Problems - MSRImentioned in allusions to times long forgotten. Problem 5. Prove or disprove: There...

Problems

1

Year 1993 Olympiad

Level A

Problem 1. Denote by S(x) the sum of the digits of a positive integer x.Solve:

(a) x + S(x) + S(S(x)) = 1993.(b) x + S(x) + S(S(x)) + S(S(S(x))) = 1993.

Problem 2*. Suppose n is the sum of the squares of three positive integers.Prove that n2 is also the sum of the squares of three positive integers.

Problem 3. A red and a blue poker chip are stacked, the red one on top.Suppose one can carry out only the following operations: (a) adding twochips of the same color to the stack together, in any position; and (b) re-moving any two neighboring chips of the same color. After finitely manyoperations, is it possible to end up with only two chips left, the blue one ontop of the red one?

Problem 4. At the court of Tsar Gorokh, the royal astrologer built a clock Adaptation

remarkably similar to modern (analog) ones, with hands for hours, minutes,and seconds, all moving smoothly around the same pont. He calls a momentof time lucky if the three hands of his clock, counting clockwise from thehour hand, appear in the order hours/minutes/seconds, and unlucky if theyappear in the order hours/seconds/minutes. Is the amount of lucky time ina 24-hour day more or less than the amount of unlucky time?

Remark. Tsar Gorokh (“King Pea”) is a character from Russian folklore, oftenmentioned in allusions to times long forgotten.

Problem 5. Prove or disprove: There is a finite string of letters of thealphabet such that (a) there are no two identical adjacent substrings, and(b) a pair of identical adjacent substrings appears as soon as one adds anyletter of the alphabet at the beginning or at the end of the string.

Problem 6. A circle centered at D passes through points A, B, and theexcenter O of the triangle ABC relative to side BC (that is, O is the centerof the circle tangent to BC and to the extensions of sides AB and AC).Prove that A, B, C, and D lie on a circle.

4 PROBLEMS

Level B

Problem 1. For two distinct points A and B in the plane, find the locusof points C such that the triangle ABC is acute and the value of its angleat A is intermediate among the triangle’s angles.

Problem 2. Let x1 = 4, x2 = 6, and define xn for n > 3 to be the leastnonprime greater than 2xn−1 − xn−2. Find x1000.

Problem 3. A paper triangle with angles of 20◦, 20◦, and 140◦ is cut alongone of its bisectors into two triangles; one of these triangles is also cut alongone of its bisectors, and so on. Can we obtain a triangle similar to the initialone after several cuts?

Problem 4. In Pete’s class, there are 28 students besides him. Each of the28 has a different number of friends in the class. How many friends doesPete have in this class?

Problem 5. To every pair of numbers x and y we assign a number x ∗ y.Find 1993 ∗ 1935 if it is known that

x ∗ x = 0 and x ∗ (y ∗ z) = (x ∗ y) + z for any x, y, z.

Problem 6. Given a convex quadrilateral ABMC with \BAM = 30◦,\ACM = 150◦, and AB = BC, prove that AM is the bisector of \BMC.

Level C

Problem 1. In the decimal representation of two numbers A and B, theminimal periods have lengths 6 and 12, respectively. What are the possibil-ities for the length of the minimal period of A + B?

Problem 2. The grandfather of Baron von Munchhausen built a castlewith a square floor plan. He divided the castle into 9 equal square areas,and placed the arsenal in the center square. The Baron’s father divided eachof the remaining 8 areas into 9 equal square halls and built a greenhouse ineach central hall. The Baron himself divided each of the 64 empty halls into9 equal square rooms and placed a swimming pool in each of the centralrooms. He then furnished the other rooms lavishly and connected each pairof adjacent furnished rooms by a door, locking all other doors.

The Baron boasts that he can tour all his furnished rooms, visiting eachexactly once and returning to the starting point. Can this be true?

Problem 3. From any point on either bank of a river one can reach theother bank by swimming a distance of no more than 1 km.

(a) Is it always possible to pilot a boat along the whole length of the riverwhile remaining within 700 m of both banks?

(b) * Same question with 800m.

YEAR 1993 OLYMPIAD 5

Remark. Both the answer and the degree of difficulty of the problem depend on adaptation

what additional assumptions are made. Naturally the boat is to be considered apoint. The original problem had a note saying so, and also that “the river joinstwo round lakes, each 10 km in radius, and the river banks consist of straight linesegments and arcs of circle.” But in spite of this precision, ambiguities remain.Can there be islands in the river? And does “within 700m”refer to the straight-line distance or the swimming distance?(See figure.) Warning: Part (b) is surprisingly difficult, unlessislands are allowed, in which case both parts are easy.

Problem 4. Given real numbers a and b, define pn = [2{an+b}], where {x}denotes the fractional part of x and [x] the integer part.

(a) Can all possible 4-uples of 0s and 1s occur as substrings of the sequencep0, p1, p2, . . . , if we are allowed to vary a and b?

(b) Can all possible 5-uples of 0s and 1s occur?

Problem 5. In a botanical classifier, a plant is identified by 100 features.Each feature can either be present or absent. A classifier is considered tobe good if any two plants have less than half of their features in common.Prove that a good classifier cannot describe more than 50 plants.

Problem 6. On the side AB of a triangle ABC, a square is constructedoutwards; let its center be O. Let M and N be the midpoints of AC andBC, and let the lengths of these sides be a and b. Find the maximum of thesum OM + ON as the angle ACB varies.

Level D

Problem 1. Knowing that tanα + tanβ = p and cot α + cotβ = q, findtan(α + β).

Problem 2. The unit square is divided into finitely many smaller squares,not necessarily of the same size. Consider the small squares that overlap(possibly at a corner) with the main diagonal. Is it possible for the sum oftheir perimeters to exceed 1993?

Problem 3. We are given n points in the plane, no three of which lie on aline. Through each pair of points a line is drawn. What is the least possiblenumber of pairwise nonparallel lines among these lines?

Problem 4. We start with a number of boxes, each with some marbles inthem. At each step, we select a number k and divide the marbles in eachbox into groups of size k with a remainder of less than k; we then removeall but one marble from each group, leaving the remainders intact.

Is it possible to ensure that in 5 steps each box is left with a singlemarble, if initially each box has at most (a) 460 marbles, (b) 461 marbles?

Problem 5. (a) It is known that the domain of a function f is the segment[−1, 1], and f(f(x)) = −x for all x; also, the graph of f is the union of finitelymany points and straight line segments. Draw a possible graph for f .

6 PROBLEMS

(b) Is it possible to draw the graph of f if the domain of f is (−1, 1)?the whole real line?

Problem 6*. A fly lives inside a regular tetrahedron with edge a. Whatis the shortest length of a flight the fly could make to visit every face andreturn to the initial spot?

Year 1994 Olympiad

Level A

Problem 1. A cooperative gets apple and grape juices in identical jugs andmakes a mixed drink, which it packages in liter bottles. One jug of applejuice was enough for exactly 6 bottles of the mix, and one jug of grape juicefor exactly 10 bottles of the mix. Then the recipe was changed, and one jugof apple juice is now sufficient for exactly 5 bottles of the mix. How manybottles of mix can is one jug of the grape juice good for now? (The drink isnot diluted with water.)

Problem 2. A student multiplying two three-digit numbers noticed that ifhe wrote the two numbers next to one another, the resulting six-digit numberwould be seven times greater than the product. Find the two factors.

Problem 3. In a triangle ABC, let P and Q be the bases of the perpen-diculars dropped from B to the bisectors of A and C. Prove that PQ ‖ AC.

Problem 4. Four grasshoppers sit at the vertices of a square. Every nowand then one of them hops over another, landing at a point symmetric, withrespect to the jumped-over grasshopper, to where it jumped from. Provethat at no time can the grasshoppers occupy the vertices of a bigger squarethan the original one.

Problem 5. The royal astrologer considers a moment in time favorable ifthe hour, the minute and the second hands of the clock all lie on the sameside of some diameter of the clock face. All other times are consideredunfavorable. The hands all turn smoothly around the same point. Is theamount of favorable time in a 24-hour day more or less than the amount ofunfavorable time?

Problem 6. Two people play a game on a piece of graph paper, 19 × 94squares in size. Each, in turn, colors a square of any desired size, so long asits edges coincide with lines of the grid and no part of it has been coloredyet. The player who colors the last square wins. Who is the winner underoptimal play, and what is the winning strategy?

8 PROBLEMS

Level B

Problem 1. Is there a nonconvex pentagon no two of whose five diagonalsintersect, other than at a vertex?

Problem 2. Sue starts with a line segment of length k, and Leo with one oflength l. First Sue divides her segment into three parts, then Leo divides hisinto three parts. If it is possible to build two triangles from the six segmentsobtained, Leo wins; otherwise Sue wins. Depending on the ratio k/l, whichplayer can be sure to win, and what should the winning strategy be?

Problem 3. Prove that the equation

x2 + y2 + z2 = x3 + y3 + z3

has infinitely many integer solutions.

Problem 4. Two circles intersect at the points A and B. Tangents aredrawn to both circles through A. The tangents intersect the circles at thepoints M and N . The lines BM and BN intersect the circles again at thepoints P (on BM) and Q (on BN). Prove that MP = NQ.

Problem 5*. Dropping one of the digits of a certain natural number leavesanother number that divides the first. The dropped digit is not the leftmostone. What is the highest possible value for the first number, assuming itdoes not end in 0?

Problem 6. In a variation on the game of Battleship, ten ships must beplaced in a 10 × 10 square of graph paper: one ship has dimensions 1 × 4,two are 1 × 3, three are 1 × 2, and four are 1 × 1. The ships cannot toucheven at a corner, but they can be anchored at the sides of the square.

(a) Prove that if the ships are placed from largest to smallest, there isalways room for all of them, regardless of where earlier ones were placed.

(b)* Show by an example that if the ships are laid down from smallestto largest, this may not be the case.

Level C

Problem 1. A student multiplying two seven-digit numbers noticed thatif she wrote the two numbers next to one another, the resulting 14-digitnumber would be three times greater than the product. Find the two factors.

Problem 2. An infinite sequence of numbers xn, where n runs over positiveintegers, is defined by the condition

xn+1 = 1 − |1 − 2xn|,where 0 6 x1 6 1.

Prove that the sequence is eventually periodic if and only if x1 is rational.

Problem 3. Each of the 1994 members of the Parliament of Dunces hasslapped exactly one colleague in the face. Prove that it is possible to draw


from this Parliament a committee whose 665 members have not slapped oneanother.

Problem 4. Let D be a point on side BC of triangle ABC. Circles areinscribed in the triangles ABD and ACD, and a common outer tangent tothe circles (distinct from BC) is drawn. Let its intersection with AD becalled K. Prove that the length of AK does not depend on the position ofD on BC.

Problem 5. Consider an arbitrary polygon, not necessarily convex.(a) Is there always a chord of the polygon that divides it into two pieces

of equal area?(b) Prove that the polygon can be divided by a chord into pieces whose

areas are each at least one-third the total area of the polygon.(By a chord of a polygon we mean a line segment whose endpoints

belong to the polygon’s perimeter while the segment itself lies entirely onthe polygon, including its perimeter.)

Problem 6*. Can a polynomial P (x) have a negative coefficient if all thepowers P n(x), for n > 1, only have positive coefficients?

Level D

Problem 1. Devise a polyhedron such that no three of its faces all havethe same number of edges.

Problem 2. See Problem 2.

