to n − 3, n − 2, and n −1, then an + bn + cn will also be an integer, by(****).

In particular, we can conclude that a2003 + b2003 + c2003 and a2004 + b2004 + c2004 are both

integers, say a2003 + b2003 + c2003 = M and a2004 + b2004 + c2004 = N .

Step 3: Final conclusion

Since each of a and b is between 0 and 1, then a2003 > a2004 and b2003 > b2004 , so

a2003 + b2003 > a2004 + b2004.

Since a is less than 56 , then a is less than 0.9, so a2 < 0.81, so a4 < 0.81( )2

< 0.7, so

a8 < 0.7( )2< 0.5 , so a16 < 0.5( )2

< 0.25 .

Similarly, since b is less than 56 , then b16 < 0.25.

Therefore, a16 + b16 < 0.5, and since each of a and b is less than 1, then

a2004 + b2004 < a2003 + b2003 < a16 + b16 < 0.5.

Therefore, since c 2003 = M − a2003 + b2003( ) and a2003 + b2003 < 0.5, then the closest

integer to c 2003 is M and the distance between them is a2003 + b2003. Similarly, the closest

integer to c 2004 is N and the distance between them is a2004 + b2004.

But a2004 + b2004 < a2003 + b2003, so c 2004 is closer to N than c 2003 is to M, as required.



presents the

Canadian Open



Solutions

c©2004 Canadian Mathematical Society

2004 COMC Solutions Page 2 of 19

Part A

1. If x + 2y = 84 = 2x + y, what is the value of x + y?

Solution 1

Since x + 2y = 84 and 2x + y = 84, then adding these two equations together, we obtain

3x + 3y = 168 or x + y = 56.

Solution 2

Since x + 2y = 84, then x = 84− 2y.

Substituting into the second equation, we get

2(84− 2y) + y = 84

168− 3y = 84

84 = 3y

y = 28

Therefore, x = 84− 2(28) = 28 and so x + y = 56.

Solution 3

Since 2x + y = 84, then y = 84− 2x.

Substituting into the first equation, we get

x + 2(84− 2x) = 84

168− 3x = 84

84 = 3x

x = 28

Therefore, y = 84− 2(28) = 28 and so x + y = 56.

Solution 4

SInce these two expressions are identical when x is replaced by y and y is replaced by x, then

x = y.

Therefore, 3x = 84 or x = 28 and so y = 28.

Thus, x + y = 56.

Answer: 56


2. Let S be the set of all three-digit positive integers whose digits are 3, 5 and 7, with no digit

repeated in the same integer. Calculate the remainder when the sum of all of the integers in S

is divided by 9.

Solution 1

We can write down the elements of S: 357, 375, 537, 573, 735, 753.

The sum of these elements is 357 + 375 + 537 + 573 + 735 + 753 = 3330.

Since 3330 is divisible by 9 (because the sum of its digits is divisible by 9), the remainder when

we divide by 9 is 0.

Solution 2

There are six numbers formed with the three given numbers.

Two of these numbers have a 3 in the 100s position, two have a 5 in the 100s position, and two

have a 7 in the 100s position.

The same can be said about the distribution of numbers in the 10s and units positions.

Therefore, the sum of the six numbers is

2(3 + 5 + 7)(100) + 2(3 + 5 + 7)(10) + 2(3 + 5 + 7)(1) = 3330

The remainder is 0 when 3330 is divided by 9.

Answer: 0

3. In the diagram, point E has coordinates (0, 2), and B lies on

the positive x-axis so that BE =√

7. Also, point C lies on the

positive x-axis so that BC = OB. If point D lies in the first

quadrant such that ∠CBD = 30◦ and ∠BCD = 90◦, what is the

length of ED?

Ox

y

B

E

D

CSolution

In order to find the length of ED, we will try to find the coordinates of D. Let the coordinates

of B be (b, 0).

Since BE =√

7 and the coordinates of E are (0, 2), then√(b− 0)2 + (0− 2)2 =

√7

b2 + 4 = 7

b2 = 3

Since the point B lies on the positive x-axis, then b =√

3 (not −√

3), so B has coordinates

(√

3, 0).


(Note that it would have also been possible to find the coordinates of B by using Pythagoras

– OE2 + OB2 = EB2 so OB2 = 7− 4 = 3 so OB =√

3 and so B has coordinates (√

3, 0).)

Since BC = OB, then C has coordinates (2√

3, 0).

Since ∠BCD = 90◦ and D lies in the first quadrant, then D has coordinates (2√

3, d), with

d > 0.

Since 4DBC has ∠BCD = 90◦ and ∠CBD = 30◦, then it is a 30◦ − 60◦ − 90◦ triangle. Since

CB =√

3 (and CB is opposite the 60◦ angle), then DC (which is opposite the 30◦ angle) has

length 1.

Therefore, D has coordinates (2√

3, 1).

Ox

y

E 0,2( )

C 2 3, 0( )

D 2 3, 1( )

B 3, 0( )

Thus, ED =√

(2√

3− 0)2 + (1− 2)2 =√

12 + 1 =√

13.

Answer:√

13

4. A function f(x) has the following properties:

i) f(1) = 1

ii) f(2x) = 4f(x) + 6

iii) f(x + 2) = f(x) + 12x + 12

Calculate f(6).

Solution 1

Using property ii) with x = 1,

f(2) = 4f(1) + 6 = 4(1) + 6 = 10

since f(1) = 1 by property i).


f(4) = 4f(2) + 6 = 4(10) + 6 = 46


Using property iii) with x = 4,

f(6) = f(4) + 12(4) + 12 = 46 + 48 + 12 = 106

Therefore, the value of f(6) is 106.

Solution 2

Using property iii) with x = 1,

f(3) = f(1) + 12(1) + 12 = 1 + 12 + 12 = 25

since f(1) = 1 by property i).


f(6) = 4f(3) + 6 = 4(25) + 6 = 106


Solution 3

Working backwards,

f(6) = 4f(3) + 6 (by property ii) with x = 3)

= 4(f(1) + 12(1) + 12) + 6 (by property iii) with x = 1)

= 4f(1) + 4(24) + 6

= 4(1) + 102 (by property i))


Answer: f(6) = 106

5. The Rice Tent Company sells tents in two different sizes, large and small. Last year, the Com-

pany sold 200 tents, of which one quarter were large. The sale of the large tents produced one

third of the company’s income. What was the ratio of the price of a large tent to the price of

a small tent?

Solution

Since the Rice Tent Company sold 200 tents, of which one quarter were large, then they sold

50 large tents and 150 small tents last year.

Let L be the price of a large tent and S the price of a small tent.

Then their income from large tents was 50L and from small tents was 150S.

Their total income last year was 50L + 150S.


From the given information,

50L =1

3(50L + 150S)

150L = 50L + 150S

100L = 150SL

S=

150

100=

3

2

Therefore, the ratio of the price of a large tent to the price of a small tent was 3 : 2.

Answer: 3 : 2

6. In the diagram, a square of side length 2 has semicircles drawn on

each side. An “elastic band” is stretched tightly around the figure.

What is the length of the elastic band in this position?

Solution

Label the four vertices of the square as W , X, Y , Z, in clockwise order.

Label the four midpoints of the sides of the square (that is, the centres of the four semicircles)

as M , N , O, P , in clockwise order, starting with M being the midpoint of WX.

In each semicircle, join the centre to the two points on that semicircle where the band just

starts (or stops) to contact the circle. Label these eight points as A, B, C, D, E, F , G, and H

in clockwise order, starting with A and B on the semicircle with centre M .

W X

Z Y

M

N

O

P

A B

C

D

EF

G

H

By symmetry, the four straight parts of the band will be equal in length (that is, BC =

DE = FG = HA) and the four arc segments of the band will be equal in length (that is,

AB = CD = EF = GH).

Therefore, the total length of the band is 4(Length of arc AB) + 4(Length of BC).

Now, BC will actually be tangent to the two semicircles (with centres M and N) where it

initially just touches them.

Thus, MB and NC are both perpendicular to BC.


Since MB = NC = 1 (because they are radii of the semicircles and each semicircle has diame-

ter 2), then MBCN must actually be rectangle, so BC is equal and parallel to MN .

Since M and N are the midpoints of the sides of the square of side length 2, then MY = Y N =

1, so MN =√

2, so BC =√

2.

Next, we determine the length of AB. Previously, we saw that MBCN is a rectangle, so

BC was parallel to MN . Similarly, HA is parallel to PM .

But PM is perpendicular to MN , so HA is perpendicular to BC.

Therefore, ∠AMB = 90◦, ie. AB is one-quarter of the circumference of a circle with radius 1

or 14(2π(1)) = 1

2π.

Therefore, the total length of the band is 4(12π) + 4(

√2) = 2π + 4

√2.

Answer: 2π + 4√

2

7. Let a and b be real numbers, with a > 1 and b > 0.

If ab = ab anda

b= a3b, determine the value of a.

Solution 1

Since ab = ab anda

b= a3b, then multiplying these two equations together, we get

a2 = ab · a3b = a4b.

Dividing both sides by a2 (since a 6= 0), we get a4b−2 = 1.

Since a > 1, then 4b− 2 = 0 or b = 12.

Substituting back into the first equation, we get 12a = a1/2 =

√a or a = 2

√a.

Squaring both sides gives a2 = 4a or a2 − 4a = 0 or a(a− 4) = 0.

Since a > 1, then a = 4.

Solution 2

Since ab = ab, then, dividing both sides by a which is not equal to 0, we get b = ab−1.

Sincea

b= a3b, then a = ba3b = ab−1a3b = a4b−1.

Comparing exponents, we get 1 = 4b− 1 or b = 12.

Substituting b = 12

into ab = ab, we have 12a = a1/2 or a1/2 = 2 or a = 4. So a = 4.

Solution 3

Since a > 1 and b > 0, we can take logarithms of both sides of both equations.

In the first equation, using log rules on log(ab) = log(ab

)gives log(a) + log(b) = b log(a).

In the first equation, using log rules on log(

ab

)= log

(a3b

)gives log(a)− log(b) = 3b log(a).

Adding these two new equations gives 2 log(a) = 4b log(a) or (4b− 2) log(a) = 0.

Since a > 1, then log(a) > 0, so we must have 4b− 2 = 0 or b = 12.


Substituting this back into the first log equation gives log(a) + log(

12

)= 1

2log(a) or

12log(a) = − log

(12

)= log(2) or log(a) = 2 log(2) = log(4), so a = 4.

Answer: 4

8. A rectangular sheet of paper, ABCD, has AD = 1 and AB = r, where 1 < r < 2. The paper is

folded along a line through A so that the edge AD falls onto the edge AB. Without unfolding,

the paper is folded again along a line through B so that the edge CB also lies on AB. The

result is a triangular piece of paper. A region of this triangle is four sheets thick. In terms of

r, what is the area of this region?

Solution

Start with the rectangular sheet of paper, ABCD, with A in the top left and B in the bottom

left.

Fold AD across so that AD lies along AB. Let D′ be the point were D touches AB and let E

be the point on DC where the fold hits DC.

Since AD′ is the old AD, then AD′ = 1.

Since D′E is perpendicular to D′A (since DC was perpendicular to AD) then D′E is parallel

to BC, so D′E = 1 as well.

A D

B C

E1

r −1

′D

We can also conclude that D′B = EC = r − 1, since AB has length r.

Next, we fold the paper so that BC lies along AB. Unfold this paper and lay it flat so

that we can see the crease.

Since BC is folded onto AB, then the crease bisects ∠ABC, that is the crease makes an angle

of 45◦ with both AB and BC.

Suppose that the crease crosses D′E at X and AE at Y .


A

B C

E′DX

Y

Now when the paper had been folded the second time (before we unfolded it!), the only way to

obtain a region four sheets thick was to fold a region two sheets thick on top of a region which

is also two sheets thick.

Since 4XY E is the only part of the paper “below” the second crease which is two sheets thick,

and, when the second fold is made, it lies entirely over another region which is two sheets thick,

then the desired area is the area of 4XY E.

Since ∠ABX = 45◦, then 4BD′X is isosceles and right-angled, so D′X = D′B = r − 1.

Thus, EX = 1−D′X = 1− (r − 1) = 2− r.

Since ∠D′XB = 45◦, then ∠Y XE = 45◦. Also, since 4AD′E is right isosceles, then ∠Y EX =

45◦, so 4XY E is isosceles and right-angled.

Y

X E

s s

2 − r

Suppose XY = s. Then√

2s = XE = 2− r or 2s2 = (2− r)2.

The area of 4XY E is 12s2 or 1

4(2− r)2.

Therefore, the area of the desired region is 14(2− r)2.

Answer: 14(2− r)2


Part B

1. The points A(−8, 6) and B(−6,−8) lie on the circle x2 + y2 = 100.

(a) Determine the equation of the line through A and B.

Solution

First, we determine the slope of the line segment AB. The slope is6− (−8)

−8− (−6)= −7.

We could now proceed to find the equation of the line in several different ways.

Using the point-slope form, we obtain y − 6 = −7(x− (−8)) or y = −7x− 50.

(b) Determine the equation of the perpendicular bisector of AB.

Solution

Since the slope of AB is −7, then the slope of the perpendicular bisector of AB is 17.

Also, the perpendicular bisector passes through the midpoint of AB, which is(1

2((−8) + (−6)),

1

2(6 + (−8))

)= (−7,−1).

Therefore, the equation of the perpendicular bisector is y− (−1) = 17(x− (−7)) or y = 1

7x.

(c) The perpendicular bisector of AB cuts the circle at two points, P in the first quadrant

and Q in the third quadrant. Determine the coordinates of P and Q.

Solution 1

y

x

P

Q

A

B

We want to find the points of intersection of the circle x2 + y2 = 100 and the line y = 17x.

From the equation of the line, x = 7y. Substituting this into the equation of the circle we


obtain

(7y)2 + y2 = 100

49y2 + y2 = 100

50y2 = 100

y2 = 2

y = ±√

2

Since x = 7y, then if y =√

2, then x = 7√

2, and if y = −√

2, then x = −7√

2.

Since P is in the first quadrant, then P has coordinates (7√

2,√

2).

Since Q is in the third quadrant, then Q has coordinates (−7√

2,−√

2).

Solution 2

We want to find the points of intersection of the circle x2 + y2 = 100 and the line y = 17x.

Substituting y = 17x into the equation of the circle we obtain

x2 +

(1

7x

)2

= 100

x2 +1

49x2 = 100

50

49x2 = 100

x2 = 98

x = ±√

98 = ±7√

2

Since y = 17x, then if x = 7

√2, then y =

√2, and if x = −7

√2, then y = −

√2.

Since P is in the first quadrant, then P has coordinates (7√

2,√

2).

Since Q is in the third quadrant, then Q has coordinates (−7√

2,−√

2).

(d) What is the length of PQ? Justify your answer.

Solution 1

The points P and Q are joined by the line y = 17x, which passes through the origin.

Since the origin is the centre of the circle, then PQ must be a diameter of the circle.

Since the circle has equation x2 + y2 = 100 = 102, then its radius is 10, so its diameter is

20.

Therefore, PQ = 20.

Solution 2

Since we know that P (7√

2,√

2) and Q(−7√

2,−√

2), then we can determine the distance


PQ by direct calculation:

PQ =

√(7√

2−(−7√

2))2

+(√

2−(−√

2))2

=

√(14√

2)2

+(2√

2)2

=

√(√2)2

[142 + 22]

=√

2[196 + 4]

=√

400

= 20

Therefore, the length of PQ is 20.

2. (a) Determine the two values of x such that x2 − 4x− 12 = 0.

Solution

Factoring the given equation x2 − 4x− 12 = 0, we obtain (x− 6)(x + 2) = 0.

Therefore, the two solutions are x = 6 and x = −2.

(b) Determine the one value of x such that x−√

4x + 12 = 0. Justify your answer.

Solution

We first eliminate the square root by isolating it on one side and squaring:

x−√

4x + 12 = 0

x =√

4x + 12

x2 = 4x + 12

x2 − 4x− 12 = 0

(x− 6)(x + 2) = 0

Therefore, the two possible solutions are x = 6 and x = −2. (Since we have squared both

sides, it is possible that we have introduced an extraneous root, so we should verify both

of these.)

If x = 6, then 6−√

4(6) + 12 = 6−√

36 = 0.

If x = −2, then (−2)−√

4(−2) + 12 = −2−√

4 = −4 6= 0.

Therefore, the one value of x that solves the equation is x = 6.


(c) Determine all real values of c such that

x2 − 4x− c−√

8x2 − 32x− 8c = 0

has precisely two distinct real solutions for x.

Solution

We start by attempting to solve this equation and then seeing what conditions on c arise.

Since 8x2 − 32x− 8c = 8(x2 − 4x− c), we let T = x2 − 4x− c.

Then the equation is

T −√

8T = 0 (∗)

T =√

8T

T 2 = 8T

T (T − 8) = 0

Therefore, T = 0 or T = 8. We can check that neither root is extraneous, so x2−x−c = 0

or x2 − 4x− c = 8.