Problem 3. A spherical cherry of radius r is dropped into a goblet whoseaxial cross-section is the graph of the function y = x4. What is the largestvalue of r that allows the cherry to touch the very bottom of the goblet? (Inother words, what is the radius of the largest disc contained in the domainy > x4 and touching the origin?)

Problem 4. A convex polyhedron has nine vertices, one of which is A. Thetranslations that send A into each of the other vertices form eight congruentpolyhedra. Prove that at least two of these eight polyhedra have an interiorpoint in their intersection.

Problem 5*. The extensions of the sides AB and CD of a convex polygonABCD intersect at a point P ; the extensions of the sides BC and ADintersect at Q. Consider three pairs of bisectors: those of the outer angles ofthe quadrilateral at vertices A and C, those of the outer angles at verticesB and D, and those of the outer angles at vertices P and Q of the trianglesQAB and PBC. Prove that if each of these three pairs of lines intersects,the intersection points are collinear.

Problem 6. Prove that for any k > 1, there exists a power of 2 such thatamong its last k digits, there are at least as many 9s as other digits. (Forexample: 212 = . . . 96, 253 = . . . 992.)

Year 1995 Olympiad

Level A

Problem 1. Ben Franklin spent a penny a day for a loaf of bread and atankard of ale. When prices went up by 20%, he started buying half a loafof bread and the same ale for a penny. Will a penny be enough to buy theale if prices rise by 20% again?

Problem 2. Prove that all the numbers of the form 10017, 100117, 1001117,and so on are divisible by 53.

Problem 3. Consider a convex quadrilateral and a point O inside it suchthat \AOB = \COD = 120◦, AO = OB and CO = OD. Let K, L and Mbe the midpoints of the sides AB, BC and CD, respectively. Prove that (a)KL = LM ; (b) the triangle KLM is equilateral.

Problem 4. To manufacture a closed box of volume at least 1995 cubicunits in the shape of a parallelepiped we have (a) 962, (b) 960, (c) 958square units of material. Is the material sufficient?

Problem 5. Roads lead from a city to several nearby villages; there is nodirect communication between villages. A truck loaded with goods to bedelivered to all the villages starts out from the city. The amount of fuelspent on a leg of the trip is proportional to the current load weight andthe distance. Suppose the weight of each item in the load is equal, in someunits, to the distance from the city to the item’s destination. Prove that thefuel cost does not depend on the order in which the deliveries are made.

Problem 6. A straight line cuts a triangle AKN off of the regular hexagonABCDEF so that AK + AN = AB. Find the sum of the angles at whichthe segment KN is seen from the vertices of the hexagon, i.e., find

\KAN + \KBN + \KCN + \KDN + \KEN + \KFN.

Level B

Problem 1. Prove that if we insert any number of digits 3 between the 0sin 12008, we get a number divisible by 19.

12 PROBLEMS

Problem 2. Consider an equilateral triangle ABC. For an arbitrary pointP inside the triangle, consider the intersection A′ of the line AP with BCand the intersection C ′ of CP with BA. Find the locus of points P forwhich the segments AA′ and CC ′ have equal length.

Problem 3. A strip is a rectangle of size 1 × k, where k is some naturalnumber. For what integers n can one cut a 1995×n rectangle into strips allof different lengths?

Problem 4. Suppose ab = cd, where a, b, c, d are natural numbers. Cana + b + c + d be a prime?

Problem 5. We start with four identical right triangles. In one move we cancut one of the triangles along the altitude perpendicular to the hypotenuseinto two triangles. Prove that, after any number of moves, there are twoidentical triangles among the whole lot.

Problem 6. A team of geologists took 80 cans of food on an expedition.The weights of the cans were all different and were listed precisely in aninventory. After a while the labels became unreadable and only the cookknew which can was which. She boasted that she could prove the identityof all the cans simultaneously using only the inventory and a few weighingoperations. Her balance has two pans and a pointer that shows the differencein weight between the contents of the pans.

(a) Show how the cook can prove her claim using four weighings.(b) Can she do it in only three?

Level C

Problem 1. The number sinα is known. What is the largest number ofpossible values for (a) sin α

2 ? (b)* sin α3 ?

Problem 2. See Problem 58.9.2. REF 58.9.2

Problem 3. The diagonals of a trapezoid ABCD meet at a point K. Lettwo circles be constructed, each having one of the lateral sides of the trape-zoid as a diameter. Supposing that K lies outside both circles, prove thatthe tangents from K to these circles have equal lengths.

Problem 4. See Problem 58.9.5. REF 58.9.5

Problem 5*. Prove that if a, b and c are integers such that a/b+ b/c+ c/aand a/c + c/b + b/a are also integers, then |a| = |b| = |c|.Problem 6. A panel has a number of buttons and a number of light bulbs.Each button is connected to some of the lights, and pressing the button flipsthe state of the lights it’s connected with from on to off or vice versa. Allthe lights are on at the start.

It is known that, for any set of lights, there is a button that flips an oddnumber of lights from this set. Prove that one can switch off all the lightsby pressing buttons.


Level D

Problem 1. Prove that

|x| + |y| + |z| 6 |x + y − z| + |x − y + z| + |−x + y + z|for all real numbers x, y, z.

Problem 2. Is it possible to color the edges of an n-angled prism with threecolors so that all three colors occur among the edges of any face and amongthe edges meeting at any vertex? Answer for (a) n = 1995 and (b) n = 1996.

Problem 3. In a triangle ABC, consider the median AA1, the bisectorAA2, and a point K on AA1 such that KA2 ‖ AC. Prove that AA2 ⊥ KC.

Problem 4*. (a) Divide the interval [−1, 1] into black and white intervalsso that the integral of any linear function over intervals of one color is thesame for both colors.

(b) Same problem with quadratic trinomials instead of linear functions.

Problem 5. Find the largest value of n such that there are two-sided infinitestrings satisfying the following conditions:

• Any subtring of B of length at most n is contained in A.• A is periodic with minimal period 1995, but B is not (it may be ape-

riodic or have a different period).

Problem 6*. Prove that there exist infinitely many nonprime values of nsuch that 3n−1 − 2n−1 is divisible by n.

Problem 7. Is there a polyhedron and a point outside it such that fromthis point none of the polyhedron’s vertices is visible?

Year 1996 Olympiad

Level A

Problem 1. If a + b2/a = b + a2/b, is it true that a = b?

Problem 2. Ten iron weights are arranged around a circle. A small bronzeball is placed between each pair of neighboring weights. The mass of eachball equals the difference of the masses of the neighboring weights. Provethat the balls can be placed on the two pans of a balance so as to bring itinto equilibrium.

Problem 3. Each intersection of a square grid is home to a gardener, andthere are flowers planted everywhere. Each flower must be looked after bythe three gardeners closest to it. Draw the area that the gardener at theorigin must look after.

Problem 4. The side BC of an equilateral triangle ABC is divided intothree equal parts by the points K and L, and point M divides the side ACin the ratio AM : MC = 1 : 2. Prove that the sum of angles AKM and ALMis 30◦.

Problem 5. A rook is at the corner of an n × n chessboard. It makesn2 successive moves (of any length), alternating between horizontal andvertical motion. For what values of n can this sequence of moves be chosenso that the rook stops on all the squares of the board and finally returns toits original position? (Note that each move must be followed by one in aperpendicular direction.)

Problem 6. (a) Eight students were given eight problems to work on. Itturns out that each problem was solved by five students. Prove that thereare two students such that each problem was solved by at least one of them.

(b) If each problem was solved by four students, show by a counterex-ample that two students satisfying the condition in part (a) need not exist.

Level B

Problem 1. Prove that any convex polygon has at most 35 angles less than170◦.

16 PROBLEMS

Problem 2. Suppose the numbers a, b, and c satisfy the inequalities

|a − b| > |c|, |b − c| > |a|, |c − a| > |b|.

Prove that one of these numbers is the sum of the other two.

Problem 3. Consider two points A and B on the circle circumscribed toa triangle ABC (the circumcircle), and assume the tangents to the circlethrough A and B meet at a point M . Choose a point N on the side BC sothat the line MN is parallel to the side AC. Prove that AN = NC.

Problem 4. The integers from 1 to n are written in a row. The samenumbers are written under them in a different order. Is it possible to writethe second row so that all the sums of pairs of corresponding numbers inthe two rows are perfect squares when (a) n = 9; (b) n = 11; (c) n = 1996?

Problem 5. Two points A and B on a circle divide it into two arcs. Con-sider all the chords joining a point on one arc AB to a point on the other.What is the locus of midpoints of these chords?

Remark. The statement is ambiguous on whether the endpoints A and B shouldbe regarded as part of the arcs. For definiteness, assume they are not.

Problem 6. Ali Baba and a thief must share a trove of 100 gold coins.They arrange them into 10 piles of 10 coins each. Ali Baba takes four cups,places them next to four piles of his choice, and transfers from each of thesepiles into the corresponding cup some number of coins —at least one, butnot the entire pile. The number of moved coins need not be the same forthe four piles.

After that, the thief must permute the cups without leaving them all inplace, and then move the coins from each cup onto the pile the cup is nextto now.

This process is repeated: Ali Baba again places the empty cups near anyfour of the ten piles, and so on. At any time, Ali Baba can take three pilesof his choice and leave; his opponent gets all the rest.

What is the greatest number of coins that Ali Baba can walk away withif the thief also works to get as many coins as possible?

Level C

Problem 1. The positive numbers a, b, and c satisfy the equation a2 +b2−ab = c2. Prove that (a − c)(b − c) 6 0.

Problem 2. A hundred points are marked on a piece of paper in a 10×10 Adaptation

square array. How many lines not parallel to either side of the square arerequired if we want lines going through all 100 points?


Problem 3. Divide side BC of an equilateral triangle ABCinto n equal parts with points P1, P2, . . . , Pn−1 as in thefigure, so BP1 = P1P2 = · · · = Pn−1C. Choose a pointM on the side AC so that AM = BP1. Prove that\AP1M+\AP2M+· · ·+\APn−1M = 30◦ if (a) n = 3;(b) n is an arbitrary positive integer.

A

P1 P2 Pn−1B C

M

Problem 4. A rook is on a corner square of an m × n chessboard. Twoplayers take turns moving it, either straight across or up and down, anynumber of squares each time; but the rook is not allowed to land on, oreven cross, a square on which it has already landed or through which it haspassed. The player who cannot make a move loses. Which player can forcea to win: the one who plays first or the second one? Describe a winningstrategy.

Problem 5. Two laws are in force in a certain country:

(1) A person may play basketball only if he/she is taller than most ofhis/her neighbors.

(2) A person gets free bus rides only if she/he is shorter than most ofher/his neighbors.

By a person’s “neighbor” is meant anyone whose house (regarded as a point)lies inside a certain circle whose center is the person’s house and whose radiusis chosen by the person — and a different choice of radius is allowed for thepurposes of the first law and the second law!

Is it possible that 90% of people or more are allowed to play basketballand that 90% or more are allowed free travel?

Problem 6. Prove that for any polynomial P (x) of degree n with positiveinteger coefficients, there exists an integer k such that P (k), P (k + 1), . . . ,P (k+1996) are composite numbers if (a) n = 1; (b) n is an arbitrary positiveinteger.

Level D

Problem 1. See Problem 1 on page 16.

Problem 2. Find a polynomial with integer coefficients having5√

2+√

3 +5√

2−√

3 as a root.

Problem 3. A point is chosen on each of eight evenly spaced parallel planes.Can these points be the vertices of a cube?

Problem 4. Prove that there exist infinitely many positive integers n withthe property that n is the sum of two perfect squares, but n − 1 and n + 1are not.

Problem 5. A point X lies outside two disjoint circles ω1 and ω2, andthe tangent segments drawn from X to ω1 and ω2 are equal. Prove that theintersection of the diagonals of the quadrilateral formed by the four tangency

18 PROBLEMS

points coincides with the intersection of the two internal common tangentsof ω1 and ω2.

Problem 6. The 2n possible strings of length n containing only the numbers1 and −1 form a table with 2n rows and n columns. Suppose that a numberof entries in the table are replaced by 0. Prove that one can select certainrows so that their sum (defined as the string obtained by adding each columnseparately) is a string of zeros.

Answers

Year 1993 Olympiad

Level A

1. (a) No solutions. (b) x = 1963. 3. No. 4. The amounts of luckyand unlucky time are the same. 5. There is always such a string.

Level B

1. The locus is given by the shaded domains and solid A B

curves in the figure. (The points on the dotted curves do notbelong to the locus.) 2. x1000 = 1

2 · 1000 · 1003 = 501500.3. No. 4. Pete has 14 friends. 5. 58.

Level C

1. 4 or 12. 2. Yes. 3. 700 m is not always possible, but 800 m is(if islands are not allowed). 4. (a) Yes. (b) No. 6. The maximum

equals 12(1 +

√2)(a + b) and is attained when \ACB = 135◦.

Level D

1. pq/(q−p), unless p = q, in which case the answer iseither 0 (if p = 0) or undefined (if p 6= 0). 2. Yes.3. n lines if n > 2; one line if n = 2. 4. (a) Yes.(b) No. 5. (a) A possible solution is shown in figure.(b) No. 6.

4√10

a.

1

2

−1

2

1

2

−1

2

−1

1

−1

1

0

Year 1994 Olympiad

Level A

1. Fifteen bottles. 2. 143 and 143. 5. There is more favorable timethan unfavorable time. 6. The first player wins.

Level B

1. Yes. 2. If k > l, Leo wins; otherwise Sue wins. 5. The number180625 becomes 17 times smaller if 8 is stricken out.

Level C

1. 3333334 and 1666667. 5. (a) Not always. 6. Yes.

Level D

3. r = 34

3√

2.

Year 1995 Olympiad

Level A

1. Yes. 4. Yes in all three cases.

Level B

2. The locus is the union of the altitude of △ABC fromvertex B with an arc of circle of measure 120◦ lying inside△ABC; see figure. 3. For n 6 998 and for n > 3989.4. No. A C

B

Level C

1. (a) 4; (b) 3.

Level D

2. (a) yes; (b) no. 5. n = 1994. 7. Yes.

Year 1996 Olympiad

Level A

1. Yes, it is true. 3. 5. For even n.

Level B

4. (a) Yes; (b) no; (c) yes. 5. Let O be the centerof the circle. Consider the circles with diameters AOand BO (see figure). The desired locus is the interior ofthese two circles minus their intersection. 6. 72 coins.

A B

O

Level C

2. 18. 4. For m = n = 1 player 2 wins. In all other cases, player 1 wins.5. Yes, it is possible.

Level D

2. x5 − 5x3 + 5x − 4. 3. Yes, they can.

Year 1993 Olympiad

Level A

1. (a) Use divisibility by 3. (b) Estimate x from below and find the remain-der after division of x by 9.2. Rewrite the expression (a2 + b2 + c2)2 in a different form.3. Consider all pairs of chips with a red chip above a blue chip.4. Favorable and unfavorable times are interchanged by reflection.5. Use induction on the number of letters in the alphabet.6. Prove that \ADB = \ACB.

Level B

2. Write down xn − xn−1 for n small.3. From what triangles can a triangle with the initial angles be obtained?4. Consider the most friendly and the least friendly of Pete’s classmates.6. Consider the points symmetric to B with respect to AM .

Level C

1. The digits of A + B must repeat in blocks of 12.4. Draw the points {an + b} on the circle of unit circumference.5. Estimate the total number of distinctions between all possible pairs ofplants with respect to all features.

Level D

4. Let there be 1, 2, . . . , n stones in the boxes. Express the maximal numberof stones after one move in terms of n and k. Investigate for which k thisnumber is minimal.5. The graph of such a function would be left unchanged by a 90◦ rotationaround the origin.6. Let P and Q be the midpoints of the sides KL and MN of a spacequadrilateral KLMN . Then

PQ 6 12(KN + LM).

Year 1994 Olympiad

Level A

1. Express the volume of the bottle in terms of the volume of the jug.2. If x and y are the three-digit numbers to be found, then 7xy = 1000x+y.3. Extend the perpendiculars until they intersect the line AC.4. The grasshoppers can only hop to points in a square grid.5. After an integer number of hours, the second hand and the minute handreturn to their positions.6. Use the axial symmetry of the rectangle.

Level B

2. Leo should try to make isosceles triangles.3. Set z = −x.4. It suffices to prove that the triangles AQN and AMP are congruent.Prove that they are similar and that the arcs AQ and MA are equal.6. (b) To prove that it is possible to place the next 1 × 3 ship, draw 8auxiliary 1 × 3 ships in such a way that a 1 × 4 or 1 × 3 ship would touchnot more than two of these auxiliary ships.

Level C

1. Let x and y be the 7-digit numbers required. Then 3xy = 107x + y.2. (a) Observe how the denominators of the terms of the sequence change.(b) Fix n. One can write xn+k = bkxn + ak. What is |bk|?3. Prove that some member of Parliament was slapped at most once.4. AK = 1

2(AB + AC − BC).5. (b) Draw a chord and move it so it remains parallel to itself. (If the chordgoes through a concave vertex of the polygon, it may split the polygon intomore than two parts; do not forget to account for this case.)6. Note that the condition “for all n > 1” can be replaced by “for n = 2and n = 3”. If P has a term with a zero coefficient while all the coefficientsof P 2 and P 3 are positive, it is possible to “jiggle” P a bit so it satisfies thecondition of the problem.

34 HINTS

Level D

1. Try to construct a polyhedron with 6 faces.3. What is the smallest value of r for which the a circle of radius r withcenter (0, r) intersects the curve y = x4 other than at the origin?4. Show that the polyhedron obtained by applying a homothety with centerA and factor 2 contains the eight translated polyhedra.5. Use Fact 25.6. It is enough to find m such that 2m is just a bit less than a power of 10.For this, it suffices to find n such that 2n + 1 is divisible by a sufficientlylarge power of 5.

Year 1995 Olympiad

Level A

2. Consider the difference between consecutive numbers.3. Consider a 120◦ rotation about the point O.5. You can either compare the fuels costs for deliveries having almost thesame route, or deduce an explicit expression for the cost for any delivery.6. Rotate all the angles of interest so their vertices coincide with A.

Level B

1. Consider the difference between consecutive numbers.2. For a fixed A′, there exist precisely two points C ′ such that AA′ = CC ′.3. Start with a similar problem for the 5 × n rectangle.5. The order in which we cut the triangles does not matter.6. (a) The cook should first put the 27 lightest cans on the left pan.(b) Use the pigeonhole principle.

Level C

1. (b) Express sin 3x in terms of sinx.3. Use the tangent-secant theorem (power of a point) for an appropriatesecant.5. Let p be a prime divisor of a. Reduce the fractions to the least commondenominator, and consider the highest powers of p dividing the numeratorand denominator.6. Use induction. (Assume it’s been proved that all the bulbs, except anarbitrary fixed one, can be switched off.)

Level D

1. |a + b| 6 |a| + |b|.3. Complete the triangle BAC to a parallelogram and use similarity.4. If f is a polynomial of degree n, then f(x) − f(x−d) is a polynomial ofdegree n − 1, for any fixed d 6= 0.6. Look for n of the form 3b − 2b.

Year 1996 Olympiad

Level A

1. A nonzero factor can be canceled from both sides of an equation.2. Let mi be the mass of the ith weight; then (m1−m2)+(m2−m3)+ · · ·+(m9 − m10) + (m10 − m1) = 0.3. Consider a flower and find out which gardeners look after it.6. (a) Consider two cases: (1) there is a student who has solved six problems,and (2) each student has solved at most five problems.

Level B

1. Use the formula for the sum of angles of a convex polygon.3. Use Fact 15 (page 124).4. (c) Reduce the problem to a similar one with a smaller n.6. First, prove that the thief can always ensure that there are no piles of lessthan four coins. Then show that Ali Baba can ensure that there are sevenpiles of at most four coins each.

Level C

2. Consider the 36 unit squares adjacent to the sides of the big square.3. Express the angles in question in terms of angles at the points A and M .5. Consider a country with 10 inhabitants whose houses are placed along astraght line.6. If k divides x − y, then k divides P (x) − P (y).

Level D

2. Compute5√

2 +√

3 · 5√

2 −√

3.3. Write the equations of parallel planes drawn through the vertices of theunit cube in such a way that the distances between any two neighboringplanes are the same.5. Use the law of sines and Fact 17 (page 124).

Solutions

Year 1993 Olympiad

Level A

Problem 1. (a) Since x, S(x) and S(S(x)) have the same remainder upondivision by 3 (by Fact 6), the sum x+S(x)+S(S(x)) is divisible by 3. Since1993 is not divisible by 3, there is no solution.

(b) Clearly x < 1993. It is easy to see that 1899, 1989 and 999 have thelargest sum of digits among the numbers from 1 to 1993. Thus S(x) 6 27.Further, S(S(x)) 6 S(19) = 10 and S(S(S(x))) 6 9. The equation to besolved implies that

x = 1993 − S(x) − S(S(x)) − S(S(S(x))) > 1993 − 27 − 10 − 9 = 1947.

As in part (a), we know that x, S(x), S(S(x)) and S(S(S(x))) leave thesame remainders when divided by 9; let’s call this remainder r. Then 4r hasremainder 4 when divided by 9 (because that’s the remainder of 1993). Inother words, 4r − 4 is divisible by 9, which implies that r − 1 is divisible by9 (since 4 and 9 are coprime; see Fact 9). So r = 1.

Now list the numbers from 1947 through 1993 that have remainder 1upon division by 9: they are 1954, 1963, 1972, 1981 and 1990. We canverify directly that only 1963 satisfies the equation.

Problem 2. Let n = a2 + b2 + c2. Expand the square and rearrange:

(a2+b2+c2)2 = a4+b4+c4+2a2b2+2a2c2+2b2c2

= (a4+b4+c4+2a2b2−2a2c2−2b2c2)+(2ac)2+(2bc)2

= (a2+b2−c2)2+(2ac)2+(2bc)2.

We can assume that a > b > c > 0, so a2 + b2 − c2 > 0. Thus we haveexpressed n2 as the sum of squares of three positive integers.

Remarks. 1. Try to prove a similar statement for the sum of four and more squares.

2. For the sum of two squares the analogous statement is not true: (12 + 12)2 = 4cannot be represented as the sum of two squares of positive integers, althoughan analogous identity is true:

(a2 + b2)2 = (a2 − b2)2 + (2ab)2.

42 SOLUTIONS

For more on sums of two squares, see the solution to Problem 4 on page 4 andRemark 2 thereto.

3. A famous theorem of Lagrange says that any positive integer can be representedas the sum of four integer squares. (An integer square, or simply a square, meansthe same as the square of an integer. This includes 0 as well as the squares ofpositive integers.)