Let’s look at these last two equations.

First, we look at x2 − 4x− c = 0. The discriminant of this quadratic equation is (−4)2 −4(−c) = 16 + 4c. Therefore, this equation has

• zero solutions if 16 + 4c < 0, so c < −4,

• exactly one solution if 16 + 4c = 0, so c = −4, and

• two distinct solutions if 16 + 4c > 0, so c > −4.

Next, we look at x2 − 4x − c = 8 or x2 − 4x − (c + 8) = 0. The discriminant of this

quadratic equation is (−4)2 − 4(−(c + 8)) = 48 + 4c. Therefore, this equation has

• zero solutions if 48 + 4c < 0, so c < −12,

• exactly one solution if 48 + 4c = 0, so c = −12, and

• two distinct solutions if 48 + 4c > 0, so c > −12.

We see also that any value of x that satisfies one of these two equations cannot satisfy the

other, since we cannot have both x2 − 4x − c = 0 and x2 − 4x − c = 8. (In other words,

no roots overlap between these two equations.)

We make a table to combine our observations:

c < −12 c = −12 −12 < c < −4 c = −4 c > −4

x2 − 4x− c = 0 0 solutions 0 solutions 0 solutions 1 solution 2 solutions

x2 − 4x− c = 8 0 solutions 1 solution 2 solutions 2 solutions 2 solutions

Total solutions 0 solutions 1 solution 2 solutions 3 solutions 4 solutions

Therefore, for there to be exactly two distinct solutions, we must have −12 < c < −4.


3. A map shows all Beryl’s Llamaburgers restaurant locations in North America. On this map,

a line segment is drawn from each restaurant to the restaurant that is closest to it. Every

restaurant has a unique closest neighbour. (Note that if A and B are two of the restaurants,

then A may be the closest to B without B being closest to A.)

(a) Prove that no three line segments on the map can form a triangle.

Solution 1

We start by assuming that three line segments on the map do form a triangle, and show

that this is in fact impossible.

Notice that if restaurants X and Y are joined by a line segment, then either X is the

closest restaurant to Y or Y is the closest restaurant to X (or both).

Assume that A, B and C are the three points on the map connect by segments.

B

A C

To begin, we focus on the segment joining A to B. Let’s assume that A is the closest

restaurant to B. (It doesn’t matter which direction we assume here.) This means that C

is not the closest restaurant to B, so BA < BC.

But B and C are connected and C is not the closest restaurant to B. Therefore, B is the

closest restaurant to C, which means CB < CA.

But C and A are also connected and A is not the closest restaurant to C. Therefore, C is

the closest restaurant to A, which means AC < AB.

But this means that BA < BC, BC < AC and AC < BA. This cannot be the case.

Therefore, it is impossible for three line segments to form a triangle.

Solution 2

We prove this by showing that constructing a triangle is impossible.

We start by considering two locations A and B and the line segment AB.

Since A and B are connected, we can assume without loss of generality that A is closest

to B. (The case B closest to A involves interchanging A and B, and the case of A and B

closest to each other is included in the case of A closest to B.)

If A is closest to B and we add a new location C which is connected to B, then B

must be closest to C (since C can’t also be closest to B along with A).


If we join C to A, then either C is closest to A or A is closest to C.

But A can’t be closest to C since B is closest to C.

Therefore, we must have C closest to A.

But then AC is shorter than AB, along with AB being shorter than BC (since A is closest

to B), which means that AC is shorter than BC or A is closer to C than B is, which isn’t

true. This is a contradiction.

Therefore, we can’t construct a triangle.

(b) Prove that no restaurant can be connected to more than five other restaurants.

Solution

We start by assuming that one restaurant can be connected to six others and show that

this is impossible. From this we can conclude that no restaurant can be connected to

more than five other restaurants (for if it could be joined to 8 others, say, then we could

consider six of them only and reach a contradiction).

Assume that restaurant A can be connected to restaurants B, C, D, E, F , and H, where

these restaurants are listed in clockwise order of their line segments joining to A.

B

C

DE

F

H

A

Consider restaurants A, B and C.

We know that B and C are both connected to A and both cannot be the closest neighbour

to A. Thus, A must be the closest neighbour to one of these, say B. (It doesn’t matter

which we choose).

Since A is the closest restaurant to B, then BA < BC.

Now consider the line joining C to A.

If C is the closest neighbour to A, then AC < AB, so AC < AB < BC.

If A is the closest neighbour to C, then CA < CB so CA < CB and BA < CB.

In either case, BC is the (strictly) longest side in 4ABC, and so must be opposite the

(strictly) largest angle.

Since the angles in a triangle add to 180◦, then if there is a largest angle, then this angle

must be larger than 60◦. Therefore, from the above reasoning, ∠BAC > 60◦.


But we can reapply this reasoning to conclude that ∠CAD, ∠DAE, ∠EAF , ∠FAH,

and ∠HAB are each greater than 60◦. But the sum of these six angles is 360◦, since they

will form a full circle around A, and six angles, each greater than 60◦, cannot add to 360◦.

So we have a contradiction.

Therefore, it is impossible for a restaurant to be connected to more than five other restau-

rants.

4. In a sumac sequence, t1, t2, t3, . . ., tm, each term is an integer greater than or equal to 0.

Also, each term, starting with the third, is the difference of the preceding two terms (that is,

tn+2 = tn− tn+1 for n ≥ 1). The sequence terminates at tm if tm−1− tm < 0. For example, 120,

71, 49, 22, 27 is a sumac sequence of length 5.

(a) Find the positive integer B so that the sumac sequence 150, B, . . . has the maximum

possible number of terms.

Solution

Suppose that we have a sumac sequence with t1 = 150 and t2 = B. Let’s write out the

next several terms (assuming that they exist) in terms of B:

t3 = 150−B t4 = 2B − 150 t5 = 300− 3B t6 = 5B − 450t7 = 750− 8B t8 = 13B − 1200 t9 = 1950− 21B t10 = 34B − 3150

In order to maximize the length of this sumac sequence, we would like to choose B so that

as many terms as possible starting from t1 are non-negative. (When we reach the first

negative “term”, we know that the sequence terminated at the previous term.)

For t2 ≥ 0, B ≥ 0.

For t3 ≥ 0, 150−B ≥ 0 or B ≤ 150.

For t4 ≥ 0, 2B − 150 ≥ 0 or B ≥ 75.

For t5 ≥ 0, 300− 3B ≥ 0 or B ≤ 100.

For t6 ≥ 0, 5B − 450 ≥ 0 or B ≥ 90.

For t7 ≥ 0, 750− 8B ≥ 0 or B ≤ 7508

= 9368.

For t8 ≥ 0, 13B − 1200 ≥ 0 or B ≥ 120013

= 92 413

.

For t9 ≥ 0, 1950− 21B ≥ 0 or B ≤ 195021

= 921821

.

Therefore, since B must be a positive integer, if we choose B = 93, then we can ensure

that each of t1 through t8 are non-negative. This will maximize the number of terms

starting from the beginning, since B must satisfy 92 413≤ B ≤ 936

8in order for at least the

first eight terms to be non-negative. (Note that t9 will in fact be negative when B = 93.)

When we set B = 93, we obtain the sumac sequence 150, 93, 57, 36, 21, 15, 6, 9.


(b) Let m be a positive integer with m ≥ 5. Determine the number of sumac sequences of

length m with tm ≤ 2000 and with no term divisible by 5.

Solution

We begin our solution by making some observations about sumac sequences.

• A sumac sequence is completely determined by its first two terms. This is true since

the first two terms give us the third, the second and third give us the fourth, and so

on. The sequence will terminate when the “next term” would be negative.

• In a sumac sequence, since for every (valid) n we have tn+2 = tn − tn+1, then tn =

tn+1 +tn+2. This means that we can “reverse engineer” a sumac sequence – if we know

terms (n + 1) and (n + 2), then we can determine term n. Thus, if we know the final

two terms in a sumac sequence, then we can determine all of the previous terms.

• From the first observation, the first two terms of a sumac sequence completely de-

termines the sequence. Is the same true of the last two terms? No. When we start

looking at a sumac sequence from the back, every new term as we proceed towards

the front will always be non-negative (since we are adding non-negative terms). Thus,

there is no “stopping condition” as there is when we work forwards. (For example, 3,

1, 2 is a sumac sequence ending with 1, 2, as is 4, 3, 1, 2.)

• However, if we know the final two terms and the length of the sequence, this completely

determines the sumac sequence (and we will always be able to find such a sequence).

Now we proceed. Let m be a fixed positive integer with m ≥ 5.

Suppose that t1, t2, . . . , tm−1, tm is a sumac sequence of length m.

Because we are given a condition on the final term of the sequence, we will examine the

sequence from the back.

Let x = tm and y = tm−1. Note that x, y and m determine the sequence.

Since x and y are the last two terms in the sumac sequence, then tm−1 − tm = y − x < 0

or x > y.

Since we have m fixed, we would like to determine how many sumac sequences we can

form with tm = x ≤ 2000, tm−1 = y < x and no term divisible by 5.

Let’s write out the last five terms of the sequence (in reverse order): x, y, x + y, x + 2y,

2x + 3y. (Since m ≥ 5, we know that there are at least five terms in the sequence.)

Since we want no term divisible by 5, let us consider x and y modulo 5 to see what hap-

pens. (There are 25 possible pairs for (x, y) modulo 5.)

Since no term is divisible by 5, then we don’t want x ≡ 0 (mod 5) or y ≡ 0 (mod 5). This

cuts us down to 16 possibilities for (x, y).

We make a table of these possibilities to determine which pairs can be eliminated simply

by looking at the last five terms. (All entries in the table are modulo 5. In any given row,

we stop if we reach a 0, since this possibility can then be discarded.)


x y x + y x + 2y 2x + 3y

1 1 2 3 0

1 2 3 0

1 3 4 2 1

1 4 0

2 1 3 4 2

2 2 4 1 0

2 3 0

2 4 1 0

3 1 4 0

3 2 0

3 3 1 4 0

3 4 2 1 3

4 1 0

4 2 1 3 4

4 3 2 0

4 4 3 2 0

So the only possible pairs for (x, y) modulo 5 are (1, 3), (2, 1), (3, 4) and (4, 2).

If we start with (x, y) = (1, 3) modulo 5, then the terms in the sequence modulo 5 are 1,

3, 4, 2, 1, 3, 4, 2, 1, . . ., ie. the terms cycle modulo 5 with no terms divisible by 5.

This similar cycling will happen with each of the other 3 pairs, so each of these 4 pairs

give no terms divisible by 5.

So for each of these pairs, we need to determine the number of pairs of non-negative

integers (x, y) with x ≤ 2000, y < x and congruent to the appropriate things modulo

5. Each such pair will give a sumac sequence of length m ≥ 5 with no term divisible by

5. (Since the divisibility of the terms is independent of length, this also means that the

number of such sequences will be independent of m!)

Case 1: (x, y) congruent to (1, 3) modulo 5

Here x can take the values 1996, 1991, . . ., 6, 1.

If x = 1996, y can be 1993, 1988, . . ., 8, 3. (399 possibilities)


This pattern continues, with one fewer possibility each time x decreases by 5, until we

reach x = 6, where y = 3 is the only possibility.

Thus, there are 399 + 398 + · · ·+ 2 + 1 possibilities for this case.























Therefore, overall there are

2(399+398+· · ·+2+1)+2(400+399+· · ·+2+1) = 399(400)+400(401) = 400(800) = 320 000

possibilities. Therefore, there are exactly 320 000 sumac sequences of length m with no

term divisible by 5 and with tm ≤ 2000.



presents the

Canadian Open



Solutions



Part A

1. Determine the value of 102 − 92 + 82 − 72 + 62 − 52 + 42 − 32 + 22 − 12.

Solution 1

Using differences of squares,

102 − 92 + 82 − 72 + 62 − 52 + 42 − 32 + 22 − 12

= (10− 9)(10 + 9) + (8− 7)(8 + 7) + (6− 5)(6 + 5) + (4− 3)(4 + 3) + (2− 1)(2 + 1)

= 1(10 + 9) + 1(8 + 7) + 1(6 + 5) + 1(4 + 3) + 1(2 + 1)

= 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

= 55

(We can get the answer 55 either by computing the sum directly, or by using the fact that

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 12(10)(11) = 55.)

Solution 2

Computing directly,

102 − 92 + 82 − 72 + 62 − 52 + 42 − 32 + 22 − 12

= 100− 81 + 64− 49 + 36− 25 + 16− 9 + 4− 1

= 19 + 15 + 11 + 7 + 3 (computing difference of each pair)

= 55

Answer: 55

2. A bug in the xy-plane starts at the point (1, 9). It moves first to the point (2, 10) and then to

the point (3, 11), and so on. It continues to move in this way until it reaches a point whose

y-coordinate is twice its x-coordinate. What are the coordinates of this point?

Solution 1

The bug starts at (1, 9) and each time moves 1 unit to the right and 1 unit up.

Thus, after k moves, the bug will be at the point (1 + k, 9 + k).

When its y-coordinate is twice its x-coordinate, we have 9 + k = 2(1 + k) or 9 + k = 2 + 2k

or k = 7.

When k = 7, the bug is at point (1 + 7, 9 + 7) = (8, 16), and the bug stops here.

Solution 2

The bug starts at (1, 9) and each time moves 1 unit to the right and 1 unit up.

Thus, at any point to which the bug moves, the y-coordinate will be 8 more than


the x-coordinate, so every such point is of the form (n, n + 8).

For the y-coordinate to be twice the x-coordinate, n + 8 = 2n or n = 8.

When n = 8, the bug is at the point (8, 16), and the bug stops here.

Solution 3

We write out the sequence of points to which the bug moves and stop when we get to a point

where the y-coordinate is twice the x-coordinate:

(1, 9), (2, 10), (3, 11), (4, 12), (5, 13), (6, 14), (7, 15), (8, 16)

Thus, the bug stops at (8, 16).

Answer: (8, 16)

3. If ax3 + bx2 + cx + d = (x2 + x − 2)(x − 4) − (x + 2)(x2 − 5x + 4) for all values of x, what is

the value of a + b + c + d?

Solution 1

We use the fact that a + b + c + d = a(13) + b(12) + c(1) + d, so a + b + c + d must be equal to

the right side of the given equation with x set equal to 1.

Thus,

a + b + c + d = (12 + 1− 2)(1− 4)− (1 + 2)(12 − 5 + 4) = 0(−3)− 3(0) = 0

Solution 2

We simplify the right side of the given equation by factoring the two quadratic polynomials:

(x2 + x− 2)(x− 4)− (x + 2)(x2 − 5x + 4) = (x− 1)(x + 2)(x− 4)− (x + 2)(x− 1)(x− 4)

= 0

Therefore, ax3 + bx2 + cx + d = 0 for all values of x. (In other words, ax3 + bx2 + cx + d = 0 is

the zero polynomial, so all of its coefficients are equal to 0.)

Therefore, a = b = c = d = 0, so a + b + c + d = 0.

Solution 3

We expand and simplify the right side:

ax3 + bx2 + cx + d = (x2 + x− 2)(x− 4)− (x + 2)(x2 − 5x + 4)

= x3 + x2 − 2x− 4x2 − 4x + 8− (x3 + 2x2 − 5x2 − 10x + 4x + 8)

= x3 − 3x2 − 6x + 8− (x3 − 3x2 − 6x + 8)

= 0

Therefore, each of the coefficients of ax3 + bx2 + cx + d are 0, so a = b = c = d = 0 and


thus a + b + c + d = 0.

Answer: 0

4. A fractionp

qis in lowest terms if p and q have no common factor larger than 1.

How many of the 71 fractions1

72,

2

72, . . . ,

70

72,71

72are in lowest terms?

Solution 1

First, we note that 72 = 23 × 32.

For one of the fractionsa

72to be in lowest terms, then a and 72 have no common factors.

In other words, a cannot be divisible by 2 or 3 (since 2 and 3 are the only prime numbers which

are divisors of 72).

How many of the positive integers from 1 to 71 are not divisible by 2 or 3?

Of these integers, 35 of the them are divisible by 2 (namely, 2, 4, 6, . . ., 70).

Also, 23 of them (namely, 3, 6, . . ., 69) are divisible by 3.

But some numbers are counted twice in these lists: all of the multiples of both 2 and 3 (ie. the

multiples of 6). These are 6, 12, . . ., 66 – that is, 11 numbers in total.

So the number of positive integers from 1 to 71 which are divisible by 2 or 3 is 35+23−11 = 47

(11 is subtracted to remove the double-counted numbers).

So the number of positive integers from 1 to 71 which are not divisible by 2 or 3 is 71−47 = 24.

Therefore, 24 of the 71 fractions1

72,

2

72, · · · ,

70

72,71

72are irreducible.

Solution 2

First, we note that 72 = 23 × 32.

For one of the fractionsa

72to be in lowest terms, then a and 72 have no common factors.