4. The number 7 is not a sum of three squares. It turns out that a natural numberis not representable as a sum of three squares if and only if it is of the form(8k + 7) · 4m.

5. Suppose that instead of specific numbers we consider sums of squares of ar- rearranged the exposition

a bit – the beginning didnot sound vary natural

bitrary variables. An identity going back at least to the third-century Greekmathematician Diophantus of Alexandria says that

(a2 + b2)(c2 + d2) = (ac + bd)2 + (ad − bc)2,

which of course can also be written as

(a21 + a2

2)(b21 + b2

2) = (a1b1 + a2b2) + (a1b2 − a2b1)2.

This elegant equality was extended by Euler in the eighteenth century to thecase of four pairs of variables:

(a21+b2

1+c21+d2

1)(a22+b2

2+c22+d2

2)

= (a1a2−b1b2−c1c2−d1d2)2+(a1b2+c1d2+a2b1−c2d1)

2

+(a1c2+a2c1+b2d1−b1d2)2+(a1d2+a2d1+b1c2−b2c1)

2.

A similar identity also exists representing

(a21 + a2

2 + · · · + a2k)(b2

1 + b22 + · · · + b2

k) (∗)as a sum of squares of polynomials when k = 8. However, for no other k —apart from the already discussed k = 2, 4, 8 and the trivial case k = 1— is theproduct (∗) writable as a sum of squares of polynomials! The proof of this factis far from elementary; to intrigue the reader, we mention only that the identityfor k = 2 is related to complex numbers, for k = 4 to quaternions, and fork = 8 to Cayley numbers, also called octonions. More about this in the bookCantor-Solodovnikov. CITE

Cantor-Solodovnikov

Problem 3. Consider all pairs of chips in the stack, not necessarily adja-cent. There are four possibilities for such pairs, in terms of the orderingof colors: RR, RB (meaning a red chip above a blue chip), BR and BB.For instance, in the stack RBRB there are three RB pairs (first and secondchips, first and last, third and last).

The reader can check that the parity (see Fact 23) of the number of RBpairs cannot change under our operations. For example, suppose we inserttwo red chips at a spot that has k blue chips below it. It is easy to see thatthis adds exactly 2k new RB pairs, so the parity remains the same. Theother operations — insertion of two blue chips and removal of two blue ortwo red chips— can be analyzed similarly.

Now, in the initial position, there is exactly one RB pair. In the desired I merged the secondsolution with remark 4.

The solution as originally

given was not very

illuminating, but itcomplements remark 4

beautifully.


final position, there are none. Since 1 and 0 have different parities, it’s notpossible to reach one state from the other.

Remarks. 1. One could equivalently consider the number of red chips below whichthere is an odd number of blue chips.

2. This problem comes up in a proof of an amusing and geometrically intuitivefact: Two continuous paths inside a square, each of which joins a pair of oppo-site vertices, must intersect. We sketch the proof in the case where the pathsare polygonal. (This will imply the general case of continuous paths by approx-imation, but a rigorous argument to justify this reduction is not so simple.)

So let’s consider two nonintersecting polygonal paths as described, one blueand one red, inside a square. We make the square’s sides vertical and horizontal.Now move a vertical line continuously from the left edge of the square to theright edge, keeping track of the intersections of this line with the red and bluepaths. The pattern of red and blue intersection points corresponds to the stackof chips in the problem. (Since the paths don’t intersect we can assume thatnone of the polygonal segments is vertical: just jiggle a vertex a tiny bit if it is.Therefore the intersection of either path with a vertical line is a finite set.)

The intersection points always appear and disappear in pairsof a single color, as in the problem. (See the diagram on the rightfor an appearing pair and a disappearing one.) Therefore if on theleft edge of the square we have the red dot above the blue dot, wecannot have the inverse situation on the right edge, contradictingthe assumption that the paths join opposite corners.

3. The geometric result just proved is closely related to the Jordan curve theorem,which states that every closed plane curve without self-intersections divides theplane into two parts, an inside and an outside. This intuitively clear statementis difficult to prove, since the curve may have no smooth pieces. But in the caseof polygonal closed curves an argument very similar to the one above providesa proof.

4. A more general way to find invariants is with the help of transformation groups.Consider two maps (transformations) of the real line, r and b, defined by r(x) =1−x (this is a reflection in the point 1 of the line) and b(x) = −1−x (a reflectionin the point −1). Obviously, composing r with r brings each point of the lineback to itself, and so gives the identity map; in group notation, r2 = 1. Similarly,b2 = 1. We can also compose the two reflections with one another. For instance,applying b, then r, has the overall effect of a translation: r(b(x)) = 1−(−1−x) =x + 2.

Now let’s associate to red chips R the reflection r, and to blue chips B thereflection b. For any stack of chips, imagine reading off the colors from thebottom of the stack up and applying consecutively the reflection correspondingto each color. The overall effect of all these reflections r and b, in the givenorder, is again a map of the real line.

The initial stack in the problem, RB, corresponds to applying b then r; this,as we saw, gives the translation x 7→ x + 2. But if we consider the reverse stackBR, the composition gives a different result altogether:

b(r(x)) = −1 − (1 − x) = x − 2.

The composition of transformations is not, in general, commutative!

44 SOLUTIONS

Now the key observation is this: two adjacent chips of the same color canceleach other out from the point of view of composition: their combined effect,as we have seen, is the identity. This means that inserting or removing twoadjacent chips of the same color has no effect on the overall map represented bythe stack of chips. Therefore no sequence of such insertions of removals can leadfrom the stack RB to the stack BR, as they represent different overall maps.

5. (Continuation of the previous remark, for more advanced readers familiar withgroup theory language.) The set of compositions is the group of transformationsof the real line generated by r and b, and these generators satisfy the relationsr2 = 1 and b2 = 1. One can also define an abstract group with generatorsr and b— these letters now being regarded just as symbols— subject to therelations r2 = 1 and b2 = 1, meaning that whenever bb or rr appear in aproduct of generators, they can be erased. This abstract group is the same asthe group of transformations we defined. (It can be checked that the group oftransformations has no additional relations beyond those in the abstract group.)It is also isomorphic to the semidirect product of Z and Z/2.

Problem 4. There are two key ideas: relating a reflection (or inversion) intime with a reflection in the pattern of the hands of the clock, and observingthat the mirror image of a lucky pattern is unlucky.

Consider the position of the hands at two distinct times: T secondsbefore noon and T seconds after noon. The patterns formed by the hands ofthe clock at those moments are mirror images of each other, with respect tothe axis of symmetry formed by a vertical diameter. (Why?) For example,here is a clock showing 01:15:22 PM (T = 5422 seconds after noon) andanother showing 10:44:38 AM (the same number of seconds before noon):

12

9

6

12

3

6

Now, a moment’s thought shows that reflection interchanges lucky andunlucky patterns of hands. Thus, to each lucky moment before noon therecorresponds an unlucky moment after noon. An interval of lucky moments inthe morning ismatched by an interval of unlucky moments in the afternoonor evening, and both intervals have the same length. Similarly, an interval ofunlucky moments in the morning is matched by an equal interval of luckymatched in the afternoon or evening. Hence the total amount of lucky timein the day equals the total amount of unlucky time.

Remarks. 1. A similar situation is considered in Problem 57.8.5. See especially REF 57.8.5

the Remark 2 following the solution to that problem.

2. It is not enough to argue that each lucky pattern of hands has a mirror imagethat is an unlucky pattern of hands, because the mirror image might conceiv- I integrated the content of

Remark 3 with the

solution above; perhaps

the matter is intuitively

obvious, but it’s also aneasy consequence of the

correspondence T 7→ −T

established in the proof!

ably not correspond to an actual moment in time. Some combinations of hand


positions simply cannot occur on a properly functioning clock. (Find examples!)So one must establish a correspondence between actual intervals of time.

Problem 5. Consider the sequence of strings

A, A‘BA, A‘BACA‘BA, A‘BACA‘BADA‘BACA‘BA, . . .

each string being obtained from the previous one by writing it twice andinserting the first unused letter between the two copies.

When we run out of letters, the last string will provide an affirmativeanswer to the problem. We will prove this by complete induction (Fact 24).For convenience, we will denote the n-th string of the sequence by Zn, andthe n-th letter of the alphabet by xn, so the sequence of strings is definedby the properties

Z1 = x1 and Zn+1 = Znxn+1Zn.

The “multiplication” on the right-hand side simply means that we are con-catenating (writing one after the other) the string denoted by Zn, the letterxn+1, and again the string Zn, is this order.

Obviously this process stops when we run out of letters, that is, aftern = 26. We claim that for any n 6 26, (a) the string Zn doesn’t haveidentical adjacent substrings, but (b) a pair of identical adjacent substringsappears as soon as one writes any one of the first n letters of the alphabeteither at the beginning or at the end of Zn.

Base of the induction. For n = 1, the statement is obvious.

Induction step. Suppose (a) and (b) are true for Z1, . . . , Zn−1, andconsider the n-th string, Zn = Zn−1xnZn−1.

Suppose Zn has two identical adjacent substrings. They cannot containthe central letter xn, since there is only one copy of it. Therefore they lieboth to the left or both to the right of the central letter; that is, they’reidentical adjacent substrings of Zn−1. But such substrings cannot exist, bythe induction assumption. This proves statement (a) for Zn.

To prove (b), again we proceed by cases. Suppose we write after Zn oneof the first n letters of the alphabet. If the letter we wrote is the centralletter xn, the result is two copies of Zn−1xn. If we wrote any other letter xk,the string now ends with Zn−1xk, with k < n. But the induction assump-tion says that Zn−1xk contains two identical adjacent substrings somewhere.Either way, we’ve found the desired identical substrings. The argument alsoapplies if write a letter to the left of Zn instead of to the right.

Having proved the induction, we now just take n = 26, so all the lettersof the alphabet are allowed. The word Zn thus constructed has 226 − 1letters, or more than 67 million!

Remark. This construction is important in combinatorics and the theory of semi-groups. Try to prove that in any infinite sequence of letters of the alphabet,there must be somewhere a string in the pattern of Zn, for any n. By a stringin the pattern of Zn we mean one that is obtained from Zn by replacing each

46 SOLUTIONS

letter xi by a fixed nonempty string Xi; for instance “abracadabra” is a stringin the pattern of Z2 (take X1 = abra and X2 = cad).

Problem 6. The circle tangent to BC and to the extensions of sides ABand AC is an excircle of the triangle ABC. Its center O is the intersection ofthe bisectors of the angle A and the exterior angle at vertex B; see Fact 16.

A

B

C

D

O

α/2α/2

γ

β

Let α, β, γ be the angles of △ABC. Then\BAO = α/2 and \CBO = (α + γ)/2, and

\ABO = β + \CBO = 90◦ + β/2.

From △AOB we see that

\AOB = 180◦ − α/2 −(90◦ + β/2

)

= 90◦ − α/2 − β/2 = γ/2,

since α + β + γ = 180◦. On the other hand,we have \AOB = 1

2\ADB, because this isthe angle inscribed in the circle centered at D. Hence \ADB = γ.

Thus, \ADB = \ACB. By the converse of the theorem on inscribedangles, the points A, B, C and D lie on one circle.

Remarks. 1. The points O, C, D lie on a straight line since \AOC = β/2 and\AOD = β/2 (prove it).

2. The statement is true for the inscribed circle as well (prove it).

Level B

Problem 1. We first draw the perpendicular to ABthrough the endpoint A. Clearly, \A < 90◦ if andonly if B and C lie on the same side of this line.Applying the same argument to B, we see that thelocus of points C such that \A < 90◦ and \B < 90◦

is the strip bounded by the perpendiculars to AB atboth A and B. (See figure on the right.)

A B

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

A B

Next we construct the circle with diameter AB.From Fact 14 we know that the points C outside thiscircle satisfy \ACB < 90◦. Taking the intersectionwith the strip already found, we see that the locus ofpoints C such that the ABC is acute is the shadedset to the left, not including the boundary.

Finally we study the condition that the angle Ais intermediate between the others, that is, either

\B 6 \A 6 \C or \C 6 \A 6 \B. (1)

Since the greater angle of a triangle subtends the longer side, the condition\B 6 \A 6 \C is equivalent to

AC 6 BC 6 AB.


The points equidistant from A and B are those on the perpendicular bisectorL of the segment AB; therefore AC 6 BC if and only if C is in the half-planedetermined by L and containing A.

At the same time, BC 6 AB if and only if C lies in the circle of centerin B and radius AB. Thus, the locus of points satisfying AC 6 BC 6 ABis the shaded set in (a) below:

A B

(a)

A B

(b)

A B

(c)

Similarly, the condition \C 6 \A 6 \B is equivalent to AB 6 BC 6 AC,and the corresponding locus is depicted in diagram (b) above. The union ofthe sets in (a) and (b), shown in diagram (c), is therefore the locus of pointsC satisfying (1). There remains to draw the intersection of this locus andthe one calculated immediately before (1).

Problem 2. We look at the first several terms of the sequence, in searchof patterns. We find x3 = 9, the least nonprime greater than 2x2 − x1 = 8; The original omitted the

first two terms of thesequence of differences,

but a person would

normally look at the whole

sequence, 2,3,5,6,7; andthen there is not much

justification for (∗). I’m

supplying a heuristic path

that fully motivates theconjecture.

then x4 = 14 is the least nonprime greater than 2x3−x2 = 12 (because 13 isprime). Continuing we obtain x5 = 20, x6 = 27, and so on, where each time— after that unfortunate 13 —the least nonprime greater than 2xn−1−xn−2

appears to be just 2xn−1 − xn−2 + 1, a composite number already.So we will conjecture that, for n > 4, the number 2xn−1 − xn−2 + 1

is composite and so equals xn. If this is so, we can write xn − xn−1 =xn−1−xn−2+1 for n > 4; that is, the differences between successive elementsof the series form an arithmetic progression:

x4 = 14= x3 + 5, x5 = 20= x4 + 6, x6 = 27= x5 + 7, . . . .

or, in compact form,

xn = xn−1 + n + 1 for n > 4. (∗)If the differences form an arithmetic progression, the numbers xn themselvescan be calculated as the sum of an arithmetic progression. That is,

xn = x4 +6+7+ · · ·+(n+1)

= 2+3+4+5+6+7+ · · ·+(n+1)︸︷︷︸

n terms

= 12n

(2×2+(n−1)

)= 1

2n(n+3).

Here we broke x4 down in order to extend the arithmetic progression back-wards, making the calculation less messy; but of course the formula for thesum of an arithmetic progression could have been applied earlier.

48 SOLUTIONS

Subject to our conjecture, then, we conclude that xn = 12n(n + 3), so

x1000 = 501500.There remains to prove our conjecture: if n > 4, then 2xn−1 − xn−2 + 1

is a composite number and so equals xn. We do this simply by testing ourformula for xn to see if it always gives a composite number for n > 4. Thisis indeed so: if n is even, xn = 1

2n(n + 3) has a factor n/2 > 1, and if n isodd, it has a factor (n + 3)/2 > 1.

Remark. It may seem that our argument was circular, in that we used an expressionderived conjecturally in order to prove the conjecture. But if fact the argumentis valid because we actually found a formula for the terms xn of our sequence(it doesn’t matter how) and then we proved that the formula gives a sequencesatisfying the conditions of the problem, namely, xn is the least nonprime greaterthan 2xn−1 − xn−2. (Our construction says that xn, for n > 4, is the leastinteger greater than 2xn−1 − xn−2; and this integer is nonprime.)

Problem 3. Assume that at some step we get a triangle similar to theinitial one. All its angles are multiples of 20◦.

Lemma. All angles of the preceding triangle, and, more generally, of allpreceding triangles are multiples of 20◦.

Proof. Let the triangle with angles α, β, γ be a child of thetriangle with angles α1, β1, γ1, obtained by cutting alongan angle bisector. Arrange the labeling so that α1 = 2αand β = β1, as in the figure. Since the sum of theangles of any triangle is 180◦, we deduce that

α α

γ1γβ=β1

γ1 = 180◦ − α1 − β1 = (α + β + γ) − 2α − β = γ − α.

Clearly, then, if α, β, γ are multiples of 20◦, so are α1, β1, γ1, the anglesof the parent triangle (see Fact 5). But then the angles of the grandparenttriangle are also multiples of 20◦, and so on. ˜

This, however, cannot be true even after the first cut of the initial trian-gle: if we start with an angle of 20◦, we get an angle of 10◦, and if we startwith the angle of 140◦ we get an angle of 70◦. The contradiction shows thatit is impossible to get a triangle similar to the initial one.

Remarks. 1. For any positive integer n, it is possible to find an initial triangle sothe construction leads to a similar triangle after n appropriate bisector cuts,and no sooner.

2. For a similar situation, see Problem 64.9.4. REF 64.9.4

Problem 4. A classmate of Pete’s can have between 0 and 28 friends.Of these 29 possibilities, we know that 28 occur. Thus, either there is aclassmate who has 28 friends, or there is a classmate who has no friends.But if a classmate is friends with everyone, then everyone has at least onefriend. So it’s not possible for both 0 and 28 to occur: either the tally offriends is 1, . . . , 28 for the various classmates, or it is 0, . . . , 27.


Denote the classmate of Pete’s with the most friends by A and the onewith the least friends by B. In the first case just considered, A is everybody’sfriend, while B has only one friend, A. In the second case B has no friends,while A is friends with everyone except B. In either case, A is a friend ofPete’s, and B is not.

Now let’s send A and B to another class. Then Pete is left with 26classmates and everybody has one fewer friend in this class than before.Thus each classmate still has a different number of friends in this class.

We again send the classmate with most friends and the one with leastfriends to another class. We can keep doing this until we have sent away 14pairs of classmates. Each pair included exactly one friend of Pete’s, so Petehad 14 friends in his class.

Remarks. 1. Several ideas work together toward the solution: friendship relationis assumed to be a symmetric relation, it’s useful to look at extreme cases, andwe are able to apply inductive descent.

2. There is one very short, but wrong solution: Replace all friendships by non-friendships, and all nonfriendships by friendships. Then Pete’s classmates willagain each have a different number of friends, and hence Pete will again have x “hence Pete will again

have x friends”: Why??friends. Therefore

x = 28 − x.

Where is the mistake? If we proved that the problem has a unique solution, allwould be OK. Nevertheless, this argument can help in guessing the answer.

3. Solve the same problem if Pete has 27 classmates.

4. This is an extension of a problem you may have heard before: Prove that inany group of more than one person, there are two people with the same numberof friends within the group (the number can be zero).

Problem 5. In the second identity set y = z. Then we get

(x ∗ y) + y = x ∗ (y ∗ y) = x ∗ 0.

Thus, x ∗ y = x ∗ 0− y. It remains to compute x ∗ 0. For this, set x = y = zin the second identity; we get

x ∗ 0 = x ∗ (x ∗ x) = (x ∗ x) + x = 0 + x = x.

Thus, x ∗ y = x ∗ 0 − y = x − y, so 1993 ∗ 1935 = 1993 − 1935 = 58.

Remark. Check that with x ∗ y = x − y, both identities are indeed verified.

Problem 6. First solution. (See figure.) Let B′ be thereflection of B in the line AM . Since AB = AB′ and\BAB′ = 2\BAM = 60◦, the triangle ABB′ isequilateral. Hence, the points A, C, B′ lie on thecircle of center B and radius AB. Since the inscribedangle is half the central angle, it follows that \ACB′ = 30◦.Then, since \MCA = 150◦, the points C, B′, M lie on the same line. Now,by construction, AM is the bisector of \BMB′, and so also of \BMC.

A

B

M

C

B′

60◦

30◦

50 SOLUTIONS

Second solution. Through the points A, C and M , draw acircle with center O (see figure). Since \ACM = 150◦,the arc AM has measure 60◦, so the triangle MAOis equilateral. Point B lies on a symmetry axis of△MAO, since \BAM = 30◦. It follows that

\AMB = \AOB = 12\AOC = \AMC,

where the second equality is a corollary of thecongruence △ABO = △CBO. A

B

M

C

O

Level C

Problem 1. The least period length of a decimal divides any other periodlength of that decimal; see Fact 4. (We regard any terminating decimal ashaving minimal period length 1).

Lemma. If k is a (not necessarily minimal) period length for each of twodecimals P and Q, then k is also a period length for P + Q and P − Q.

Proof. Recall (Fact 13) that a recurring decimal P with period k can bewritten in the form

P =X

10l(10k − 1),

where X is an integer. Similarly, we can write

Q =Y

10m(10k − 1),

where Y is an integer. Without loss of generality we may assume that l > m.We obtain

P ± Q =X ± 10l−mY

10l(10k − 1),

where again the numerator is an integer, so the decimals corresponding toP + Q and P − Q are recur with periods of length k. ˜

Now we can solve the problem. We know that A has least period 6 andB has least period 12. The lemma implies that 12 is a period length ofA+B, so the divisors of 12 are the candidates for the least period of A+B.But 6 cannot be a period length of A+B, otherwise B = (A+B)−A wouldhave a period of length 6, contradicting the assumption. Hence the leastperiod of A + B cannot be a divisor of 6.

Two possibilities remain for the least period of A + B: 12 and 4. Bothoptions are possible:

A = 0.(000001), B = 0.(000000000001), A + B = 0.(000001000002);

A = 0.(000001), B = 0.(011100110110), A + B = 0.(0111).

Remarks. 1. How would someone come up with this last example? By workingbackwards! Take any decimal of least period 4 and call it A + B. Now subtractany decimal B of least period 6. The result has least period 12. (Why?)


2. We can find all possible least periods of a sum of two decimals. Let m, n, k bethe least periods of the decimals P , Q and P + Q, respectively. Then k divideslcm(m,n), by the proof of the lemma; but at the same time m divides lcm(n, k),and n divides lcm(m, k). Let p1, . . . , ps be all the prime divisors of lcm(m,n),and write

m = pα1

1 . . . pαs

s , n = pβ1

1 . . . pβs

s ,

where the exponent are allowed to be 0; see Fact 10. The preceding argumentsimply that k = pγ1

1 . . . pγs

s , where

γi =

{max(αi, βi) if αi 6= βi,

any number from 0 through αi if αi = βi.

The opposite is also true: any such k can be a least period of a sum of twodecimals whose least periods are m and n, respectively.

Problem 2. We can certainly make a tour of the eightrooms in any hall, going clockwise, say. We can also com-bine tours of two adjacent halls (top diagram) by using twoof the three doors that separate the two halls (middle dia-gram).

Now the key observation is that we can always add ahall to our tour, so long as it is shares a wall with a hallalready on the tour. For example, the bottom diagram addsa third hall to the first two. We can continue adding one

Problem 2: I redrew themiddle diagram so it’s

compatible with the

bottom one.