Since the only primes which are divisors of 72 are 2 and 3, then a and 72 have no common

factors when a is not divisible by 2 or 3.

Look at the first few fractions in the list:

1

72,

2

72,

3

72,

4

72,

5

72,

6

72,

7

72,

8

72,

9

72,10

72,11

72,12

72

From this list, the ones which are in lowest terms are

1

72,

5

72,

7

72,11

72

So the 1st and 5th of each of the two sets of 6 fractions above are in lowest terms.

This pattern will continue, so if we include the fraction72

72(which we know is not in lowest

terms) at the end of the list, we obtain 12 sets of 6 fractions, and 2 fractions out of each set

will be in lowest terms, giving 12× 2 = 24 fractions in lowest terms.


(Why does this pattern continue? Each of the fractions in the list can be written in one

of the following forms:

6k + 1

72,6k + 2

72,6k + 3

72,6k + 4

72,6k + 5

72,6k + 6

72

Since the numerators 6k + 2, 6k + 4 and 6k + 6 are divisible by 2 and the numerator 6k + 3 is

divisible by 3, then none of the fractions with these as numerators is in lowest terms.

Also, 6k+1 and 6k+5 are never divisible by 2 or 3, so these corresponding fractions are always

in lowest terms.

Thus, 2 out of each set of 6 fractions is in lowest terms.)

Answer: 24

5. An office building has 50 storeys, 25 of which are painted black and the other 25 of which are

painted gold. If the number of gold storeys in the top half of the building is added to the

number of black storeys in the bottom half of the building, the sum is 28. How many gold

storeys are there in the top half of the building?

Solution 1

Let G be the number of gold storeys in the top half of the building.

Then there are 25−G black storeys in the top half of the building.

Since there are 25 black storeys in total, then the number of black storeys in the bottom half

of the building is 25− (25−G) = G.

Since the sum of the number of gold storeys in the top half of the building and the number of

black storeys in the bottom half of the building is 28, then G + G = 28, or G = 14.

Thus, there are 14 gold storeys in the top half of the building.

Solution 2

Let G and g be the number of gold storeys in the top and bottom halfs of the building, and

B and b the number of black storeys in the top and bottom halfs of the building.

Then G + B = 25 and g + b = 25, looking at the top and bottom halfs of the building.

Also, G + g = 25 and B + b = 25, since 25 of the storeys are painted in each colour.

Also, G + b = 28 from the given information, or b = 28−G.

Since B + b = 25, then B + 28−G = 25, so B = G− 3.

Since G + B = 25, then G + G− 3 = 25 or 2G = 28 or G = 14.

Thus, there are 14 gold storeys in the top half of the building.

Answer: 14


6. In the grid shown, each row has a value as-

signed to it and each column has a value as-

signed to it. The number in each cell is the

sum of its row and column values. For exam-

ple, the “8” is the sum of the value assigned to

the 3rd row and the value assigned to the 4th

column. Determine the values of x and y.

3 0 5 6 −2

−2 −5 0 1 y

5 2 x 8 0

0 −3 2 3 −5

−4 −7 −2 −1 −9

Solution 1

First, we label the values assigned to the five columns A, B, C,D,E and the values assigned to

the five rows a, b, c, d, e.

Look at the sub-grid0 1

x 8.

Since the 0 is in row 2 and column 3, then 0 = b + C.

Similarly, 1 = b + D, 8 = c + D and x = c + C.

But then 0 + 8 = (b + C) + (c + D) = (c + C) + (b + D) = x + 1 or x = 7.

In a similar way, we can show by looking at the sub-grid1 y

8 0that we must have

1 + 0 = y + 8 or y = −7.

Thus, x = 7 and y = −7.

(In fact, in any sub-grid of the formp q

r s, we must have p + s = q + r.)

Solution 2



Suppose that we try A = 0.

Looking at the “3” in the first row and first column, A + a = 3, so a = 3.

Since a = 3 and the entry in the first row and second column is 0, then a + B = 0, or B = −3.

Similarly, C = 2, D = 3 and E = −5.

Since A = 0 and the entry in the second row and first column is −2, then b + A = 0, then

b = −2.

Since y = b + E, then y = −2 + (−5) = −7.

Since A = 0 and the entry in the third row and first column is 5, then c + A = 5, so c = 5.

Since x = c + C, then x = 5 + 2 = 7.

Thus, x = 7 and y = −7.

Solution 3



If we choose five entries from the table which include one from each row and one from each


column, then the sum of these entries is constant no matter how we choose the entries, as it is

always equal to A + B + C + D + E + a + b + c + d + e.

Here are three ways in which this can be done (looking at the bolded numbers):

3 0 5 6 −2

−2 −5 0 1 y

5 2 x 8 0

0 −3 2 3 −5

−4 −7 −2 −1 −9

3 0 5 6 −2

−2 −5 0 1 y

5 2 x 8 0

0 −3 2 3 −5

−4 −7 −2 −1 −9

3 0 5 6 −2

−2 −5 0 1 y

5 2 x 8 0

0 −3 2 3 −5

−4 −7 −2 −1 −9

Therefore, 3 + (−5) + 2 + 8 + (−9) = (−4) + (−3) + x + 1 + (−2) = 3 + y + 2 + (−2) + 3 or

−1 = x− 8 = y + 6.

Thus, x = 7 and y = −7.

Solution 4



Consider the first two entries in row 1.

We have 3 = A + a and 0 = B + a.

Subtracting these, we obtain 3 = 3− 0 = (A + a)− (B + a) = A−B.

Notice that whenever we take entries in columns 1 and 2 from the same row, their difference

will always equal A−B, which is equal to 3.

Similarly, since the difference between the 0 and the 5 in the first row is 5, then every entry in

column 3 will be 5 greater than the entry in column 2 from the same row.

Thus, x = 2 + 5 = 7.

Also, since the difference between the 6 and the −2 in the first row is 8, then every entry in

column 5 will be 8 less than the entry in column 4 from the same row.

Thus, y = 1− 8 = −7.

Therefore, x = 7 and y = −7.

Answer: x = 7 and y = −7


7. In the diagram, the semi-circle has centre O and diameter AB. A ray of light leaves point P

in a direction perpendicular to AB. It bounces off the semi-circle at point D in such a way

that ∠PDO = ∠EDO. (In other words, the angle of incidence equals the angle of reflection

at D.) The ray DE then bounces off the circle in a similar way at E before finally hitting the

semicircle again at B. Determine ∠DOP .

A B

E

D

P O

Solution 1

Join D and E to O, and let ∠DOP = x.

Since DP ⊥ AB, then ∠PDO = 90◦ − x.

Since the angle of incidence equals the angle of reflection at D, then ∠EDO = ∠PDO = 90◦−x.

A B

E

D

P Ox

90 x

90 x

90 x

90 x

90 x

Since DO and EO are both radii, then DO = EO, so 4EDO is isosceles, and so

∠DEO = ∠EDO = 90◦ − x.

Since the angle of incidence equals the angle of reflection at E, then ∠DEO = ∠BEO = 90◦−x.

Since EO and BO are both radii, then EO = BO, so 4BEO is isosceles, and so

∠EBO = ∠BEO = 90◦ − x.

Consider quadrilateral PDEB.

We have ∠DPB = 90◦, ∠PDE = (90◦ − x) + (90◦ − x) = 180◦ − 2x,

∠DEB = (90◦ − x) + (90◦ − x) = 180◦ − 2x, and ∠EBP = 90◦ − x.

Since the sum of the angles in the quadrilateral is 360◦, then

90◦ + 180◦ − 2x + 180◦ − 2x + 90◦ − x = 360◦ or 540◦ − 5x = 360◦ or 5x = 180◦ or x = 36◦.

Therefore, ∠DOP = x = 36◦.

Solution 2

Join D and E to O, and let ∠DOP = x.

Since DP ⊥ AB, then ∠PDO = 90◦ − x.



A B

E

D

P Ox

90 x

90 x

90 x

90 x

90 x2x 2x


∠DEO = ∠EDO = 90◦ − x. Also, ∠DOE = 180◦ − 2(90◦ − x) = 2x.



∠EBO = ∠BEO = 90◦ − x. Also, ∠EOB = 180◦ − 2(90◦ − x) = 2x.

Since POB is a straight line, then ∠POD + ∠DOE + ∠EOB = 180◦ or x + 2x + 2x = 180◦ or

5x = 180◦ or x = 36◦.

Therefore, ∠DOP = x = 36◦.

Solution 3

Reflect the diagram across AB to complete the circle and form the pentagon DEBE ′D′. (Note

that DPD′ is a straight line since ∠DPO = ∠D′PO = 90◦.)

E

D

A BO

D

E

Since DO, EO, BO, E ′O and D′O are all radii, then DO = EO = BO = E ′O = D′O.

Let ∠DOP = x. Since DP ⊥ AB, then ∠PDO = 90◦ − x.

By reflection, ∠D′OP = ∠DOP = x, so ∠DOD′ = 2x.



∠DEO = ∠EDO = 90◦ − x. Also, ∠DOE = 180◦ − 2(90◦ − x) = 2x.



∠EBO = ∠BEO = 90◦ − x. Also, ∠EOB = 180◦ − 2(90◦ − x) = 2x.


Therefore, the triangles DOE, EOB, BOE ′, E ′OD′ and D′OD are all congruent by side-angle-

side. Therefore, pentagon DEBE ′D′ is a regular pentagon.

Thus, ∠DOD′ = 15(360◦) = 72◦ since the central angles of each of the five sides of the pentagon

are equal.

Since 4DOD′ is isosceles and OP is perpendicular to DD′, then ∠POD = 12∠DOD′ = 36◦.

Thus, ∠POD = 36◦.

Answer: ∠DOP = 36◦

8. The number 18 is not the sum of any 2 consecutive positive integers, but is the sum of consec-

utive positive integers in at least 2 different ways, since 5 + 6 + 7 = 18 and 3 + 4 + 5 + 6 = 18.

Determine a positive integer less than 400 that is not the sum of any 11 consecutive positive

integers, but is the sum of consecutive positive integers in at least 11 different ways.

Solution

Suppose that the positive integer N is the sum of an odd number of consecutive integers, say

2k + 1 consecutive integers. Then for some integer a,

N = (a− k) + (a− (k − 1)) + · · ·+ (a− 1) + a + (a + 1) + · · ·+ (a + k) = (2k + 1)a

Thus, 2k+1 is a divisor of N (ie. the number of integers in the representation is a divisor of N).

Next, suppose that N is the sum of an even number of consecutive integers, say 2k consecutive

integers. Then for some integer b,

N = (b−k)+(b−(k−1))+· · ·+(b−1)+b+(b+1)+· · · (b+(k−1)) = 2kb−k = k(2b−1) =1

2(2k)(2b−1)

Thus, k is a divisor of N and 2k is not a divisor of N (since 2b−1 is odd and so has no factor of 2).

We would like to find a positive integer N which is not the sum of 11 consecutive positive

integers (and so is not a multiple of 11) but is the sum of consecutive positive integers in 11

different ways.

Let’s consider the number of integers in each of the ways in which we write N as the sum

of consecutive integers. Note that if N is the sum of m consecutive positive integers, then N

is at least 1 + 2 + · · · + m. We make a table of what properties N must have for N to be the

sum of m consecutive integers for m = 2 to m = 10:


m N at least Property of N

2 3 Divisible by 1, not by 2

3 6 Divisible by 3


5 15 Divisible by 5


(ie. divisible by 3, not by 2)

7 28 Divisible by 7


9 45 Divisible by 9


(ie. divisible by 5, not by 2)

How can we combine as many of these as possible? If we make N at least 55 and divisible

by 5, 7 and 9 and not divisible by 2, then N will be the sum of 2, 3, 5, 6, 7, 9 and 10 con-

secutive positive integers (7 representations in total). In this case, N must be divisible by

5 × 7 × 9 = 315. So following this line of thought, if N is less than 400, then we must have

N = 315. Now, 315 is also

• the sum of 15 consecutive positive integers since 315 is divisible by 15 and is at least 120

(ie. 1 + 2 + 3 + · · ·+ 15),

• the sum of 14 consecutive positive integers since 315 is divisible by 7, not by 14, and is at

least 105 (ie. 1 + 2 + 3 + · · ·+ 14)

• the sum of 18 consecutive positive integers since 315 is divisible by 9, not by 18, and is at

least 171 (ie. 1 + 2 + 3 + · · ·+ 18)

• the sum of 21 consecutive positive integers since 315 is divisible by 21 and is at least 231

(ie. 1 + 2 + 3 + · · ·+ 21)

So 315 is the sum of consecutive positive integers in at least 11 ways, and is not the sum

of 11 consecutive positive integers. (In fact, 315 is the unique answer, but we are not asked to

justify this.)

(Note: A good way to write a solution to this problem would be to first figure out in rough

that 315 was the answer, and then begin the solution by claiming that 315 is the answer. We

could then demonstrate that 315 works by showing that it can be represented in the correct

number of ways. While this approach is perfectly correct, it would not give much of a clue as

to how the answer was obtained.)

Answer: 315


Part B

1. A line with slope −3 intersects the positive x-axis at A and the positive y-axis at B. A second

line intersects the x-axis at C(7, 0) and the y-axis at D. The lines intersect at E(3, 4).

x

D

B

O A C (7, 0)

E (3, 4)

y

(a) Find the slope of the line through C and E.

Solution

Since C has coordinates (7, 0) and E has coordinates (3, 4), then the slope of the line

through C and E is0− 4

7− 3=−4

4= −1

(b) Find the equation of the line through C and E, and the coordinates of the point D.

Solution 1

Since the line through C and E has slope −1 and passes through the point (7, 0), then

the line has equation y − 0 = (−1)(x− 7) or y = −x + 7.

From the equation of the line, y = 7 is the y-intercept of the line.

Since D is the point where this line crosses the y-axis, then D has coordinates (0, 7).

Solution 2

Since the line through C and E has slope −1 and passes through the point (3, 4), then

the line has equation y − 4 = (−1)(x− 3) or y = −x + 7.


Since D is the point where this line crosses the y-axis, then D has coordinates (0, 7).

(c) Find the equation of the line through A and B, and the coordinates of the point B.

Solution

Since the line through A and B has slope −3 and passes through the point E(3, 4), then


the line has equation y − 4 = (−3)(x− 3) or y = −3x + 13.


Since B is the point where this line crosses the y-axis, then B has coordinates (0, 13).

(d) Determine the area of the shaded region.

Solution 1

The area of the shaded region is the sum of the areas of 4DOC and 4BDE.

4DOC is right-angled at O, so the area of 4DOC is 12(DO)(OC) = 1

2(7)(7) = 49

2.

We can consider 4BDE as having base BD of length 13− 7 = 6 and height equal to the

distance of E from the y-axis (a distance of 3).

Therefore, the area of 4BDE is 12(6)(3) = 9.

Thus, the area of the shaded region is 492

+ 9 = 672.

Solution 2

The area of the shaded region is the sum of the areas of 4BOA and 4AEC.

4BOA is right-angled at O, so the area of 4BOA is 12(BO)(OA).

Point A is the point where the line y = −3x+13 crosses the x-axis, so it has x-coordinate

which satisfies −3x + 13 = 0, ie. x = 133.

Therefore, the area of 4BOA is 12(13)

(133

)= 169

6.

We can consider 4AEC as having base AC of length 7− 133

= 83

and height equal to the

distance of E from the x-axis (a distance of 4).

Therefore, the area of 4AEC is 12(4)

(83

)= 16

3.

Thus, the area of the shaded region is 1696

+ 163

= 2016

= 672.

Solution 3

Drop perpendiculars from E to point X on the x-axis and to point Y on the y-axis.

x

D

B

O A C (7, 0)

E (3, 4)

y

X

Y

Then Y has coordinates (0, 4), X has coordinates (3, 0), and OXEY is a rectangle.

The area of the shaded region is thus the sum of the areas of 4BY E, rectangle OXEY

and 4EXC.


Since 4BY E is right-angled at Y , its area is 12(BY )(Y E) = 1

2(13− 4)(3− 0) = 27

2.

The area of rectangle OY EX is 3× 4 = 12.

Since 4EXC is right-angled at X, its area is 12(EX)(XC) = 1

2(4− 0)(7− 3) = 8.

Therefore, the area of the shaded region is 272

+ 12 + 8 = 672.

2. (a) Determine all possible ordered pairs (a, b) such that

a− b = 1

2a2 + ab− 3b2 = 22

Solution 1

Factoring the left side of the second equation, we get 2a2 + ab− 3b2 = (a− b)(2a + 3b).

Since a− b = 1, we get (1)(2a + 3b) = 22 or 2a + 3b = 22.

So we now have a− b = 1 and 2a + 3b = 22.

Adding 3 times the first equation to the second equation, we get 5a = 25 or a = 5.

Substituting back into the first equation, we get b = 4.

Thus, the only solution is (a, b) = (5, 4).

Solution 2

From the first equation, a = b + 1.