The remark in the original

didn’t seem right as

written, if “a hall” means

eight rooms (as in theproblem). Example:

[0, 3]×[0, 3] ∪ [3, 6]×[2, 5].

One could make the

remark true by addingmore conditions, but I

don’t think it would add

insight.

hall at a time, until the tour includes all the halls of thecastle.

Problem 3. The answer to (a) is that it is notalways possible; see figure on the right for a

counterexample. In it, AC = 1000 m, AB > 1400 m,CD = 1m. The segment AB divides the river intotwo parts, and boat going down the river end to endmust cross this segment at some point. The distance from anypoint of AB to one of the banks exceeds 700 m.

Part (b) turned out to be unexpectedly difficult, unlessislands are allowed, in which case a counterexample is nothard to find (see figure on the right; the river has width justunder 1 km).

We (the authors of the book) believe that the intention of the authorsof the problem was to not allow islands, and we were at first unable thesolve it. A contest was announced, and a solution was found through thejoint efforts of A. Akopyan, V. Kleptsyn, M. Prokhorova, and the authors.It turned out to be more an example of research-level mathematics problemthan an Olympiad problem!

The solution is given in full in Appendix B, but the main steps— eachof which requires a whole argument in itself —are these: as follows:

• Show that no disc of radius 750 m lies entirely in the river.

52 SOLUTIONS

• Deduce that if a point in the river is not within 750 m of one bank, itlies within 750 m of the other bank.

• Use the (possibly very complicated) boundary of the set of points thatlie within 750 m of the left bank to prove that there is a path thatremains within 800 m of both banks.

Problem 4. Since we’re taking fractional parts of numbers, it’s a goodidea to visualize the real line “rolled up” into a circle of unit length, sonumbers with the same fractional part correspond to the same point on thecircle. (See figure on the right, and compare the remark after the solution ofProblem 60.10.6.) It is clear that pn = 0 if xn = {an + b} REF 60.10.6

lies on the upper semicircle [0, 12), and pn = 1 if xn lies on

the lower semicircle. Moreover, xn = {an + b} is the pointon the circle obtained from {b} by n consecutive rotationsthrough the arc {a}.

0

ba+b

2a+b

1

2

(a) A rational value of a leads to a periodic pattern for the sequence pn.(Why?) Trying out b = 0 and the simplest possible fractions for a, we seethat all 4-tuples of 0s and 1s can occur: The table in the original

omits 0100• The “bigon” with a = 12 gives the sequence 010101. . . , so 0101 can

occur. (We only list 4-tuples starting with 0, since we know the com-plementary ones can be obtained by choosing b = 1

2 instead of b = 0).

• The equilateral triangle with a = 13 gives the sequence 001001. . . , so

we get 0010 and 0100.• The square with a = 1

4 gives 0011 and 0110.

• The octagon with a = 18 gives 0000, 0001, 0011 again, and 0111.

(b) Let’s look at runs of consecutive 0s and 1s in the sequence p0, p1, p2, . . . . I’ve rewritten the solution

based on the original’s

Remark, because such asolution is not difficult,

and seems more

enlightening than the

previous analysis of cases.Also it allows an

immediate answer to an

alternative interpretation

of part (a) - See Remark 2below.

We will show that all such runs have either the same length or lengthsdiffering by 1, except that the first run may be short (it can start in themiddle, so to speak). In particular, the string 00010 can never occur for anya and b, because it would mean the sequence has a run of 0s of length atleast 3, but also a run of 1s of length 1.

The reason the runs have almost uniform length is that the spacingbetween xn and xn+1 around the circle is the same for all n. More precisely, if

1

2i> |a| >

1

2(i+1), for some n > 1,

a change from 0 to 1 must be followed by a change from 1 to 0 after exactlyi or i+1 entries, because the number of contiguous arcs of length a that canfit in a semicircle is either i−1 or i, depending on where the first endpointstarts. (We also need the fact that the semicircle is open at one end andclosed at the other).

This proves our claim for a ∈ (0, 1). We needn’t consider values of aoutside [0, 1) because only the fractional part of a matters. Finally, fora = 0, there is only one run, of infinite length. So our claim is proved in allcases.


Remarks. 1. This problem is relevant in symbolic dynamics; see Smale-horseshoe. CITE Smale-horseshoe

2. The original Olympiad problem said “. . . the sequence determined by some aand b”, which is perhaps ambiguous: the question might be whether all possible4-uples of 0s and 1s can occur for some fixed choice of a and b. Our solutionto part (b) shows that the answer is no: any values of a and b that allow thestring 0000 cannot allow the string 0100.

Problem 5. Let m be the number of plants in a certain good classifier. Letus estimate the total number S of distinctions between all pairs of plantswith respect to all features. There are 1

2m(m−1) pairs of plants, and each

pair differs in at least 51 features, so S > 51 · 12m(m−1).

There is another way to look at S. Let mi be the number of plantshaving feature i. The number of pairs of plants that can be distinguished bymeans of feature i is (m−mi)mi. Summing over all the features, we obtainagain the total number S of distinctions:

S =

100∑

i=1

(m−mi)mi.

Now, the arithmetic mean of m−mi and mi is 12m, so the inequality between

the arithmetic and geometric means (Fact 26) gives (m − mi)mi 6 14m2.

Therefore S 6 100 · 14m2 = 25m2. Thus

51 · 12m(m−1) 6 S 6 25m2, (1)

which after simplification gives 12m(m−51) 6 0. Hence m 6 51.

It remains to prove that m 6= 51. If m = 51, we have a strict inequalitymi(m − mi) < 1

4m2 (since we have an integer on the left and a fraction onthe right). That means the second inequality in 1 is strict, implying m < 51.This contradiction implies that a good classifier cannot describe more than50 plants.

Remarks. 1. One might be tempted to guess that a good classifier can describe 50plants, but with a bit more work one can show that this is far from true. Whatwe do is extend the classifier with one extra feature, which we declare presentif and only if an even number of the original 100 features were present. Thisnew classifier has 101 features and any pair differs by at least 52 of them: if apair differs in 51 of the original features, it must also differ in the new feature.(Why?)

Now the same arguments as above yield

52 · 12m(m−1) 6 S 6 101 · 1

4m2,

which leads to m 6 34. Thus, the new classifier (and hence the initial one) candescribe at most 34 plants.

2. This problem is related to error correcting codes. Replace plants by messagesand descriptions of features by length-n strings of bits (0s and 1s). The classi-fier— which is now a collection of m strings— is said to be a code of length n.The minimum number d of differences between two sequences in the code is thecode distance; in our problem d = 51.

54 SOLUTIONS

If we take a message in the code and distort it arbitrarily by flipping nomore than 1

2 (d − 1) positions, we are still able to recover the original message,simply by selecting the message that shares the most bits with the distortedone. (There will be at most 1

2 (d−1) differences, whereas the comparison with

any other string will give at least d− 12 (d−1) > 1

2 (d−1) distances.) This is whythe code is said to be error-correcting.

One important open problem of code theory is finding the maximal size(number of different messages) of a length-n error-correcting code with codedistance d, for arbitrary n and d. A famous result in this direction is thePlotkin–Levenshtein theorem, which establishes an upper bound (called thePlotkin bound) in the case d > 1

2n, and provides certain natural conditions thatguarantee the bound can be achieved. In our problem we have n = 100 andd = 51, so Plotkin’s bound applies, and its value is 34. This bound is achievable:there does exist a code with 34 messages.

Problem 6. Let the remaining vertices of the square onside AB be D and E, so the square is ABDE, and set γ =\ACB. (See figure on the right.) Applying the intercepttheorem to the triangle ADC we see that CD = 2OM ;similarly, CE = 2ON . Therefore it suffices to find themaximum of CD + CE = 2(OM + ON).

First solution. On side BC of △ABC, construct asquare CBD′E′ outwards. (See figure below.) TrianglesABD′ and DBC have two sides and the included angle equal, so CD = AD′.

γ

A B

C

DE

NM

O

A B

C

DE

D′

E′

c

c

a

a√

2

b

In the triangle ACD′, two sides are known: AC = b and CD′ = a√

2.Moreover, \ADC = γ + 45◦. The side AD′ attainsits maximal value when the triangle degenerates intoa segment, so

max(CD) = max(AD′) = b + a√

2

for γ = 135◦. Similarly, we have max(CE) = a+b√

2,again for γ = 135◦. Thus, each of OM and ONattains its maximum when γ = 135◦, and so doestheir sum:

max(OM + ON) =1 +

√2

2(a + b).

Second solution. Let \CAB = α, \ABC = β, c = AB, d = CD, e = CE.The law of cosines for △ABC gives

c2 = a2 + b2 − 2ab cos γ.

Next we apply the law of cosines to △AEC:

e2 = b2 + c2 − 2bc cos(90◦ + α) = b2 + c2 + 2bc sin α,

since cos(90◦ + α) = − sin α. Substituting c2 from the first formula into thesecond, we get

e2 = 2b2 + a2 − 2ab cos γ + 2bc sin α.


The law of sines for △ABC implies that sinα = (a/c) sin γ. Therefore

e2 = 2b2 + a2 + 2ab (sin γ − cos γ).

Similarly, d2 = 2a2 + b2 + 2ab (sin γ − cos γ).Hence both e and d attain their maximum values when sin γ−cos γ does,

which is to say, when γ = 135◦. Hence the maximum of e+d is also attainedfor γ = 135◦, and it equals 1

2(1 +√

2)(a + b).

Level D

Problem 1. If tan(α + β) is defined, then

tan (α + β) =tanα + tanβ

1 − tanα · tanβ=

p

1 − tanα · tan β. (1)

The product of tangents is related with p and q as follows:

q =1

tanα+

1

tanβ=

tanα + tanβ

tanα · tanβ=

p

tanα · tanβ. (2)

We deduce from (2) that either p and q are both zero or they are bothnonzero; therefore we have only these cases to consider:

1. If p = 0 = q, then (1) implies tan(α + β) = 0. We have to verify herethat the denominator of (1) does not vanish. Indeed, if it did, we’d have

tanα = − tanβ, so 1 − tanα · tanβ = 1 + tan2 α > 0.

2. If p 6= 0, q 6= 0 and p 6= q, then (2) implies tanα · tanβ = p/q, so (1)implies tan(α + β) = pq/(q − p).

3. If p 6= 0, q 6= 0 but p = q, then tan(α + β) is not defined.

Problem 2. We will construct a subdivision into squares satisfying thecondition fo the problem. We first divide the unit square into four equalsquares. The little squares that intersect the main di-agonal only at a vertex will be called level-1 squares.We subdivide each of the remaining squares into fourequal squares of side 1

4 . The little squares of side 14

that intersect the main diagonal only at a vertex willbe called level-2 squares. We continue in this way (seefigure) until we have 500 levels of squares.

12

2

3

3

3

3

1

2

1

4

1

8 1

2

2

3

3

3

3

There are 2k squares at level k, each with side 2−k. Hence the totalperimeter of all level-k squares is 4, and the total perimeter of all squaresintersecting the diagonal is 4 · 500 > 1993.