Substituting into the second equation, we obtain

2(b + 1)2 + (b + 1)(b)− 3b2 = 22

(2b2 + 4b + 2) + (b2 + b)− 3b2 = 22

5b = 20

b = 4

Substituting back into the first equation, we get a = 5, so the only solution is (a, b) = (5, 4).

Solution 3

From the first equation, b = a− 1.

Substituting into the second equation, we obtain

2a2 + a(a− 1)− 3(a− 1)2 = 22

2a2 + (a2 − a)− (3a2 − 6a + 3) = 22

5a = 25

a = 5

Substituting back into the first equation, we get b = 4, so the only solution is (a, b) = (5, 4).


(b) Determine all possible ordered triples (x, y, z) such that

x2 − yz + xy + zx = 82

y2 − zx + xy + yz = −18

z2 − xy + zx + yz = 18

Solution 1

If we add the second equation to the third equation, we obtain

y2 − zx + xy + yz + z2 − xy + zx + yz = −18 + 18

y2 + 2yz + z2 = 0

(y + z)2 = 0

y + z = 0

z = −y

Substituting back into the three equations, we obtain

x2 + y2 = 82

2xy = −18

−2xy = 18

Thus, x2 + y2 = 82 and xy = −9.

Therefore, (x + y)2 = x2 + 2xy + y2 = 82 + (−18) = 64, so x + y = ±8.

If x + y = 8, then y = 8− x and so since xy = −9, then x(8− x) = −9 or x2− 8x− 9 = 0

or (x− 9)(x + 1) = 0 so x = 9 or x = −1.

Since x + y = 8, then if x = 9, we have y = −1 and z = −y = 1.

Since x + y = 8, then if x = −1, we have y = 9 and z = −y = −9.

If x+y = −8, then y = −8−x and so since xy = −9, then x(−8−x) = −9 or x2+8x−9 = 0

or (x + 9)(x− 1) = 0 so x = −9 or x = 1.

Since x + y = −8, then if x = −9, we have y = 1 and z = −y = −1.

Since x + y = −8, then if x = 1, we have y = −9 and z = −y = 9.

Therefore, the four solutions are (x, y, z) = (9,−1, 1), (−1, 9,−9), (−9, 1,−1), (1,−9, 9).


Solution 2

If we add the first equation to the second equation, we obtain

x2 − yz + xy + zx + y2 − zx + xy + yz = 82− 18

x2 + 2xy + y2 = 64

(x + y)2 = 64

x + y = ±8

Similarly, adding the first equation to the third equation, we obtain x2 + 2xz + z2 = 100

or x + z = ±10.

Also, adding the second equation to the third equation, we obtain y2 + 2yz + z2 = 0

or y + z = 0, and so z = −y.

Using x + z = ±10 and z = −y, we obtain x− y = ±10.

Thus, we have x + y = ±8 and x− y = ±10.

If x + y = 8 and x − y = 10, then adding these equations, we get 2x = 18 or x = 9

and so y = −1 and z = −y = 1.

If x + y = 8 and x − y = −10, then adding these equations, we get 2x = −2 or x = −1

and so y = 9 and z = −y = −9.

If x + y = −8 and x − y = 10, then adding these equations, we get 2x = 2 or x = 1 and

so y = −9 and z = −y = 9.

If x + y = −8 and x− y = −10, then adding these equations, we get 2x = −18 or x = −9

and so y = 1 and z = −y = −1.

Therefore, the four solutions are (x, y, z) = (9,−1, 1), (−1, 9,−9), (−9, 1,−1), (1,−9, 9).

3. Four tiles identical to the one shown, with a > b > 0,

are arranged without overlap to form a square with

a square hole in the middle.a

a

b

b

(a) If the outer square has area (a + b)2, show that the area of the inner square

is (a− b)2.

Solution 1

Each tile can be split into two right-angled triangles along a diagonal, each with legs of

lengths a and b.

The area of each of these triangles is 12ab, so the area of each tile is ab.


If the outer square has area (a + b)2 and this is partially covered with four tiles each of

area ab, then the area of the leftover portion (ie. the square hole) is

(a + b)2 − 4ab = a2 + 2ab + b2 − 4ab = a2 − 2ab + b2 = (a− b)2

Solution 2

If the outer square has area (a + b)2, then the side length of the outer square is a + b.

In order to get a side length of a + b, we need to line up the “a” side of a tile with the “b”

side of a second tile, as shown.

a

a

b

b a

b

b

(Note that the tiles do fit together in this way, since each is a quadrilateral with two right

angles, so the remaining two angles add to 180◦, that is, a straight line.)

We can complete the square as follows:

a

b

b a

b

b

bb

a

ba

Now the inner hole is clearly a rectangle (as it has four right angles) and is in fact a square

as its four sides are all of length a− b (as each of its sides are the remaining portion of a

line segment of length a when a segment of length b is cut off from one end).

Since the inner square has side length a− b, then its area is (a− b)2.

(b) Determine the smallest integer value of N for which there are prime numbers a and b such

that the ratio of the area of the inner square to the area of the outer square is 1 : N .

Solution

From (a), the ratio of the area of the inner square to the area of the outer square is(a− b)2

(a + b)2.

We would like to find integers N for which there are prime numbers a and b such

that(a− b)2

(a + b)2=

1

N(and in fact find the minimum such N).

Taking the positive square root of both sides, we obtaina− b

a + b=

1√N

.


Since the left side is a rational number (since a and b are integers), then√

N must be

rational, so N must be a perfect square.

Suppose N = k2, for some positive integer k.

Thus, we havea− b

a + b=

1

kor a + b = k(a− b) or (k − 1)a = (k + 1)b.

Since we would like to find the smallest value of N which works, then we try to find

the smallest value of k which works.

Does k = 1 work? Are there prime numbers a and b so that 0 = 2b? No, since this means

b = 0.

Does k = 2 work? Are there prime numbers a and b so that a = 3b? No, since here a is a

multiple of 3, so the only possible prime value of a is 3, which would make b = 1, which

is not a prime.

Does k = 3 work? Are there prime numbers a and b so that 2a = 4b (ie. a = 2b)? No,

since here a is a multiple of 2, so the only possible prime value of a is 2, which would make

b = 1, which is not a prime.

Does k = 4 work? Are there prime numbers a and b so that 3a = 5b?

Yes: a = 5 and b = 3.

Therefore, the smallest value of k which works is k = 4, so the smallest value of N which

works is N = 16.

(c) Determine, with justification, all positive integers N for which there are odd integers

a > b > 0 such that the ratio of the area of the inner square to the area of the outer

square is 1 : N .

Solution

Suppose that N is a positive integer for which there are odd integers a > b > 0 such that(a− b)2

(a + b)2=

1

N.

Then, as in (b), N must be a perfect square, say N = k2, for some positive integer k.

Since a and b are odd, then set a = 2A + 1 and b = 2B + 1, for some integers A and B.

Thus we have(2A− 2B)2

(2A + 2B + 2)2=

1

k2or

A−B

A + B + 1=

1

kor k(A−B) = A + B + 1.

If A and B have the same parity (ie. both even or both odd), then A − B is even so the

left side is even and A + B + 1 is odd, so the right side is odd. Since we cannot have an

odd number equal to an even number, then this cannot happen.

Thus, A and B must have opposite parity (ie. one even and the other odd). In this case,

A−B is odd and A + B + 1 is even. Since k(A−B) = A + B + 1, then k is even.

Therefore, N must be an even perfect square.

We must now check if every even perfect square is a possible value for N .


Suppose N = (2m)2.

Using our substitutions from above, can we find integers A and B so that

2m(A−B) = A + B + 1?

If A = m and B = m−1, then A−B = 1 and A+B+1 = 2m, so 2m(A−B) = A+B+1.

So if a = 2A + 1 = 2m + 1 and b = 2B + 1 = 2m− 1, then(a− b)2

(a + b)2=

1

(2m)2=

1

N.

Therefore, the positive integers N which have the required property are all even perfect

squares.

4. Triangle ABC has its base on line segment PN and vertex A on line PM . Circles with centres

O and Q, having radii r1 and r2, respectively, are tangent to the triangle ABC externally and

to each of PM and PN .

P N

M

A

B C

E

D

GF

KL

QO

(a) Prove that the line through K and L bisects the perimeter of triangle ABC.

Solution

We must show that KB + BC + CL = KA + AL.

Since BK and BF are tangents to the left circle from the same point B, then BK = BF .

Since CL and CG are tangents to the right circle from the same point C, then CL = CG.

Since AK and AD are tangents to the left circle from the same point A, then AK = AD.

Since AL and AE are tangents to the right circle from the same point A, then AL = AE.

Therefore, KB + BC + CL = FB + BC + CG = FG and KA + AL = DA + AE = DE.

Now FG = PG− PF and DE = PE − PD.

Since PE and PG are tangents to the right circle from the same point P , then PE = PG.

Since PD and PF are tangents to the left circle from the same point P , then PD = PF .

Therefore, FG = PG− PF = PE − PD = DE, so KB + BC + CL = KA + AL, ie. the

line through K and L bisects the perimeter of triangle ABC.

(b) Let T be the point of contact of BC with the circle inscribed in triangle ABC.

Prove that (TC)(r1) + (TB)(r2) is equal to the area of triangle ABC.

Solution 1


Let I be the centre of the circle inscribed in 4ABC, T be the point of contact of this

circle with BC, and r the radius of this circle.

Join O to K and B, and I to B and T .

P N

M

A

B C

E

D

GF

KL

Q

O

T

I

(Note that the circle with centre I is not necessarily tangent to AB at K or AC at L.)

Note that OK is perpendicular to KB and IT is perpendicular to BC.

Now OB bisects ∠FBK and IB bisects ∠KBC, since the circles with centres O and I

are tangent to FB and BK, and BA and BC, respectively.

Now ∠KOB = 90◦−∠KBO = 90◦− 12∠FBK = 1

2(180◦−∠FBK) = 1

2∠KBC = ∠IBT ,

so 4OKB is similar to 4BTI.

Therefore,BK

KO=

IT

TBor

BK

r1

=r

TBor r1 =

(TB)(BK)

r.

Similarly, r2 =(TC)(LC)

r.

Therefore,

(TC)(r1) + (TB)(r2) =(TC)(TB)(BK)

r+

(TB)(TC)(LC)

r=

(TB)(TC)

r(BK + LC)

Let BC = a, AB = c, AC = b, and let s be the semi-perimeter of 4ABC (that is, s is

half of the perimeter of 4ABC).

Now, from (a), since KB + BC + LC = s, then BK + LC = s−BC = s− a.

Therefore, (TC)(r1) + (TB)(r2) =(TB)(TC)

r(s− a).

We can now focus entirely on 4ABC.

Let X and Y be the points where the circle with centre I is tangent to sides AB and AC,

respectively.


A

B C

XY

T

I

Using tangent arguments as in (a), we see that AX = AY , BX = BT and CY = CT .

Since AX + AY + BX + BT + CY + CT = 2s, then BT + AY + Y C = s,

so TB = s− (AY + Y C) = s− AC = s− b.

Similarly, TC = s− c.

Therefore, (TC)(r1) + (TB)(r2) =(s− b)(s− c)(s− a)

r=

s(s− a)(s− b)(s− c)

sr.

Let |4ABC| denote the area of 4ABC.

Then s(s− a)(s− b)(s− c) = |4ABC|2 by Heron’s formula.

Also,

sr =1

2r(AB + BC + AC) =

1

2(IX)(AB) +

1

2(IT )(BC) +

1

2(IY )(AC) (∗)

since IX, IT and IY are all radii of the circle with centre I.

Since IX, IT and IY are perpendicular to AB, BC and AC, respectively, then the three

terms on the right side of (∗) are the areas of 4IAB, 4IBC and 4ICA, respectively,

and so their sum is |4ABC|, ie. sr = |4ABC|.

Thus, (TC)(r1) + (TB)(r2) =|4ABC|2

|4ABC|= |4ABC|, as required.

Solution 2

Join O to F and B and Q to C and G.

Since the circle with centre O is tangent to PB and AB at F and K, then OF is perpen-

dicular to PB and OB bisects ∠FBK.

Similarly, QG is perpendicular to CN and QC bisects ∠GCL.

Extend AB and AC through B and C, respectively, and construct the circle which is

tangent to AB extended, BC, and AC extended, and lies outside 4ABC. This circle is

called an excircle of 4ABC.

The centre of this excircle, which we label U , is on the angle bisector of the angle formed

by AB extended and BC, as the circle is tangent to these two lines, so U lies on OB

extended. Similarly, U lies on QC extended.


P N

M

A

B C

E

D

GF

KL

Q

O

U

V

XY

Let AB = c, AC = b, BC = a, let s be the semi-perimeter (that is, half of the perimeter)

of 4ABC, and let |4ABC| denote the area of 4ABC.

Then the radius of the excircle, which we will denote rA is equal to|4ABC|

s− a. (See the

end of this solution for a proof of this fact.)

Let V be the point where the excircle is tangent to BC.

Then UV is perpendicular to BC.

Thus, 4OFB is similar to 4UV B and 4QGC is similar to 4UV C (since they have

opposite angles which are equal and right angles).

This tells us thatOF

FB=

UV

V Bor

r1

FB=

rA

V Band

QG

GC=

UV

V Cor

r2

GC=

rA

V C.

Therefore, since FB = BK and CG = CL and KB + BC + CL = s by (a), then

(V B)(r1)+(V C)(r2) = rA(FB+CG) = rA(KB+LC) = rA(s−BC) = rA(s−a) = |4ABC|

Now suppose that the excircle is tangent to AB extended and AC extended at X and Y ,

respectively.

Then AX = AY , and AX = AB +BX = AB +BV and AY = AC +CY = AC +CV (by

equal tangents from B and C), so AX+AY = AB+AC+BV +V C = AB+AC+BC = 2s,

ie. AX = AY = s.

Thus, V B = BX = AX − AB = s− c and similarly V C = s− b.

But TB = s− b = V C and TC = s− c = V B (see Solution 1), so

|4ABC| = (TC)(r1) + (TB)(r2)

as required.

(Why is rA =|4ABC|

s− a?


Join X, Y and V to U . Note that ∠AXU = ∠AY U = ∠BV U = ∠CV U = 90◦.

Then AXUY is a shape of the same type as in Problem 3, so its area is equal to

AX · UX = srA.

Similarly, the areas of BV UX and CV UY are rA(s− c) and rA(s− b).

Thus,

|4ABC| = Area of AXUY −Area of BV UX−Area of CV UY = rA(s− (s− c)− (s− b))

But, s − (s − c) − (s − b) = b + c − s = a + b + c − a − s = 2s − a − s = s − a, so

|4ABC| = rA(s− a), which is what we wanted to show.)

Solution 3

Let AB = c, AC = b, BC = a, and let s denote the semi-perimeter of 4ABC (that is,

half of its perimeter).

Then by (a), AK + AL = KB + BC + LC = s.

Since PM and PN are tangent to both circles, then the line through O and Q passes

through P .

Join O to D, F and K, and Q to L, E and G.

P N

M

A

B C

E

D

GF

K

L

Q

O

In each case, the centre of a circle is being joined to a point where the circle is tangent to

a line, so creates a right angle.

Therefore, 4POF is similar to 4PQG, soPF

OF=

PG

QGor r1(PG) = r2(PF ).

Each of the shapes PDOF , ADOK, BFOK, AEQL, CGQL and PEQG has two right

angles and two pairs of equal sides (ie. each is a shape as in Problem 3).

The area of each of these shapes is the product of the lengths of two of the sides which

meet at a right angle.

We use |PEQG| to denote the area of the shape PEQG, and so on.


Therefore,

|PEQG| = |4ABC|+ |PDOF |+ |ADOK|+ |BFOK|+ |AEQL|+ |CGQL|

(PG)(QG) = |4ABC|+ (PF )(OF ) + (AK)(OK) + (KB)(OK) + (AL)(QL) + (CL)(LQ)

|4ABC| = r2(PG)− r2(AL)− r2(CL)− r1(PF )− r1(AK)− r1(KB)

= r2(PG− AL− CL) + r1(−PF − AK −KB)

= r2(PG− CG− AL) + r1(−PF − AB) (equal tangents)

= r2(PC − AL) + r1(−PF − AB)

= r2(PF + FB + BC − AL) + r1(−PF − AB)

= r2(PF + BK + a− (s− AK)) + r1(−PF − AB)

(since AK + AL = s, and BK = FB by equal tangents)

= r2(PF + AK + BK + a− s) + r1(−PF − AB)

= r2(PF + AB + a− s) + r1(−PF − AB)

= r2(PF ) + r2(c + a− s) + r1(−PF − AB)

= r1(PG) + r2(a + b + c− b− s) + r1(−PF − AB) (since r1(PG) = r2(PF ))

= r2(2s− b− s) + r1(PG− PF − AB)

= r2(s− b) + r1(GF − AB)

= r2(s− b) + r1(FB + BC + CG− c)

= r2(s− b) + r1(KB + BC + CL− c) (equal tangents)

= r2(s− b) + r1(s− c)

As in Solution 1, TB = s− c and TC = s− b.

Therefore, |4ABC| = (TC)(r1) + (TB)(r2), as required.

Solution 4

Let ∠ABC = 2β, ∠ACB = 2γ and ∠MPN = 2θ.

Then ∠PAB = 2β − 2θ and ∠MAC = 2γ + 2θ, using external angles in 4PAB and

4PAC. Also, ∠ABP = 180◦ − 2β.

Since the circle with centre O is tangent to AP and AK, then O lies on the bisector of

∠PAK, so ∠KAO = β − θ. Similarly, ∠LAQ = γ + θ and ∠KBO = 90◦ − β.

Since 4OKB is right-angled at K (since AB is tangent to the circle with centre O at K),

then ∠KOB = β.

Thus, tan(∠KAO) = tan(β − θ) =KO

AKand tan(∠KOB) = tan(β) =

KB

KO.


Therefore,

AB = AK + KB

AB =KO

tan(β − θ)+ KO tan(β)

AB = r1

[1 + tan(β) tan(θ)

tan(β)− tan(θ)+ tan(β)

](since KO = r1)

AB = r1

[1 + tan(β) tan(θ)

tan(β)− tan(θ)+

tan2(β)− tan(β) tan(θ)

tan(β)− tan(θ)

]AB = r1

[1 + tan2(β)

tan(β)− tan(θ)

]r1 =

AB(tan(β)− tan(θ))

1 + tan2(β)

r1 =AB(tan(β)− tan(θ))

sec2(β)

r1 = AB sin(β) cos(β)− AB cos2(β) tan(θ)

r1 = 12AB sin(2β)− AB cos2(β) tan(θ)

But AB sin(2β) is the length of the height, h, of 4ABC from A to BC.

Thus r1 = 12h− AB cos2(β) tan(θ).

Similarly, r2 = 12h + AC cos2(γ) tan(θ).

Since the circle with centre I is tangent to AB and BC, then I lies on the angle bi-

sector of ∠ABC, so ∠IBT = β, so tan(β) =IT

TB.

Thus, TB =IT

tan(β)=

r

tan(β).

Similarly, TC =r

tan(γ).

Therefore,

r1(TC) + r2(TB) = TC[

12h− AB cos2(β) tan(θ)

]+ TB

[12h + AC cos2(γ) tan(θ)

]= 1

2h(TC + TB) + tan(θ) [−TC · AB cos2(β) + TB · AC cos2(γ)] (∗)

The first term on the right side of (∗) equals 12h(BC) which equals the area of 4ABC.

Considering the second factor of the second term, we obtain

TB · AC cos2(γ)− TC · AB cos2(β)

=r

tan(β)AC cos2(γ)− r

tan(γ)AB cos2(β)

=r

2 tan(β) tan(γ)

(2AC tan(γ) cos2(γ)− 2AB tan(β) cos2(β)

)=

r

2 tan(β) tan(γ)(2AC sin(γ) cos(γ)− 2AB sin(β) cos(β))

=r

2 tan(β) tan(γ)(AC sin(2γ)− AB sin(2β))


But AC sin(2γ) = AB sin(2β) = h, so this second factor equals 0, so the second term of

the right side of (∗) equals 0.

Therefore, r1(TC) + r2(TB) equals the area of 4ABC, as required.



presents the

Canadian Open



Supported by:

Solutions


2006 COMC Solutions Page 2

Part A

1. What is the value of(1 + 1

2

) (1 + 1

3

) (1 + 1

4

) (1 + 1

5

)?

Solution 1(1 + 1

2

) (1 + 1

3

) (1 + 1

4

) (1 + 1

5

)=

(32

) (43

) (54

) (65

)=

(�32

)(�4�3

)(�5�4

)(6

�5

)(simplifying numerators and denominators)

= 62

= 3

Solution 2 (1 + 1

2

) (1 + 1

3

) (1 + 1

4

) (1 + 1

5

)=

(32

) (43

) (54

) (65

)= 360

120

= 3

2. If f(2x + 1) = (x− 12)(x + 13), what is the value of f(31)?

Solution 1

Since f(2x + 1) = (x− 12)(x + 13), then

f(31) = f(2(15) + 1) = (15− 12)(15 + 13) = 3(28) = 84

Solution 2

If w = 2x + 1, then x =w − 1

2.

Since f(2x + 1) = (x− 12)(x + 13), then

f(w) =

(w − 1

2− 12

)(w − 1

2+ 13

)=

(w − 25

2

)(w + 25

2

)Therefore,

f(31) =

(31− 25

2

)(31 + 25

2

)= 3(28) = 84


3. In 4ABC, M is the midpoint of BC, as shown. If

∠ABM = 15◦ and ∠AMC = 30◦, what is the size of

∠BCA?

MB

A

C

Solution

Since ∠AMC = 30◦, then ∠AMB = 180◦ − ∠AMC = 150◦.

Since ∠ABM = 15◦ and ∠AMB = 150◦, then ∠BAM = 180◦ − ∠ABM − ∠AMB = 15◦.

Since ∠ABM = ∠BAM , then BM = MA.

Since BM = MA and BM = MC, then MA = MC, so ∠MAC = ∠MCA.

Thus, ∠MCA = 12(180◦ − ∠AMC) = 75◦.

Therefore, ∠BCA = ∠MCA = 75◦.

4. Determine all solutions (x, y) to the system of equations

4

x+

5

y2= 12

3

x+

7

y2= 22

Solution 1

Subtracting 5 times the second equation from 7 times the first equation, we obtain

7

(4

x+

5

y2

)− 5

(3

x+

7

y2

)= 7(12)− 5(22)

13

x= −26

x = −12

Substituting x = −12

into the first equation, we obtain4

−12

+5

y2= 12 or −8 +

5

y2= 12 or

5

y2= 20 or y2 = 1

4.

Therefore, y = ±12.

Thus, the solutions are(−1

2, 1

2

)and

(−1

2,−1

2

).

(We can check by substitution that both of these solutions work.)


Solution 2

Subtracting 4 times the second equation from 3 times the first equation, we obtain

3

(4

x+

5

y2

)− 4

(3

x+

7

y2

)= 3(12)− 4(22)

−13

y2= −52

y2 = 14

y = ±12

Substituting y = ±12

into the first equation, we obtain4

x+

5(±1

2

)2 = 12 or4

x+ 20 = 12 or

4

x= −8 or x = −1

2.

Thus, the solutions are(−1

2, 1

2

)and

(−1

2,−1

2

).

(We can check by substitution that both of these solutions work.)

5. In 4ABC, BC = 4, AB = x, AC = x + 2, and cos(∠BAC) =x + 8

2x + 4.

Determine all possible values of x.

Solution

Using the cosine law in 4ABC,

BC2 = AB2 + AC2 − 2(AB)(AC) cos(∠BAC)

42 = x2 + (x + 2)2 − 2x(x + 2)x + 8

2x + 4

16 = x2 + x2 + 4x + 4− x(x + 8)

0 = x2 − 4x− 12

0 = (x− 6)(x + 2)

Therefore, x = 6 or x = −2.

Since AB = x, then x must be positive, so x = 6.

6. Determine the number of integers n that satisfy all three of the conditions below:

• each digit of n is either 1 or 0,

• n is divisible by 6, and

• 0 < n < 107.

Solution 1

Since 0 < n < 107, then n is a positive integer with fewer than 8 digits.


Since n is divisible by 6, then n is even. Since each digit of n is either 1 or 0, then n must end

with a 0.

Since n is divisible by 6, then n is divisible by 3, so n has the sum of its digits divisible by 3.

Since each digit of n is 0 or 1 and n has at most 6 non-zero digits, then the sum of the digits

of n must be 3 or 6 (that is, n contains either 3 or 6 digits equal to 1).

Since n has at most 7 digits, we can write n in terms of its digits as abcdef0, where each

of a, b, c, d, e, f can be 0 or 1. (We allow n to begin with a 0 in this representation.)

If n contains 6 digits equal to 1, then there is no choice in where the 1’s are placed so

n = 1111110.

If n contains 3 digits equal to 1, then 3 of the 6 digits a through f are 1 (and the other 3 are

0). The number of such possibilities is

(6

3

)= 20.

Therefore, there are 20 + 1 = 21 such integers n.

Solution 2

Since 0 < n < 107, then n is a positive integer with fewer than 8 digits.

Since n is divisible by 6, then n is even. Since each digit of n is either 1 or 0, then n ends with

a 0.

Since n is divisible by 6, then n is divisible by 3, so has the sum of its digits divisible by 3.

Since each digit of n is 0 or 1 and n has at most 6 non-zero digits, then the sum of the digits

of n must be 3 or 6 (that is, n contains either 3 or 6 digits equal to 1).

If n contains 6 digits equal to 1, then n = 1111110, since n has at most 7 digits.

If n contains 3 digits equal to 1, then n has between 4 and 7 digits, and must begin with 1.

If n has 4 digits, then n must be 1110.

If n has 5 digits, then n has the form 1abc0 with 2 of a, b, c equal to 1. There are

(3

2

)= 3

such possibilities.

If n has 6 digits, then n has the form 1abcd0 with 2 of a, b, c, d equal to 1. There are

(4

2

)= 6

such possibilities.

If n has 7 digits, then n has the form 1abcde0 with 2 of a, b, c, d, e equal to 1. There are(5

2

)= 10 such possibilities.

Therefore, there are 1 + 1 + 3 + 6 + 10 = 21 possibilities for n.


7. Suppose n and D are integers with n positive and 0 ≤ D ≤ 9.

Determine n ifn

810= 0.9D5 = 0.9D59D59D5 . . . .

Solution

First, we note that 0.9D5 =9D5

999since

1000(0.9D5) = 9D5.9D5

1000(0.9D5)− 0.9D5 = 9D5.9D5− 0.9D5

999(0.9D5) = 9D5

0.9D5 =9D5

999

Alternatively, we could derive this result by noticing that

0.9D5 = 0.9D59D59D5 . . .

=9D5

103+

9D5

106+

9D5

109+ · · ·

=

9D5

103

1− 1

103

(summing the infinite geometric series)

=9D5

1000− 1

=9D5

999

Therefore,

n

810=

9D5

999999n = 810(9D5)

111n = 90(9D5)

37n = 30(9D5)

Thus, 30(9D5) is divisible by 37. Since 30 is not divisible by 37 and 37 is prime, then 9D5

must be divisible by 37.

The multiples of 37 between 900 and 1000 are 925, 962 and 999.

Thus, 9D5 must be 925, so D = 2.

So 37n = 30(925) or n = 30(25) = 750.


8. What is the probability that 2 or more successive heads will occur at least once in 10 tosses of

a fair coin?

Solution 1

For a given toss, we use T to represent a result of tails and H for heads.

There are 210 = 1024 possible sequences of outcomes when a fair coin is tossed 10 times.

Let tn be the number of sequences of n tosses of a fair coin which do not contain 2 or more

successive heads.

(So the number of sequences of length 10 that contain 2 or more successive heads is 1024− t10

which means that the desired probability is1024− t10

1024.)

Note that t1 = 2, as both T and H do not contain 2 successive H’s.

Also, t2 = 3. (These are the sequences TT , TH, HT .)

Consider a sequence of n tosses of a fair coin which do not contain 2 or more successive heads,

where n ≥ 3.

Such a sequence must begin with H or T .

If the sequence begins with H, the second toss must be T , and the last n − 2 can be any

sequence of n− 2 tosses that does not contain 2 or more successive heads. There are tn−2 such

sequences of n − 2 tosses, so there are tn−2 sequences of n tosses beginning with H and not

containing 2 or more successive heads.

If the sequence begins with T , the last n − 1 tosses can be any sequence of n − 1 tosses that

does not contain 2 or more successive heads. There are tn−1 such sequences of n− 1 tosses, so

there are tn−1 sequences of n tosses beginning with T and not containing 2 or more successive

heads.

Therefore, tn = tn−1 + tn−2, since each sequence begins with H or T .

Starting with t1 = 2 and t2 = 3, we can then generate the sequence tn for n = 1 to n = 10 by

adding the two previous terms to obtain the next term:

2, 3, 5, 8, 13, 21, 34, 55, 89, 144

So t10 = 144, so the desired probability is1024− 144

1024=

880

1024=

55

64.

Solution 2

For a given toss, we use T to represent a result of tails and H for heads.

There are 210 = 1024 possible sequences of outcomes when a fair coin is tossed 10 times.

Let us count the number of such sequences which do not contain 2 or more successive H’s, by

grouping them by the number of H’s that they contain. (Note that not containing 2 or more

successive H’s is equivalent to not containing the pair HH.)


If a sequence of length 10 consists of 0 H and 10 T ’s, there is only 1 possibility.

If a sequence of length 10 consists of 1 H and 9 T ’s, there are

(10

1

)= 10 possibilities.

If a sequence of length 10 consists of 2 H’s and 8 T ’s, then we start with

T T T T T T T T

Each of the two H’s must be placed in separate spaces. We can then eliminate any unused

spaces to obtain a sequence of 8 T ’s and 2 H’s containing no consecutive H’s (and we get all

such sequences this way). There are

(9

2

)= 36 ways of positioning the H’s, and so 36 such

sequences.

In a similar way, with 3 H’s and 7 T ’s, there are

(8

3

)= 56 such sequences.

With 4 H’s and 6 T ’s, there are

(7

4


With 5 H’s and 5 T ’s, there are

(6

5


Therefore, there is a total of 1 + 10 + 36 + 56 + 35 + 6 = 144 sequences of 10 tosses which do

not contain 2 or more successive H’s.

Thus, there are 1024− 144 = 880 sequences of 10 tosses which do contain 2 or more successive

H’s. Therefore, the probability that a given sequence contains 2 or more successive H’s is8801024

= 5564

.


Part B

1. Piotr places numbers on a 3 by 3 grid using the following rule, called “Piotr’s Principle”:

For any three adjacent numbers in a horizontal, vertical or diagonal line, the

middle number is always the average (mean) of its two neighbours.

(a) Using Piotr’s principle, determine the missing numbers

in the grid to the right. (You should fill in the missing

numbers in the grid in your answer booklet.)

3 19

8

Solution

Since the average of 3 and 19 is 12(3 + 19) = 11, then 11 goes between the 3 and 19.

The number which goes below 8 is the number whose average with 3 is 8, so 13 goes

below 8.

The average of 13 and 19, or 16, goes in the middle square.

The number which goes to the right of the 16 is the number whose average with 8 is 16,

or 24.

The number which goes below 24 is the number whose average with 19 is 24, or 29.

The number which goes between 13 and 29 is their average, which is 21.

Therefore, the completed grid is

3 11 19

8 16 24

13 21 29

.

(We can check that each line obeys Piotr’s Principle.)

Note

There are other orders in which the squares can be filled.

(b) Determine, with justification, the total of the nine num-

bers when the grid to the right is completed using Piotr’s

Principle.

x

5 23

Note

When we have the three numbers a, X, b on a line, then X is the average of a and b, so

X = 12(a + b).

When we have the three numbers a, b, X on a line, then b is the average of a and X, so

b = 12(a + X) or 2b = a + X or X = 2b− a.

These facts will be useful as we solve (b) and (c).


Solution 1

The average of 5 and 23 is 12(5+23) = 14, which goes in the square between the 5 and 23.

Since the average of the numbers above and below the 5 equals 5, then their sum is

2(5) = 10. (Note that we do not need to know the actual numbers, only their sum.)

Similarly, the sum of the numbers above and below the 14 is 2(14) = 28 and the sum of

the numbers above and below the 23 is 2(23) = 46.

Therefore, the sum of the numbers in the whole grid is 5 + 10 + 14 + 28 + 23 + 46 = 126.

Solution 2

The average of 5 and 23 is 12(5+23) = 14, which goes in the square between the 5 and 23.

Since the average of the x and the number below the 5 is 5, then the number below the 5

is 10− x.

Since the average of the x and the bottom right number is 14, then the bottom right

number is 28− x.

The average of 10− x and 28− x is 12(10− x + 28− x) = 19− x, which goes in the middle

square on the bottom row.

Since the average of 19 − x and the number above the 14 is 14, then the number above

the 14 is 2(14)− (19− x) = 9 + x.

Since the average of 28 − x and the number above the 23 is 23, then the number above

the 23 is 2(23)− (28− x) = 18 + x.

Thus, the completed grid is

x 9 + x 18 + x

5 14 23

10− x 19− x 28− x

and so the sum of the entries is

x + 9 + x + 18 + x + 5 + 14 + 23 + 10− x + 19− x + 28− x = 126.

(c) Determine, with justification, the values of x and y when

the grid to the right is completed using Piotr’s Principle.x 7

9 y

20

Solution

The centre square is the average of 9 and y and is also the average of x and 20.

Comparing these facts, 12(9 + y) = 1

2(x + 20) or 9 + y = x + 20 or x− y = −11.