Remarks. 1. A stronger result is in fact true: The unit squarecan be partitioned into squares in such a way that the sum The previous sketch was

not convincing since it’s

not possible to “enlargethe sides of the small

squares” all at once.

of perimeters of squares intersecting the main diagonal ina segment exceeds any given number. The construction isa modification of the previous one; we make the under-the-diagonal level-1 square have side 3

5 , say, instead of 12 , while

the complement is subdivided into two squares of side 25 along the diagonal,

56 SOLUTIONS

plus an irregular area that we leave aside, knowing that it can be subdividedinto tiny squares (since 3

5 is a rational number).We then repeat the procedure on the two new squares along the diagonal to

get level-2 squares, and so on. All level-k squares now have side (25 )k, instead

of (12 )k as before, and there are 2k−1 of them, instead of 2k. So now the total

perimeter of level-k squares is 2 · (45 )k; that is, it decreases in a geometric pro-

gression instead of being the same for all levels. But the ratio of the progressioncan be made arbitrarily close to 1, by replacing the number 3

5 by some rationalnumber very close to 1

2 . So the sum of perimeters can be made as large asdesired.

2. This problem arose during a lecture of the important mathematician N. N. Luzin,when he wanted to shorten the proof of a theorem of Cauchy (Luzin loved to im-provise). Luzin conjectured: Fix a curve of a bounded length in the unit squareand consider a partition of the square into little squares. The total perimeter ofthe little squares that intersect the curve is bounded by a constant depending onthe curve only. A. N. Kolmogorov, who was to become an even more famousmathematician, was at the lecture and soon constructed a counterexample.

Problem 3. For n = 2 the answer is obviously 1. So assume n > 3.

The desired number is at most n, because we can exhibit an arrangementof points generating only n pairwise nonparallel lines: the vertices of aregular n-gon,

We prove this by showing that there are as many nonparallel lines asthere are axes of symmetry of the polygon. To each side and each diagonal,we assign an axis of symmetry: the perpendicular bisector of this side ordiagonal. Two sides or diagonals have the same perpendicular bisector ifand only if they are parallel. Therefore we just need to count the axes ofsymmetry of a regular n-gon.

For n odd, each axis of symmetry passes through a unique vertex. Hence,the total number of axes of symmetry is n.

For n even, an axis of symmetry passes either through a pair of oppositevertices or through the midpoints of opposite sides. There are n/2 axes ofeither type. Hence, the total number of axes of symmetry is again n.

Now we prove the converse: The desired number is at least n. That is,for any arrangement of n points, no three of which lie on a line, we canalways find n pairwise nonparallel lines.

It is easy to find n − 1 lines: just take a point and consider all the linesfrom it to the other points. It is a bit harder to construct the n-th line.

Among the n points, take one, say O, having highest y-coordinate. Place We cannot take for

granted the existence and

properties of the convex

hull; for someone who hasnot seen them before, they

require a bit of

justification. So Isubstituted a proof that

doesn’t depend on them,

and placed the convex hull

in a remark.

the origin there. Of the remaining n− 1 points, choose A such that the rayOA makes the biggest possible angle with the positive x-axis,and B such that the ray OB makes the least possible anglewith the positive x-axis. All the rays connecting O withthe other points lie inside the angle AOB, and none ofthese rays can be parallel to the line AB. (Why?) Nowwe just take AB as the n-th line. (See figure.)

A B

O


Remarks. 1. The points A, O and B are adjacent vertices of the convex hull of then points, that is to say, the smallest convex polygon containingall the points. (The convex hull of the seven points in thepreceding figure has five vertices; see figure on the right.)It can be shown that this polygon has vertices among then points, and contains all n points.

2. The given nonlinearity condition cannot be replaced by the weaker one that thepoints do not all lie on a single line. For example, for the set consisting of thevertices of a regular 2k-gon and its center, there are only 2k nonparallel lines.

Problem 4. Clearly, the worst-case scenario is when all marble populationsoccur at the start; that is, we have n boxes—where n = 460 in part (a)and n = 461 in part (b) —and there is a box with j marbles for everyj = 1, 2, . . . , n. So from now on we assume this is the situation.

We start with the observation that a box having m = qk + r marbles(0 6 r < k) before a step with group size k with be left with q + r marblesafter that step; see Fact 7.

Lemma 1. After the first step, with group size k, there is a number f(n, k)such that the marble populations are exactly all the numbers in the interval[1, f(n,k)], and no others.

Proof. Let f(n, k) be the highest marble population in a box at the end ofthe step. We show by reverse induction on j (Fact 24) that there is a boxwith exactly j marbles, for all j = 1, . . . , f(n, k).

Suppose this is true for some j. Then there exists numbers m (startingpopulation), q (quotient) and r (remainder) such that 1 6 m 6 n, 0 6 r < k,m = qk + r, and j = q + r.

If r > 0, we look at the box that started off with m−1 marbles. If r = 0,we take the one that had m − k marbles. Either way, we have found a boxholding exactly j − 1 marbles at the end of the step. This completes theinduction step and proves the lemma. ˜

This lemma effectively reduces the situation at the end of the first stepto the original situation, with a smaller number of boxes (we just ignoreboxes with duplicate populations). There remains to select k so the newlargest population f(n, k) is as low as possible.

Lemma 2. The largest population f(n, k) after the first step is given by

f(n, k) =[n+1

k

]

+ k − 2, (1)

and its value is minimal when k =[√

n+1]+ 1. The original had “it is

easy to see” regarding the

maximality of a certainbox; but it seems to me

less obvious that other

things that you’vecarefully spelled out. So

I’ve spelled this out too.

Proof. The function “population of a box at the

k Qk n

end of the first step” grows by 1 when its argument(the initial population) grows by 1, except when theargument is 1 less that a multiple of k, in which casethe function drops by k−2 (see figure, where k = 5).

58 SOLUTIONS

Thus the maximum of the function is always achieved for an argument valuethat is 1 less than a multiple of k, and is as large as possible under thiscondition. So let Q = [(n + 1)/k] be the quotient of the division of n + 1 byk. The box that started off with Qk−1 = (Q−1)k+(k−1) marbles achievesthe maximum, and its new population is Q+k−2 marbles. This proves (1).

To find the value of k that minimizes f(n, k), we first write

f(n, k) =[n+1

k+ k

]

− 2.

The function inside the brackets decreases in the interval (0,√

n+1 ) andincreases in the interval (

√n+1, n]. Since [x] does not decrease, the function

f(n, k) attains its maximum either at k =[√

n+1]

or at k =[√

n+1]+ 1.

It remains to show that we always have

f(n, [√

n+1 ] + 1) 6 f(n, [√

n+1 ]).

Let [√

n+1 ] = s. Then

s2 6 n+1 < (s+1)2. (2)

Thanks to (1) it suffices to prove that[n+1s+1

]

<[n+1

s

]

.

Equation (2) implies that Original had (1), but I

think (2) is meant.[n+1s+1

]

6 s and[n+1

s

]

> s.

Therefore, it suffices to prove it is not possible for both sides to equal s. But[n+1

s

]

= s =⇒ n+1s

< s+1 =⇒ n+1s+1

< s =⇒[n+1s+1

]

< s.

The lemma is proved. ˜

We are now ready to find the minimum number of steps required for anystarting value of n. We simply apply repeatedly the function

g(n) = f(n, [√

n+1 ] + 1),

which represents the maximum population after one step, with an optimalchoice of k. We verify by direct computation that, after five applications,

g(g(g(g(g(460))))) = 1 but g(g(g(g(g(461))))) = 2.

The sequences of steps for n = 460 and n = 461 are as follows:

n step k n

460 1 22 461

40 2 7 41

10 3 4 11

4 4 3 5

2 5 2 3

1 2


To recapitulate, we spell out the details of the argument for n = 461.Lemmas 1 and 2 say that if the boxes start with all populations from 1through 461, then after step 1 there remain all populations from 1 throughg(461) = f(461, 22) = 41; after step 2 there remain all populations from 1through g(41) = 11; after step 3, all populations from 1 through 5; after step4, all populations from 1 through 3; and after step 5, populations 1 and 2.Thus, for n = 461 it is not always possible to be left with a single marble ineach box.

Remarks. 1. Instead of proving that

f(n, [√

n+1 ] + 1) 6 f(n, [√

n+1 ]),

one could just check case by case.

2. This problem originated in computer programming. A text containing one ormore blank spaces between words had to be processed so as to leave preciselyone blank between words. The programmers solved the problem by recursivelyapplying the operation of selecting a natural number k and replacing everygroup of k consecutive blanks by a single blank.

Problem 5. (a) A possible solution is shown in figure. The function isdefined by

f(x) =

−12 − x for − 1 6 x < −1

2 ,

x − 12 for − 1

2 6 x < 0,

0 for x = 0,

x + 12 for 0 < x 6 1

2 ,12 − x for 1

2 < x 6 1.

1

2

−1

2

1

2

−1

2

−1

1

−1

1

0

(b) We will show that a function satisfying the conditions of the problemcannot exist on the interval (−1, 1). The case of the function defined on thewhole real axis is analogous.

Suppose, to the contrary, that such a function f(x) exists. Its graphis mapped to itself under a clockwise 90◦ rotation: if (x, y) is a point onthe graph, we have y = f(x), so f(y) = −x by assumption, and the point(y,−x) also belongs to the graph; but this is precisely the image of (x, y)under the specified rotation.

By applying the 90◦ rotation repeatedly we see that the graph maps toitself also under 180◦ and 270◦ rotations.

This implies that the coordinate axes cannot intercept the graph exceptat the origin: any other intersection would have three symmetrically placedcopies, and in particular there would be two distinct intersections of thegraph with the y-axis, which is impossible.

Now consider the portion of the graph that lies within the open firstquadrant {(x, y) : x > 0, y > 0}. Recalling the assumption that the graphis a union of finitely many points and line segments, we can write thisintersection as

L1 ∪ L2 ∪ · · · ∪ Ln ∪ P1 ∪ P2 ∪ · · · ∪ Pm,

60 SOLUTIONS

where the Lk are line segments and the Pj are points. We may assume thatthe line segments Lk are pairwise disjoint and open (meaning the endpointsare excluded) and that the points Pj are distinct and do not belong to anyof the line segments Lk.

For each k and j, let Jk be the line segment obtained from Lk by aclockwise 90◦ rotation, and let Qj be the point obtained from Pj by the samerotation. All these points and segments lie on the open fourth quadrant, andin fact

J1 ∪ J2 ∪ · · · ∪ Jn ∪ Q1 ∪ Q2 ∪ · · · ∪ Qm

is precisely the intersection of the graph with the fourth quadrant. (Why?)Since we already know the graph does not intersect the positive x-axis,

we conclude that the intersection of the graph with the half-plane x > 0consists of 2n line segments (the Lk and Jk) and 2m points (the Pj and Qj).None of the line segments can be vertical. Thus the projections of all theseline segments and points on the x-axis partition the interval (0, 1) into 2nopen intervals and 2m points. But it is impossible to divide an open intervalinto an even number of subintervals using an even number of points! Wehave reached a contradiction.

Problem 6. Clearly a shortest flight touches each faceof the tetrahedron exactly once. Let the tetrahedronhave vertices ABCD, and let the shortest flight bethe space quadrilateral EFGH , where E ∈ △ABC,F ∈ △BCD, G ∈ △ABD, and H ∈ △ACD(see figure). Our job is to find the perimeter of thequadrilateral EFGH .

A

B

C

D

E

FG

H

Draw the symmetry plane of the tetrahedron containing CD and per-pendicular to AB. Let E1F1G1H1 be the reflection of EFGH in this plane(so E1 and G1 lie on the same faces as E and G, respectively, while F1 lies onthe same face as H , and H1 lies on the same face as F ). The quadrilateralsEFGH and E1F1G1H1 have the same perimeter.