The number in the top right corner gives an average of 7 when combined with x (so equals

2(7)− x = 14− x) and gives an average of y when combined with 20 (so equals 2y − 20).

Therefore, 14− x = 2y − 20 or x + 2y = 34.

Subtracting the first equation from the second, we obtain 3y = 45 or y = 15.

Substituting back into the first equation, we obtain x = 4.


We check by completing the grid. Starting with

4 7

9 15

20

gives, after some work,

4 7 10

9 12 15

14 17 20

, which does obey Piotr’s Principle.

Therefore, x = 4 and y = 15.

2. In the diagram, the circle x2+y2 = 25 intersects

the x-axis at points A and B. The line x = 11

intersects the x-axis at point C. Point P moves

along the line x = 11 above the x-axis and AP

intersects the circle at Q.

y

xA B

PQ

C

x = 11

(a) Determine the coordinates of P when 4AQB has maximum area. Justify your answer.

Solution

Since the circle has equation x2 + y2 = 25, then to find the coordinates of A and B, the

x-intercepts of the circle, we set y = 0 to obtain x2 = 25 or x = ±5. Therefore, A and B

have coordinates (−5, 0) and (5, 0), respectively.

Since 4AQB has a base AB of constant length and a variable height, then the area of

4AQB is maximized when the height of 4AQB is maximized (that is, when Q is furthest

from AB).

To maximize the height of 4AQB, we would like Q to have as large a y-coordinate as

possible. Thus, we would like Q to be at the “top” of the circle – that is, at the place

where the circle intersects the y-axis.

Since the circle has equation x2 + y2 = 25, then setting x = 0, we obtain y2 = 25 or

y = ±5, so Q has coordinates (0, 5) as Q lies above the x-axis.

Therefore, P lies on the line through A(−5, 0) and Q(0, 5). This line has slope 1 and

y-intercept 5, so has equation y = x + 5.

Since P has x-coordinate 11 and lies on the line with equation y = x + 5, then P has

coordinates (11, 16).


(b) Determine the coordinates of P when Q is the midpoint of AP . Justify your answer.

Solution

Suppose the coordinates of P are (11, p).

We will determine p so that the midpoint of PA lies on the circle. (This is equivalent to

finding P so that the point on the circle is the midpoint of P and A.)

Since A has coordinates (−5, 0), then for Q to be the midpoint of AP , Q must have coor-

dinates(

12(−5 + 11), 1

2(0 + p)

)= (3, 1

2p).

For (3, 12p) to lie on the circle,

32 +(

12p)2

= 25

14p2 = 16

p2 = 64

p = ±8

Since P must lie above the x-axis, then p = 8.

Therefore, P has coordinates (11, 8).

(c) Determine the coordinates of P when the area of 4AQB is 14

of the area of 4APC. Jus-

tify your answer.

Solution 1

Join Q to B.

y

xA B

PQ

C

x = 11

Since AB is a diameter of the circle, then ∠AQB = 90◦.

Thus, since 4AQB and 4ACP are both right-angled and share a common angle at A,

then 4AQB and 4ACP are similar.

Since the area of 4ACP is 4 times the area of 4AQB, then the sides of 4ACP are√4 = 2 times as long as the corresponding sides of 4AQB.

Since AB = 10, then AP = 2AB = 20.


We also know that AC = 16 (since C has coordinates (11, 0) and A has coordinates

(−5, 0)).

Therefore, by the Pythagorean Theorem, PC =√

AP 2 − AC2 =√

202 − 162 = 12.

Thus, P has coordinates (11, 12).

Solution 2

Let the coordinates of P be (11, p), and the coordinates of Q be (a, b). Thus, the height

of 4AQB is b.

The area of 4AQB is 12(AB)(b) = 5b since AB = 10.

The area of 4APC is 12(AC)(p) = 8p since AC = 16.

Since the area of 4AQB is 14

that of 4APC, then 5b = 2p or b = 25p.

This tells us that Q must be 25

of the way along from A to P .

Since A has x-coordinate −5 and P has x-coordinate 11, then Q has x-coordinate

−5 + 25(11− (−5)) = 7

5. Therefore, Q has coordinates

(75, 2

5p).

Since the circle has equation x2 + y2 = 25, then(75

)2+(

25p)2

= 25

4925

+ 425

p2 = 62525

4p2 = 576

p2 = 144

and so p = 12 since p > 0. Therefore, P has coordinates (11, 12).

3. (a) In the diagram, trapezoid ABCD has parallel

sides AB and DC of lengths 10 and 20 and sides

AD and BC of lengths 6 and 8. Determine the

area of trapezoid ABCD.

A B

CD

Solution 1

Extend DA and CB to meet at P .

A B

CD

P

Since AB is parallel to DC, then ∠PAB = ∠PDC and ∠PBA = ∠PCD.

Therefore, 4PAB is similar to 4PDC.


Since AB = 12DC, then the sides of 4PAB are 1

2the length of the corresponding sides of

4PDC.

Therefore, PA = AD = 6 and PB = BC = 8.

Thus, the sides of 4PDC have lengths 12, 16 and 20. Since 122 +162 = 202, then 4PDC

is right-angled at P by the Pythagorean Theorem.

Thus, the area of 4PDC is 12(12)(16) = 96.

Since 4PAB is right-angled at P , its area is 12(6)(8) = 24.

Therefore, the area of trapezoid ABCD is 96− 24 = 72.

Solution 2

Drop perpendiculars from A and B to P and Q on DC.

A B

CD P Q

Let AP = BQ = h and let DP = x.

Since AB = 10 and ABQP is a rectangle, then PQ = 10.

Since DC has length 20, then QC = 20− x− 10 = 10− x.

Using the Pythagorean Theorem in 4DPA, we obtain x2 + h2 = 62.

Using the Pythagorean Theorem in 4CQB, we obtain (10− x)2 + h2 = 82.

Subtracting the first of these equations from the second, we obtain 100 − 20x = 28 or

20x = 72 or x = 185.

Substituting back into the first equation, h2 = 36−(

185

)2= 576

25so h = 24

5.

Therefore, the area of trapezoid ABCD is 12

(245

)(10 + 20) = 72.

Solution 3

Drop perpendiculars from A and B to P and Q on DC.

A B

CD P Q

Cut out rectangle ABQP and join the two remaining pieces along the cut line.

The remaining shape is a triangle DCX with side lengths DX = 6, XC = 8 and DC =

20 − 10 = 10. Since 62 + 82 = 102, then 4DCX is right-angled by the Pythagorean

Theorem.

Since sin(∠XDC) =XC

DC=

8

10=

4

5, then the length of the altitude from X to DC is

XD sin(∠XDC) = 6(

45

)= 24

5


which is also the height of the original trapezoid.

(We could also determine the length of this altitude by calculating the area of 4XDC in

two ways: once as 12(DX)(XC) = 24 and once as 1

2h(DC) = 5h.)

Therefore, the area of the original trapezoid is 12

(245

)(10 + 20) = 72.

(Alternatively, the area of the original trapezoid is the sum of the areas of rectangle ABQP

(245× 10 = 48) and 4XDC (24), for a total of 72.)

Solution 4

Draw BX from B to X on DC so that BX is parallel to AD.

A B

CD X

Then ABXD is a parallelogram so BX = AD = 6 and DX = AB = 10.

Therefore, XC = DC −DX = 10.

Thus, 4BXC has sides of length 6, 8 and 10. Since 62 + 82 = 102, then 4BXC is right-

angled at B by the Pythagorean Theorem. The area of 4BXC is thus 12(6)(8) = 24.

Join B to D. BD divides the area of ABXD in half.

Also, BX divides the area of 4BDC in half, since it is a median.

Therefore, the areas of 4ABD, 4BDX and 4XBC are all equal.

So the area of trapezoid ABCD is 3(24) = 72.

Solution 5

Let X be the midpoint of DC. Join X to A and B.

Then AB = DX = XC = 10.

Since AB = DX and AB is parallel to DX, then AD and BX are parallel and equal, so

BX = 6.

Since AB = XC and AB is parallel to XC, then AX and BC are parallel and equal, so

AX = 6.

Therefore, the trapezoid is divided into three triangles, each of which has side lengths 6,

8 and 10.

A triangle with side lengths 6, 8 and 10 is right-angled (since 62 + 82 = 102), so has area12(6)(8) = 24.

Therefore, the area of the trapezoid is 3× 24 = 72.


(b) In the diagram, PQRS is a rectangle and

T is the midpoint of RS. The inscribed

circles of 4PTS and 4RTQ each have

radius 3. The inscribed circle of 4QPT

has radius 4. Determine the dimensions of

rectangle PQRS.

R ST

PQ

Solution 1

Let RT = a (so RS = 2a) and RQ = b.

Drop a perpendicular from T to Z on QP . By symmetry, Z is also the point of tangency

of the middle circle to QP .

Let O be the centre of the circle inscribed in 4QTP and C be the centre of the circle

inscribed in 4RTQ.

Let A, B and D be the points of tangency of the circle with centre C to QR, RT and QT ,

respectively.

Let Y be the point of tangency of the circle with centre O to QT .

Join C to A, B and D, and O to Y . These radii are perpendicular to QR, RT , QT , and

QT , respectively.

R ST

PQ Z

CA

B

DO

Y

We know that OY = OZ = 4.

Since TZ = b, then TO = b− 4.

Since QZ = a, then QY = a (equal tangents).

Since CA = CB = 3 and RACB is a rectangle (as it has three right angles), then RACB

is a square and RA = RB = 3.

Therefore, AQ = b− 3 and BT = a− 3.

By equal tangents, TD = BT = a− 3 and QD = QA = b− 3.

Now, TY = QT −QY = QD + TD −QY = (b− 3) + (a− 3)− a = b− 6.

Therefore, 4TOY is right-angled at Y with sides of length TO = b− 4, TY = b− 6 and

OY = 4.

By the Pythagorean Theorem, 42 + (b− 6)2 = (b− 4)2 or 4b = 36 or b = 9.

Therefore, TY = 9− 6 = 3 and tan(∠OTY ) =OY

TY=

4

3.

Also, 43

= tan(∠QTZ) =a

b.

Since b = 9, a = 12.


Therefore, the rectangle is 24 by 9.

Solution 2


Drop a perpendicular from T to Z on QP . By symmetry, Z is also the point of tangency

of the middle circle to QP , and QZ = a.

Since the incircle of 4QRT has radius 3, then so does the incircle of 4QZT .

Let O be the centre of the circle inscribed in 4QTP and C ′ the centre of the circle in-

scribed in 4QZT .

Since C ′ and O both lie on the angle bisector of ∠TQP , then tan(∠C ′QP ) = tan(∠OQP ).

Since C ′ is 3 units from the line TZ, then tan(∠C ′QP ) =3

a− 3and tan(∠OQP ) =

4

aso

3

a− 3=

4

aor 3a = 4a− 12 or a = 12.

We can calculate b = 9 as in Solution 1, to obtain that the rectangle is 24 by 9.

Solution 3


We calculate the areas of each of 4QRT and 4QTP in two ways: once using the stan-

dard 12bh formula and once using the less well-known Area = sr formula, where s is the

semi-perimeter of the triangle (that is, half of the perimeter) and r is the inradius (that

is, the radius of the inscribed circle).

In 4QRT , RT = a, RQ = b, QT =√

a2 + b2 and the inradius is 3, so

12ab = 1

2(a + b +

√a2 + b2)(3)

In 4QTP , QT = TP =√

a2 + b2, QP = 2a, the height is b, and the inradius is 4, so

12(2a)b = 1

2(2√

a2 + b2 + 2a)(4)

Simplifying these equations, we get the system of equations

ab = 3(a + b) + 3√

a2 + b2

ab = 4a + 4√

a2 + b2

Eliminating the ab terms, we obtain 3b− a =√

a2 + b2.

Squaring both sides, we obtain 9b2 − 6ab + a2 = a2 + b2 or 8b2 − 6ab = 0.

Since b 6= 0,a

b=

4

3.


Substituting a = 43b into the first equation yields

43b2 = 4b + 3b + 3

√169b2 + b2

43b2 = 7b +

√25b2

43b2 = 7b + 5b (since b > 0)

4b2 = 36b

b = 9

since b 6= 0. Therefore, a = 12 and the rectangle is 24 by 9.

Solution 3

We use the notation and diagram from Solution 1.

Since RS and QP are parallel, then ∠BTD = ∠Y QZ.

Since C and O are the centres of inscribed circles, then C lies on the angle bisector of

∠BTD and O lies on the angle bisector of ∠Y QZ.

Therefore, ∠BTC = 12∠BTD = 1

2∠Y QZ = ∠OQZ.

Therefore, 4BTC and 4ZQO are similar, as each is right-angled.

Thus, BT : QZ = BC : OZ = 3 : 4.

Suppose that BT = 3x and QZ = 4x.

Then RT = RB + BT = AC + 3x = 3 + 3x, since RBCA is a square.

But RT = QZ so 4x = 3 + 3x or x = 3.

Let QA = y.

Then QT = QD + DT = QA + BT by equal tangents, so QT = y + 3x = y + 9.

Since 4QRT is right-angled, then

QR2 + RT 2 = QT 2

(y + 3)2 + 122 = (y + 9)2

y2 + 6y + 9 + 144 = y2 + 18y + 81

12y = 72

y = 6

Therefore, since RT = 3 + 3x, then RS = 24 and RQ = 3 + y = 9, so the rectangle is 24

by 9.


4. (a) Determine, with justification, the fractionp

q, where p and q are positive integers and

q < 100, that is closest to, but not equal to, 37.

Solution

We would like to find positive integers p and q with q < 100 which minimizes∣∣∣∣pq − 3

7

∣∣∣∣ =

∣∣∣∣7p− 3q

7q

∣∣∣∣ =|7p− 3q|

7q

To minimize such a fraction, we would like to make the numerator small while making the

denominator large.

Since the two fractions pq

and 37

are not equal, the numerator of their difference cannot be

0. Since the numerator is a positive integer, its minimum possible value is 1.

We consider the largest possible values of q (starting with 99) and determine if 7p − 3q

can possibly be equal to 1 or −1.

If q = 99, 7p − 3q = 7p − 297, which cannot equal ±1 since the nearest multiple of 7

to 297 is 294.

If q = 98, 7p − 3q = 7p − 294, which cannot equal ±1 since 7p − 294 is always divisible

by 7.

If q = 97, 7p − 3q = 7p − 291, which cannot equal ±1 since the nearest multiple of 7 to

291 is 294.

If q = 96, 7p− 3q = 7p− 288, which equals −1 if p = 41.

If p = 41 and q = 96, the difference between the fractions3

7and

p

qis

1

7(96).

If q > 96, the numerator of|7p− 3q|

7qis always at least 2, so the difference is at least

2

7(99)>

1

7(96).

If q < 96, the difference betweenp

qand

3

7is at least

1

7(95)>

1

7(96).

So 4196

minimizes the difference, so it is the closest fraction to 37

under the given conditions.


(b) The baseball sum of two rational numbersa

band

c

dis defined to be

a + c

b + d. (A rational

number is a fraction whose numerator and denominator are both integers and whose de-

nominator is not equal to 0.) Starting with the rational numbers 01

and 11

as Stage 0, the

baseball sum of each consecutive pair of rational numbers in a stage is inserted between

the pair to arrive at the next stage. The first few stages of this process are shown below:

STAGE 0: 01

11

STAGE 1: 01

12

11

STAGE 2: 01

13

12

23

11

STAGE 3: 01

14

13

25

12

35

23

34

11

Prove that

(i) no rational number will be inserted more than once,

Solution

Consider two rational numbersa

band

c

dwhich occur next to each other at a given

stage witha

b<

c

d.

Note that this means that ad < bc or bc− ad > 0.

The rational number that will be inserted between them at the next stage isa + c

b + d.

Nowa

b<

a + c

b + d⇔ a(b + d) < b(a + c) ⇔ 0 < bc− ad

which we know to be true, and

a + c

b + d<

c

d⇔ d(a + c) < c(b + d) ⇔ 0 < bc− ad

which is again true.

Therefore,a

b<

a + c

b + d<

c

d.

This tells us that every rational number which is inserted at any given stage is strictly

between the two rational numbers on either side.

Therefore, once a given rational number is inserted, every other rational number which

is inserted must be either strictly larger or strictly smaller, as the list at each stage

must be strictly increasing.

Therefore, no rational number will be inserted more than once.


(ii) no inserted fraction is reducible,

Solution

First, we prove a lemma.

Lemma

Ifa

b<

c

dare consecutive rational numbers in a given stage, then bc− ad = 1.

Proof

At Stage 0, the two fractions obey this property.

Assume that the property holds for all fractions in Stage k.

Consider two consecutive fractionsa

b<

c

dat Stage k.

The fraction that will be inserted betweena

band

c

dat Stage k + 1 is

a + c

b + d,

givinga

b<

a + c

b + d<

c

d.

Note that b(a + c)− a(b + d) = bc− ad = 1 and c(b + d)− d(a + c) = bc− ad = 1.

This tells us that each pair of consecutive fraction at Stage k + 1 obeys this property.

Therefore, by induction, the required property holds.

Suppose then that a fractionkp

kq(k, p, q ∈ Z+) is inserted between

a

band

c

d.

By the Lemma, we must have b(kp)− a(kq) = 1 or k(bp− aq) = 1.

Therefore, k divides 1, so k = 1.

Thus, any inserted fraction can only have a common factor of 1 between its numerator

and denominator, so is irreducible. Thus, no inserted fraction is reducible.

(iii) every rational number between 0 and 1 will be inserted in the pattern at some stage.

Solution 1

We note first that every rational number of the forms1

nand

n− 1

nfor n ≥ 2 do enter

the pattern as the first and last new entry in Stage n − 1. (These rational numbers

enter between0

1and

1

n− 1, and

n− 2

n− 1and

1

1, respectively.)

Assume that there are rational numbers between 0 and 1 which are not inserted

in the pattern at the some stage.

Suppose thatp

qwith p, q ∈ Z+ and gcd(p, q) = 1 is such a rational number with

minimal denominator. (Note that all irreducible fractions with denominators 1, 2, 3

are inserted already.)


Since gcd(p, q) = 1, the Diophantine equation py − qx = 1 has solutions.

In fact, this Diophantine equation has a unique solution with 0 ≤ y < q. We can say

further that 0 < y < q since q > 1. (If y = 0, we have −qx = 1 so q would have to

be 1.)

Since 0 < y < q and qx = py − 1, then 0 < qx < pq so 0 < x < p as well.

Suppose that (x, y) = (a, b) is this unique solution.

Note that pb− qa = 1 so pb > qa soa

b<

p

q.

Consider the fractionp− a

q − b.

Its numerator and denominator are each a positive integer since 0 < a < p and

0 < b < q. Also, note that b(p− a)− a(q − b) = bp− aq = 1.

Among other things, this tells us thata

b<

p− a

q − b.

If we can prove thata

band

p− a

q − bare consecutive at some stage, then

p

qwill be

inserted between them at the next stage.

Since each ofa

band

p− a

q − bhas a denominator less than q, it appears in the pat-

tern asp

qhas the smallest denominator among those fractions which do not appear.

Note next thata

bcannot be

0

1.

(If it was, a = 0 so bp− aq = 1 gives bp = 1 so p = 1.

We know that every fraction with p = 1 enters the pattern, so p 6= 1.)

Also,p− a

q − bcannot be

1

1.

(If it was, then p − a = q − b so b(p − a) − a(q − b) = 1 gives (b − a)(q − b) = 1 so

b− a = 1.

Since p− a = q − b then q − p = b− a = 1.

But every fraction of the formn− 1

nenters the pattern, so q − p 6= 1.)

This tells us that each ofa

band

p− a

q − bactually entered the pattern at some stage.

There are now three cases: b < q − b, b > q − b and b = q − b.

• Assume that b < q − b.

Consider the point when the fractionp− a

q − bwas inserted into the pattern.

Suppose thatp− a

q − bwas inserted immediately between

m

n<

M

N.


(That is, m + M = p− a and n + N = q − b).

Note that 0 < n < q− b. (n and N cannot be 0 since denominators cannot be 0.)

Sincem

n<

p− a

q − bare consecutive fractions at this stage, we must also have

(p− a)n− (q − b)m = 1.

But since gcd(p−a, q− b) = 1, the Diophantine equation (p−a)Y − (q− b)X = 1

must have a unique solution with 0 < Y < q − b.

But (X, Y ) = (a, b) is such a solution since 0 < b < q−b and b(p−a)−a(q−b) = 1

(from above).

Therefore,a

bmust be the fraction immediately to the left of

p− a

q − bat the stage

wherep− a

q − benters, which means that

p

qwill be inserted into the pattern, contra-

dicting our assumption.

• Assume that q − b < b.

Consider the point when the fractiona

bwas inserted into the pattern.

Suppose thata

bwas inserted immediately between

m

n<

M

N(that is, m + M = a

and n + N = b).

Note that 0 < n < b. (n and N cannot be 0 since denominators cannot be 0.)

Sincem

n<

a

bare consecutive fractions at this stage, we must also have

an− bm = 1.

But since gcd(a, b) = 1, the Diophantine equation aY − bX = 1 must have a

unique solution with 0 < Y < b.

But (X,Y ) = (p − a, q − b) is such a solution since 0 < q − b < b and

b(p− a)− a(q − b) = 1 (from above).

Therefore,p− a

q − bmust be the fraction immediately to the left of

a

bat the stage

wherea

benters, which means that

p

qwill be inserted into the pattern, contradict-

ing our assumption.

• Assume that q − b = b.

In this case, q = 2b.

But bp− aq = 1 so b(p− 2a) = 1, and so b = 1 giving q = 2.

But we know that every irreducible fraction with denominator 2 does enter.

(Namely, the fraction 12.)

So this case cannot occur.


Thus, every rational number between 0 and 1 will be inserted in the pattern at some

stage.

Solution 2

Supposea

band

c

dare consecutive (irreducible) fractions at some stage.

Define S(a

b,c

d

)= a + b + c + d, the sum of the numerators and denominators of the

consecutive fractions.

We consider the minimum value of S at a given stage.

Whena + c

b + dis inserted with

a

b<

a + c

b + d<

c

d, the two new sums are

S

(a

b,a + c

b + d

)= a + b + a + c + b + d = 2a + 2b + c + d

and

S

(a + c

b + d,c

d

)= a + c + b + d + c + d = a + b + 2c + 2d

each of which is larger than S(a

b,c

d

).

So the minimum value of these sums must increase from one stage to the next.

Suppose that the fractiona

bbetween 0 and 1 with a, b ∈ Z+ and gcd(a, b) = 1 is

never inserted into the pattern.

At any given Stage, since the fractiona

bdoes not occur, it must be strictly between

two consecutive fractions, saym1

n1

<a

b<

m2

n2

. (Sincea

bnever occurs, we must be able

to find such a pair of fractions at every Stage.)

(We know that m2n1 − n2m1 = 1 from (b).)

Thus, m1b < n1a and n2a < m2b. Since each of these quantities is a positive integer,

n1a−m1b ≥ 1 and m2b− n2a ≥ 1.

Now

m2 + n2 + m1 + n1

= (m2 + n2)(1) + (m1 + n1)(1)

≤ (m2 + n2)(n1a−m1b) + (m1 + n1)(m2b− n2a)

= m2n1a + n1n2a−m1m2b−m1n2b + m1m2b + m2n1b−m1n2a− n1n2a

= a(m2n1 −m1n2) + b(m2n1 −m1n2)

= a + b

But for a fixed fractiona

b, a + b is fixed and the minimum possible value of

m2 + n2 + m1 + n1


increases from one stage to the next, so eventually a + b < m2 + n2 + m1 + n1, a

contradiction, since once this happens we cannot find two fractions in that Stage

between which to puta

b.

That is, there will be a stage beyond which we cannot find two consecutive fractions

witha

bbetween them. This means that

a

bmust actually occur in the pattern.

Therefore, every rational number between 0 and 1 will be inserted in the pattern at

some stage.



presents the

Sun Life FinancialCanadian Open Mathematics Challenge


Solutions



Part A

1. Solution 1

If a = 15 and b = −9, then

a2 + 2ab + b2 = (a + b)2 = (15 + (−9))2 = 62 = 36

Solution 2

If a = 15 and b = −9, then

a2 + 2ab + b2 = 152 + 2(15)(−9) + (−9)2 = 225− 270 + 81 = 36

Answer: 36

2. Since there are 60 seconds in a minute, the wind power generator turns30

60=

1

2of a revolution

each second.

Since a full revolution is 360◦, then the generator turns 12(360◦) = 180◦ each second.

(Alternatively, the generator turns through 30×360◦ in one minute, so through 30×360◦÷60 =

180◦ in one second.)

Answer: 180

3. Since AD = 4 and AD is perpendicular to the x-axis, then A has y-coordinate 4.

Suppose that the coordinates of A are (a, 4). (This tells us also that D has coordinates (a, 0).)

Since A lies on the line y = x + 10, then 4 = a + 10, or a = −6.

Therefore, A has coordinates (−6, 4) and D has coordinates (−6, 0).

Since ABCD is a rectangle, then AB is parallel to the x-axis, so B has y-coordinate 4.

Suppose that the coordinates of B are (b, 4). (This tells us also that C has coordinates (b, 0)

since BC is perpendicular to the x-axis.)

Since B lies on the line y = −2x + 10, then 4 = −2b + 10 so 2b = 6 or b = 3.

Therefore, B has coordinates (3, 4) and C has coordinates (3, 0).

Now the height of rectangle ABCD equals the length of AD, so is 4.

The width of rectangle ABCD equals the length of CD, which is 3− (−6) = 9.

Therefore, the area of rectangle is 9× 4 = 36.

Answer: 36

4. Solution 1

Suppose that there were 3k boys and 2k girls in the school in June, for some positive integer k.


In September, there were thus 3k− 80 boys and 2k− 20 girls in the school. Since the new ratio

is 7 : 5, then

3k − 80

2k − 20=

7

55(3k − 80) = 7(2k − 20)

15k − 400 = 14k − 140

k = 260

Therefore, the total number of the students in the school in June was 3k + 2k = 5k = 5(260),

or 1300 students.

Solution 2

Suppose that there were b boys and g girls in the school in June.

In September, there were thus b− 80 boys and g − 20 girls in the school.

From the given information, we know thatb

g=

3

2and

b− 80

g − 20=

7

5.

Eliminating fractions gives the equations 2b = 3g and 5(b−80) = 7(g−20) or 5b−400 = 7g−140

or 5b− 7g = 260.

Multiplying the second equation by 2 gives 10b− 14g = 520, and substituting 10b = 15g gives

g = 520.

Therefore, b = 32(520) = 780, so there were b + g = 780 + 520 = 1300 students in the school in

June.

Answer: 1300

5. Solution 1

When the nine numbers are placed in the array in any arrangement, the sum of the row sums

is always 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45, because each of the nine numbers appears in

exactly one row.

Similarly, the sum of the column sums is also always 45, as again each of the nine numbers

appears in exactly one column.

Therefore, the grand sum S equals 90 plus the sum of the diagonal sums, and so depends only

on the diagonal entries, labelled in the array below:

a c

e

g k

So S = 90 + (a + e + k) + (c + e + g) = 90 + 2e + a + c + g + k. To make S as large as possible,

we must make 2e + a + c + g + k as large as possible.


Since a, c, e, g, k can be any of the numbers from 1 to 9, then S is largest when e = 9 and

a, c, g, k are 5, 6, 7, 8 in some order, for example in the configuration below:

5 1 6

2 9 3

7 4 8

Therefore, the maximum possible value of S is 90 + 2(9) + 8 + 7 + 6 + 5 = 90 + 44 = 134.

Solution 2

Suppose that a, b, c, d, e, f, g, h, k represent the numbers 1 to 9 in some order, and are entered

in the array as shown:

a b c

d e f

g h k

The grand sum is thus

S = (a + b + c) + (d + e + f) + (g + h + k) + (a + d + g) + (b + e + h) + (c + f + k) +

(a + e + k) + (c + e + g)

= 4e + 3a + 3c + 3g + 3k + 2b + 2d + 2f + 2h

To make S as large as possible, we should assign the values of a, b, c, d, e, f, g, h, k so that the

largest values go to the variables with the largest coefficients in the expression for S.

In other words, S will be maximized when e = 9, a, c, g, k equal 8, 7, 6, 5 in some order, and

b, d, f, h equal 4, 3, 2, 1 in some order.

Therefore, the maximum possible value of S is

S = 4(9) + 3(8 + 7 + 6 + 5) + 2(4 + 3 + 2 + 1) = 36 + 3(36) + 2(10) = 134

Answer: 134

6. Suppose that r is the radius of the circle.

Join O to P and drop a perpendicular from P to Q on OA.

P

NA

O

Q


Since OA and PN are perpendicular to AN and PQ is perpendicular to OA, then QPNA is

a rectangle. Therefore, QP = AN = 15 and QA = PN = 9.

Since O is the centre of the circle and A and P are on the circumference, then OA = OP = r.

Since OA = r and QA = 9, then OQ = r − 9.

Since 4OQP is right-angled at Q, then, by the Pythagorean Theorem,

OP 2 = OQ2 + QP 2

r2 = (r − 9)2 + 152

r2 = r2 − 18r + 81 + 225

18r = 306

r = 17

Therefore, the radius is 17.

Answer: 17

7. From the second equation, x + y = 7− z, so after squaring both sides, we obtain

x2 + 2xy + y2 = 49− 14z + z2 (∗)

From the third equation, x2 + y2 = 133− z2, so using this and the first equation to substitute

into (∗), we get

(133− z2) + 2(z2) = 49− 14z + z2

14z = −84

z = −6

Substituting this value for z back into the first two equations, we get xy = (−6)2 = 36 and

x + y = 7− (−6) = 13.

Therefore, y = 13− x and so x(13− x) = 36 or 0 = x2 − 13x + 36.

This tells us that 0 = (x− 4)(x− 9) so x = 4 or x = 9.

If x = 4, then y = 13− x tells us that y = 9.

If x = 9, then y = 13− x tells us that y = 4.

Therefore, the solutions are (x, y, z) = (4, 9,−6), (9, 4,−6).

Answer: (4, 9,−6), (9, 4,−6)

8. In order to travel from A to B along the segments without travelling along any segment more

than once, we must always move up, down or to the right. (In other words, we can never

travel to the left without retracing our steps.) To see this, we note that if we do travel along


a segment to the left, then we must have travelled along the other horizontal segment in this

square to the right at an earlier stage. We would then need to travel back along one of these

segments to get to B, thus retracing our steps.

Any route from A to B involves exactly 9 moves to the right and some number of moves up

and down.

Any route from A to B involves exactly one more move down than moves up, as we start at the

top of the grid and end up at the bottom. Therefore, the total number of up and down moves

must be odd, as it equals (x + 1) + x = 2x + 1, where x is the total number of up moves.

There are 10 vertical segments. Any choice of an odd number of these vertical segments uniquely

determines a route from A to B, as we must start at A, travel to the top of the leftmost of

these segments, travel down the segment, travel to the right to the bottom of the next segment,

travel up it, and so on.

Therefore, the routes from A to B are in exact correspondence with choices of an odd number

of the 10 vertical segments.

We compute the number of routes using n of these segments, for n = 1, 3, 5, 7, 9. In each case,

the length of the route will be 9 + n.

For n = 1 and n = 9, the number of routes is

(10

1

)=

(10

9

)= 10.

For n = 3 and n = 7, the number of routes is

(10

3

)=

(10

7

)=

10(9)(8)

3(2)(1)= 120.

For n = 5, the number of routes is

(10

5

)=

10(9)(8)(7)(6)

5(4)(3)(2)(1)=

10(9)(8)(7)

5(4)= 2(9)(2)(7) = 252.

Therefore, the route length with the maximum number of routes is when n = 5. In this case,

the route length is 14 and the number of routes is 252.

(Instead of going through all of the above calculations, we could have remarked that among

the numbers

(10

n

), the largest occurs when n is exactly half of 10.)

Answer: Length= 14, Number of Routes= 252


Part B

1. (a) Since x− 1, 2x + 2, and 7x + 1 form an arithmetic sequence, then

(2x + 2)− (x− 1) = (7x + 1)− (2x + 2)

x + 3 = 5x− 1

4 = 4x

x = 1

so x = 1.

(b) Solution 1

Since x = 1, the first term of the sequence is 0.

Since the last term is 72, the sequence is arithmetic, and we are told that there is a middle

term, then this middle term is equal to0 + 72

2= 36.

(Note that if there was an even number of terms, there would not necessarily be a middle

term. Since we are asked to find the middle term, we can safely assume that there is one!)

Solution 2

Since x = 1, the first three terms of the sequence are 0, 4, 8.

Since the common difference is 4 and the first term is 0, the number of times that the

difference needs to be added to get to the final term of 72 is72− 0

4= 18.

Therefore, 72 is the 19th term.

The middle term is thus the 10th term, or 0 + 4(10− 1) = 36.

(c) Since y − 1, 2y + 2, and 7y + 1 form a geometric sequence, then

2y + 2

y − 1=

7y + 1

2y + 2

(2y + 2)2 = (y − 1)(7y + 1)

4y2 + 8y + 4 = 7y2 − 6y − 1

0 = 3y2 − 14y − 5

0 = (3y + 1)(y − 5)

Therefore, y = −13

or y = 5.

(d) If y = −13, the first three terms of the sequence are −4

3, 4

3, −4

3.

In this case, the common ratio between successive terms is43

−43

= −1.

Therefore, the 6th term in this sequence is −43(−1)5 = 4

3.

If y = 5, the first three terms of the sequence are 4, 12, 36.


In this case, the common ratio between successive terms is12

4= 3.

Therefore, the 6th term in this sequence is 4(35) = 4(243) = 972.

2. (a) Solution 1

Since ∠ABC = ∠BCD = 90◦, then BA and CD are parallel, so ABCD is a trapezoid.

Thus, the area of ABCD is 12(24)(9 + 18) = 12(27) = 324.

Solution 2

Since ∠ABC = 90◦, then the area of 4ABC is 12(9)(24) = 9(12) = 108.

Also, since ∠BCD = 90◦, then 4ACD has height 24.

Therefore, the area of 4ACD is 12(18)(24) = 9(24) = 216.

Thus, the area of quadrilateral ABCD is 108 + 216 = 324.

(b) Solution 1

Since BA is parallel to CD, then ∠ABD = ∠BDC.

Since ∠BEA = ∠DEC as well, then 4ABE is similar to 4CDE.

Therefore,DE

BE=

CD

AB=

18

9= 2, so DE : EB = 2 : 1, as required.

Solution 2

As suggested by the diagram, we coordinatize the diagram.

Put C at the origin, D on the positive x-axis (with coordinates (18, 0)) and B on the

positive y-axis (with coordinates (0, 24)).

Since ∠ABC = 90◦, then A has coordinates (9, 24).

Therefore, the line through C and A has slope 249

= 83

so has equation y = 83x.

Also, the line through B and D has slope −2418

= −43, so has equation y = −4

3x + 24.

Point E lies at the point of intersection of these lines, so we combine the equations to find

the coordinates of E, getting 83x = −4

3x + 24 or 4x = 24 or x = 6.

Therefore, E has y-coordinate 83(6) = 16, so E has coordinates (6, 16).

To show that DE : EB = 2 : 1, we can note that E lies one-third of the way along from B

to D since the x-coordinate of E is one-third that of D (and the x-coordinate of B is 0),

or since the y-coordinate of E is two-thirds that of B (and the y-coordinate of D is 0).

Alternatively, we could calculate the length BE (which is 10) and the length of ED (which

is 20).

Using any of these methods, DE : EB = 2 : 1.

(c) Solution 1

From (b), 4ABE is similar to 4CDE and their sides are in the ratio 1 : 2.

This also tells us that the height of 4CDE is twice that of 4ABE.

Since the sum of the heights of the two triangles is 24, then the height of 4CDE is23(24) = 16.


Therefore, the area of 4DEC is 12(18)(16) = 144.

Solution 2

From (b), the coordinates of E are (6, 16).

Therefore, the height of 4DEC is 16.

Therefore, the area of 4DEC is 12(18)(16) = 144.

(d) Solution 1

From (c), the area of 4DEC is 144.

From Solution 2 of (a), the area of 4ACD is 216.

The area of 4DAE is the difference in these areas, or 216− 144 = 72.

Solution 2

Using the coordinatization from (b), the coordinates of A are (9, 24), the coordinates of

E are (6, 16), and the coordinates of D are (18, 0).

Using the “up products and down products” method, the area of the triangle is

1

2

∣∣∣∣∣∣∣∣∣∣9 24

6 16

18 0

9 24

∣∣∣∣∣∣∣∣∣∣=

1

2|9(16) + 6(0) + 18(24)− 24(6)− 16(18)− 0(9)|

=1

2|144 + 0 + 432− 144− 288− 0|

= 72

3. (a) Alphonse should win.

If Alphonse starts by taking 1 stone, then by Rule #3, Beryl must remove at least 1 stone

and at most 2(1)− 1 = 1 stone. In other words, Beryl must remove 1 stone.

This in turn forces Alphonse to remove 1 stone, and so on.

Continuing in this way, Alphonse removes 1 stone from an odd-sized pile at each turn and

Beryl removes 1 stone from an even-sized pile at each turn. Thus, Alphonse removes the

last stone.

Therefore, Alphonse wins by removing 1 stone initially since 7 is odd.

(In fact, this argument shows that Alphonse should win whenever N is odd.)

(b) Beryl should win.

We make a chart in which the rows enumerate a possible combination of moves. Each

move indicates the number of stones removed.


A1 B1 A2 B2 A3 B3 A4 B4 Winner

1 1 1 1 1 1 1 1 B

2 2 1 1 1 1 B

2 2 2 2 B

2 2 3 1 B

3 5 B

4 4 B

5 3 B

6 2 B

7 1 B

Thus, no matter what number of stones Alphonse removes initially, there is a move that

Beryl can make which allows her to win. (There are possible combinations of moves where

Alphonse wins that are not listed in this chart.) Therefore, Beryl should win when N = 8.

Her strategy is:

• If Alphonse removes 3 or more stones, then he can remove the remaining stones in

the pile and win.

• If Alphonse removes 1 or 2 stones, then he can win by using the table above, choosing

a row for which she wins. In effect, Beryl repeats Alphonse’s move on her first turn.

This ensures that Alphonse receives a pile with an even number of stones and that he

can remove no more than 3 stones on his next turn. Thus, she can win, as the table

shows.

(c) Solution 1

We show that Beryl has a winning strategy if and only if N = 2m, with m a positive

integer.

First, if N is odd, we know that Alphonse has a winning strategy as in (a) (Alphonse

removes 1 stone, forcing Beryl to remove 1 stone, and so on).

Second, if N = 2, then Beryl wins as Alphonse must remove 1 stone to begin, so Beryl

removes the remaining stone.

Next, we show that if N = 2k, then the player who has the winning strategy for N = k

also has a winning strategy for N = 2k. This will tell us that Beryl has a winning strategy

for N = 2, 4, 8, 16, . . . (in general, for N = 2m) and that Alphonse has a winning strategy

if N = 2mq where q is an odd integer (since Alphonse wins for N = q, 2q, 4q, . . .). Since

every even integer can be written in one of these two forms, this will complete our proof.

So consider N = 2k.

• If either player removes an odd number of stones from an even-sized pile (leaving an


odd-sized pile), then they can be forced to lose, as the other player can then remove

1 stone from an odd-sized pile and force a win as in (a). So if Alphonse removes

an even number of stones to start, then Beryl should next remove an even number

of stones (so that Alphonse can’t immediately force her to lose), so the pile size will

always remain even and each player’s move will always be to remove an even number

of stones.

• Suppose that Alphonse has a winning strategy for N = k of the form a1, b1, a2, b2, . . . , aj.

Here, we mean that Alphonse removes a1 stones on his first turn and responds to

Beryl’s first move b1 by removing a2 and so on. (Of course, a2 will depend on b1

which could take a number of values, and so on.) Since these are valid moves, then

1 ≤ a1 < k, and b1 < 2a1, and a2 < 2b1, and so on.

Then 2a1, 2b1, 2a2, 2b2, . . . , 2aj will be a winning strategy for Alphonse for N = 2k

since 1 < 2a1 < 2k, and 2b1 < 2(2a1), and 2a2 < 2(2b1), and so on, so this is a valid

sequence of moves and they exhaust the pile with Alphonse taking the last stone.

In other words, to win when N = 2k, Alphonse consults his winning strategy for

N = k. He removes twice his initial winning move for N = k. If Beryl removes 2b

stones next, Alphonse then removes 2a stones, where a is his winning response to

Beryl removing b stones in the N = k game. This guarantees that he will win.

• Suppose that Beryl has a winning strategy for N = k.

By an analogous argument, Beryl has a winning strategy for N = 2k, for if Alphonse

removes 2a stones, then she removes 2b stones, where b is her winning responding

move to Alphonse removing a stones in the N = k game.

Therefore, Beryl wins if and only if N = 2m, with m a positive integer.

Solution 2

We show that Beryl has a winning strategy if and only if N = 2m, with m a positive

integer.

Suppose first that N is not a power of 2.

We can write N as a sum of distinct powers of 2, in the form N = 2k1 + 2k2 + · · · + 2kj ,

where k1 > k2 > · · · > kj ≥ 0. (In essence, we are writing N in binary.) Since N is not

itself a power of 2, then this representation uses more than one power of 2 (that is, j ≥ 2).

We will show that Alphonse has a strategy where he can always reduce the number of

powers of 2 being used, while Beryl can never reduce the number of powers of 2 being

used. This will show that Alphonse can always remove the final stone, as only he can

reduce the number of powers of 2 being used to 0.

Alphonse’s strategy is to initially remove the smallest power of 2 from the representation

of N (that is, he removes 2kj stones).


On her first turn, Beryl thus receives a pile with 2k1 +2k2 + · · ·+2kj−1 stones. By rule #3,

she must remove fewer than 2(2kj) = 2kj+1 stones. Since kj−1 > kj, then kj−1 ≥ kj + 1, so

Beryl must remove fewer than 2kj−1 stones.

When she removes these stones, the 2kj−1 will be removed from the representation of the

number of remaining stones, but will be replaced by at least one (if not more) smaller

powers of 2.

Thus, Beryl cannot reduce the number of powers of 2 in the representation.

Suppose that Alphonse thus receives a pile with 2k1 +2k2 + · · ·+2kj−2 +2d1 +2d2 + · · ·+2dh

stones, with k1 > k2 > · · · > kj−2 > d1 > d2 > · · · > dh and h ≥ 1.

This means that Beryl removed B = 2kj−1 − (2d1 + 2d2 + · · ·+ 2dh) stones.

But B > 0 and B is divisible by 2dh (since B = 2dh(2kj−1−dh − (2d1−dh + 2d2−dh + · · ·+ 20

)and each of the exponents initially were larger than dh), so B ≥ 2dh .

Therefore, Alphonse can remove 2dh stones on his turn (that is, the smallest power of 2 in

the representation of the number of remaining stones) since 2dh stones satisfies Rule #3,

and so his strategy can continue.

Therefore, Alphonse has a winning strategy if N is not a power of 2.

If N is a power of 2, then Alphonse on his first turn cannot decrease the number of

powers of 2 in the representation of N . (This is a similar argument to the one above for

Beryl’s first turn.) On Beryl’s first turn, though, she can reduce the number of powers of

2 (as in Alphonse’s second turn above).

Therefore, the roles are reversed, and Beryl can always reduce the number of powers of 2,

while Alphonse cannot. Therefore, Beryl has a winning strategy when N is a power of 2.

Therefore, Beryl has a winning strategy if and only if N is a power of 2.

4. (a) Solution 1

In t seconds, the mouse runs 7t metres and the cat runs 13t metres.

Using this, we get a triangle with the cat and mouse meeting at point P .

60

13t7t

C M

P

120


By the cosine law,

CP 2 = CM2 + MP 2 − 2(CM)(MP ) cos(∠CMP )

(13t)2 = 602 + (7t)2 − 2(60)(7t) cos(120◦)

169t2 = 3600 + 49t2 − 120(7t)(−12)

169t2 = 3600 + 49t2 + 60(7t)

120t2 − 420t− 3600 = 0

2t2 − 7t− 60 = 0

(2t− 15)(t + 4) = 0

Therefore, t = 152

or t = −4.

Since t represents a time, then t > 0, so t = 152.

Solution 2

In t seconds, the mouse runs 7t metres and the cat runs 13t metres.

Using this, we get a triangle with the cat and mouse meeting at point P . Drop a perpen-

dicular from P to N on CM extended.

60

13t7t

C M

P

60N

Since ∠PMN = 60◦, then 4PMN is a 30◦-60◦-90◦ triangle.

Therefore, MN = 12PM = 7

2t and PN =

√3MN = 7

√3

2t.

This gives us right-angled 4CPN with CP = 13t, PN = 7√

32

t, and CN = 60 + 72t.

By the Pythagorean Theorem,

CP 2 = CN2 + NP 2

(13t)2 =(60 + 7

2t)2

+(

7√

32

t)2

169t2 = 3600 + 420t + 494t2 + 147

4t2

169t2 = 3600 + 420 + 49t2

120t2 − 420t− 3600 = 0

2t2 − 7t− 60 = 0

(2t− 15)(t + 4) = 0

Therefore, t = 152

or t = −4.

Since t represents a time, then t > 0, so t = 152.


(b) Solution 1

We coordinatize the situation, as suggested in the diagram with C having coordinates

(−60, 0) and M having coordinates (0, 0).

Suppose that the cat intercepts the mouse at point P (x, y).

Since the cat runs at 13 m/s and the mouse at 7 m/s, thenCP

MP=

13

7. Thus,√

(x + 60)2 + y2√x2 + y2

=13

7

(x + 60)2 + y2

x2 + y2=

169

49

49((x + 60)2 + y2) = 169(x2 + y2)

0 = 120x2 − 2(49)(60)x− 49(602) + 120y2

0 = x2 − 49x− 49(30) + y2

Since this equation is of the form 0 = x2 + ax + y2 + by + c and there is at least one

point whose coordinates satisfy the equation (for example, setting y = 0 gives a quadratic

equation with positive discriminant), then it is the equation of a circle, so all points of

interception lie on a circle. (We could also complete the square to obtain the equation

(x− 492)2 +y2 = (91

2)2, which is the equation of the circle with centre (49

2, 0) and radius 91

2.)

Solution 2

We coordinatize the situation, as suggested in the diagram, with C having coordinates



Suppose that the cat intercepts the mouse after t seconds and that the mouse runs in the

direction θ East of North. (θ here could be negative. We can assume that −90◦ ≤ θ ≤ 90◦

to keep the situation in the upper half of the plane. If θ did not lie in this range, then

P would be in the lower half plane and we could reflect it in the x-axis and use this

argument.)

60 mC M

P(x, y)

7t13t

θ


As in (a),

CP 2 = CM2 + MP 2 − 2(CM)(MP ) cos(∠CMP )

(13t)2 = 602 + (7t)2 − 2(60)(7t) cos(90◦ + θ)

120t2 = 3600 + 120(7t) sin θ

t2 − 7t sin θ = 30

But MP 2 = 49t2 = x2 + y2 and x = 7t cos(90◦ − θ) = 7t sin θ, so

x2 + y2

49− x = 30

x2 − 49x− 49(30) + y2 = 0

so all possible points P lie on a circle, as in Solution 1.

Solution 3

We coordinatize the situation, as suggested in the diagram with C having coordinates



Suppose that the cat intercepts the mouse after t seconds and that the mouse runs in the

direction θ East of North. (θ here could be negative. We can assume that −90◦ ≤ θ ≤ 90◦

to keep the situation in the upper half of the plane. If θ did not lie in this range, then

P would be in the lower half plane and we could reflect it in the x-axis and use this

argument.)

60 mC M

P(x, y)

7t13t

θ

If the mouse decides to run due East, then it will be caught when −60+13t = 7t or t = 10,

so will be caught at B(70, 0).

If the mouse decides to run due West, then it will be caught when −60 + 13t = −7t or

t = 3, so will be caught at A(−21, 0).

The positions above the x-axis where the mouse will be caught should be exactly symmetric

with the positions below the x-axis where the mouse will be caught. Therefore, if these

positions lie on a circle, then a diameter of this circle should lie on the x-axis.

Since the only positions on the x-axis where the mouse will be caught are A(−21, 0) and


B(70, 0), then these must be endpoints of the diameter.

Therefore, the circle will have centre E with coordinates(

12(−21 + 70), 0

)=(

492, 0)

and

radius 12(70− (−21)) = 91

2.

From the diagram above, the coordinates of the point P of intersection will be

(7t cos(90◦ − θ), 7t sin(90◦ − θ)) = (7t sin θ, 7t cos θ)

If we can show that PE = 912

for every value of θ, then we will have shown that every

point of intersection lies on the circle with centre E and radius 912.

Now

PE2 =(7t sin θ − 49

2

)2+ (7t cos θ)2

= 49t2 sin2 θ − 7(49)t sin θ + 492

4+ 49t2 cos2 θ

= 49t2(sin2 θ + cos2 θ)− 7(49)t sin θ + 492

4

= 49t2 − 7(49)t sin θ + 492

4

From Solution 2, t2 − 7t sin θ = 30, so

PE2 = 49(30) + 492

4= 49

(30 + 49

4

)= 49

(1694

)= 72132

4=(

912

)2so PE = 91

2, as required.

Therefore, all points of intersection lie on a circle.

(c) From (b), we know that the points of intersection lie on the circle with diameter AB,

where A has coordinates (−21, 0) and B has coordinates (70, 0).

Suppose that the mouse is intercepted at point P1 after running d1 metres and at point

P2 after running d2 metres.

A BM

d1

d2

21 70

P1

P2

By the Intersecting Chords Theorem, d1d2 = 21(70).

By the Arithmetic Mean-Geometric Mean Inequality,

d1 + d2

2≥√

d1d2 =√

21(70) = 7√

30

Therefore, d1 + d2 ≥ 2(7√

30) = 14√

30.


(The Arithmetic Mean-Geometric Mean Inequality (known as the AM-GM Inequality)

comes from the fact that if d1 and d2 are non-negative, then (d1 − d2)2 ≥ 0.

Thus, d21 + 2d1d2 + d2 ≥ 4d1d2, so

(d1 + d2

2

)2

≥ d1d2, sod1 + d2

2≥√

d1d2.)

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