Lemma. In any space quadrilateral, the distance between the midpoints oftwo opposite edges is less than or equal to the mean of thelengths of the remaining edges.

Proof. Let KLMN be the quadrilateral and let P andQ be the midpoints of KL and MN , respectively (seefigure). Denote by R the midpoint of the diagonalLN . We have PR = 1

2KN and RQ = 12LM . Hence

PQ 6 PR + RQ = 12(KN + LM). ˜

K

L

M

N

P

Q

R

Denote the midpoints of the segments EE1, FH1, GG1, and HF1 byE2, F2, G2, and H2, respectively. These points also lie on the faces of thetetrahedron. By the lemma, the perimeter of E2F2G2H2 is no greater thanthat of EFGH . Moreover, E2 and G2 lie on the symmetry plane of the


tetrahedron containing CD; that is, they lie respectively on the mediansCT and DT of the faces ABC and ABD, where T is the midpoint of AB.

Next we repeat this symmetrization operation for the symmetry planeperpendicular to the first. That is, we reflect E2F2G2H2 in the symmetryplane of ABCD that contains AB, obtaining a quadrilateral E3F3G3H3, andthen we take the midpoints of the segments E2E3, F2H3, G2G3,and H2F3, obtaining a quadrilateral E4F4G4H4, whose verticesall belong to one of the two planes of symmetry of ABCDconsidered so far (one through AB and the other CD).

Specifically, the vertices E4 and G4 lie on CT and DT ,while F4 and H4 lie on the medians AS and BS of thefaces BCD and ACD, respectively, where S is themidpoint of CD. (See figure on the right.)

A

B

C

D

S

T

Again, the perimeter of E4F4G4H4 is no longer than that of E2F2G2H2,which as we know is no longer than that of EFGH . Hence, the perimeterof EFGH is at least 4d, where d is the distance between CT and BS.

It remains to construct a path of length 4d and find d. Let the commonperpendicular to CT and BS intersect CT at E0 and BS at F0. Let G0 bethe reflection of E0 in the plane ABS. It follows from symmetry that F0G0 isthe common perpendicular to BS and DT . Similarly we construct the pointH0 such that G0H0 is the common perpendicular to DT and AS and H0E0

is the common perpendiculars to AS and CT . The perimeter of E0F0G0H0

is 4d. We also have to prove that the bases of our common perpendicularslie on the faces of the tetrahedron, rather than on their extensions. Thiswill be checked below (we still have to calculate d).

Draw the plane through AB perpendicular to CT and takethe projection of the tetrahedron on this plane. We obtaintriangle ABD′, in which AB = a and D′T = a

√

2/3, bythe formula for the length of the altitude of the regulartetrahedron. (See figure on the right.) A B

D′

S′

T

The projection sends S to S′, the midpoint of D′T . Hence, d equals thedistance between T and the line BS′, because the common perpendicular isparallel to the plane of projection. It is also obvious that the base of theperpendicular dropped from T onto the line BS′ lies on the segment BS′

and not on its extension; hence F0 lies on the segment BS. We similarlyprove that the remaining vertices of the quadrilateral lie on the medians,not on their extensions.

In the right triangle BTS′, the legs BT = 12a and TS′ = 1

2a√

23 are

known. Hence

BS′ = 12a

√53 , d =

BT · TS′

BS′ =a√10

.

Remarks. 1. The solution implies that there are three suitable paths. (Why?)

62 SOLUTIONS

2. An analogous problem on the plane is known: A beetle crawls inside a trianglewith sides a, b, c. What is the shortest length of a path that visits each sideand returns to the initial point?

It the case of an acute triangle the answer is the path joining the basesof the altitudes; this is known as Fagnano’s problem. (See Coxeter, Chapter 4, CITE Coxeter

§ 5.) For a right or obtuse triangle the path degenerates into an altitude traveledtwice; see Problem 70 in ShklyarskiChentsovYaglom. CITE Shklyars-

kiChentsovYaglom

Year 1994 Olympiad

Level A

Problem 1. First solution. A bottle of mix used to require 16 of a jug of

apple juice and 110 of a jug of grape juice. Hence, the volume of a bottle is

equal to 16 + 1

10 = 415 of the volume of a jug.

After the change in the recipe, a bottle of drink requires 15 of a jug of a

apple juice and 1x of a jug of grape juice. Hence, the volume of the bottle is

equal to1

5+

1

x=

4

15

of the volume of the jug. So we have an equation1x

=415

− 15

=115

. Thus,x = 15.

Second solution. To make 30 bottles of the drink according to the old recipe,one needed 5 jugs of apple juice and 3 of grape juice: a total of 8 jugs. Afterthe change, 30 bottles require 6 jugs of apple juice, and hence 8−6 = 2 jugsof grape juice. Now one jug of grape juice serves 30/2 = 15 bottles.

Problem 2. Let x and y be the three-digit numbers to be found. If wewrite three zeros after x, we get 1000x. If instead we write the digits of y,we get 1000x + y. See Fact 11. Thus we have the equation

7xy = 1000x + y. (1)

First solution. Divide both sides of (1) by x:

7y = 1000 +y

x.

The number y/x is positive and smaller than 10 since y 6 999 and x > 100.Therefore

1000 < 7y < 1010.

Dividing this inequality by 7 we get

14267 < y < 1442

7 .

64 SOLUTIONS

Since y is an integer, y is equal to either 143 or to 144. Let y = 143.Substituting this value of y into (1) we deduce that

7x · 143 = 1000x + 143.

Solving this equation we find that x = 143. If y = 144, a similar equationyields x = 18, which has too few digits.

Second solution. Rewrite (1) in the form

1000x = (7x − 1)y.

It is not difficult to see that x and 7x − 1 have no common divisors exceptfor 1 and −1. Indeed, if d is a common divisor of x and 7x − 1, then ddivides 7x, and hence it divides 1 = 7x− (7x− 1); see Fact 5. But 1 is onlydivisible by 1 and −1.

Thus 7x−1 divides the product 1000x and is relatively prime to x. Thisimplies that 7x − 1 divides 1000 (see Fact 9). Since

7x − 1 > 7 · 100 − 1 = 699 >1000

2,

the only possibility is 7x − 1 = 1000. Hence x = 143. Substituting back in(1), we find y = 143.

Remark. Compare with Problem 57.10.1. REF 57.10.1

Problem 3. Extend the segments BQ and BP so theyintersect the line AC at A′ and C ′, respectively. In thetriangle ABC ′, the segment AP is both an altitude and abisector (see figure). Therefore, this triangle is isoscelesat A, and AP is also a median: BP = C ′P . Analo-gously, the segment CQ is a median of △CBA′, i.e.,BQ = QA′. Hence PQ is a midline of △A′BC ′,that is, PQ ‖ A′C ′. But then PQ ‖ AC. A

B

C

PQ

A′ C′

Problem 4. Let the initial configuration be a 1×1 square on a piece of graphpaper. Whenever a grasshopper jumps, it must land on a crossing of the grid,because the jump is symmetric with respect to a crossing and starts from acrossing (see figure; we use the fact that a square grid is centrally symmetricwith respect to any of its crossing). Another way to say thisis that the x- and y- coordinates of each grasshopper mustremain whole numbers, because the differences in x- andy-coordinates between the two grasshoppers involved in ajump are replaced by their negatives.

In particular, since the distance between two grasshoppers cannot beless than 1, at no time can their positions coincide with the corners of asmaller square than the original one.

Now suppose the grasshoppers could land on the corners of a largersquare than the original one. The reverse of a legal jump is also a legaljump, so by performing the jumps backwards, we’d be in a situation where


grasshoppers starting at the vertices of a (larger) square ended up occupyingthe vertices of a smaller square. But we know this is impossible.

Remark. If the grasshoppers had started at the vertices of some other parallelogram,they would be destined to always jump to crossings of the (generally slanted)grid made of copies of the initial parallelogram. The set of crossings of such agrid is called a lattice.

Problem 5. We start with some obvious remarks.

• Any two hands of the clock define an unfavorable sectorbetween their extensions, in the sense that if the thirdhand is in that sector, the moment is unfavorable. Thesize of this sector does not exceed 180◦ (see figure).

• After a whole number of hours, the minute and second hands will returnto the exact same position they occupy now.

• After six hours, any unfavorable moment is replaced by a favorable one(the hour hand rotates through 180◦ and moves from an unfavorablesector to a favorable one).

• The converse is not true: there are favorable moments, six hours afterwhich the time is again favorable. For example, any time between3:00:00 and 3:00:05 is favorable, but so are the times six hours later,from 9:00:00 to 9:00:05.

Now imagine the day divided into favorable and unfavorable intervals oftime. The operation of adding six hours maps all unfavorable intervals tofavorable ones of the same length (though the latter might be part of largerfavorable intervals). So the total length of favorable intervals is at least asmuch as that of unfavorable intervals. The same operation of adding sixhours also maps some favorable intervals to favorable intervals (consider the5-second interval discussed in the previous paragraph). Thus the favorableintervals add up to strictly more time than the unfavorable ones.

Remarks. 1. A similar situation is considered in Problem 56.8.4. REF 56.8.4

2. The original question asked “what is there more of, favorable time or unfavorabletime?” Obviously what we’re interested in is the total length of the intervals offavorable and unfavorable time, not whether there are more moments of eachkind of time. There are infinitely many moments of each of the two kinds, andthey can be placed in one-to-one correspondence, just as the intervals [0, 1] and[0, 2] can be placed in one-to-one correspondence.

The question of when it is possible to establish a one-to-one correspondencebetween two infinite sets is interesting and deep. (For finites sets, the answer issimple: a one-to-one correspondence exists if and only if the sets have the samenumber of elements — indeed this is one way to define the natural numbers!)Here are some example results; for details, see Settheory: CITE Settheory

(a) There is a one-to-one correspondence between the set of all integers andthe set of positive integers.

(b) There is a one-to-one correspondence between the set of all rational num-bers and that of positive integers.

66 SOLUTIONS

(c) There is no one-to-one correspondence between the set of all rational num-bers and that of all irrational numbers.

(d) There is a one-to-one correspondence between the points of a line segmentand the points of a square.

Problem 6. The first player colors the 18×18 square adjacent to the longerside of the rectangle so it’s equidistant from the right and left edges of therectangle (see figure for an analogous situation with a 7 × 14 board). Therest of the rectangle is imagined to be divided into congruent halves. Nowthe second player’s every move can be coun-tered by the first with a symmetric move; thesecond player would only be able to preventthis by coloring a square that straddles thesymmetry axis, which is impossible.

Level B

Problem 1. Let D and E be points inside a triangle ABC.Draw all 10 segments between pairs of points; they canbe divided into two non-self-intersecting polygonal linesof five segments each (see figure). Either of these twopolygonal lines can be taken as the desired pentagon;the other polygonal line is made of the diagonals ofthe pentagon. A

B

C

D

E

Problem 2. First case. If k > l, Sue wins, by securing a part longerthan the sum of all the other parts. For example, she can cut k into partsof length

l + 23(k − l), 1

6(k − l), 16(k − l).

(See figure below.) The largest of these, l+ 23(k− l), cannot serve as a side of

any triangle: by the triangle inequality the sum of the lengths of the othertwo sides should be greater than this, but the sum of the lengths of all othersegments is only l + 1

3(k − l).

k

l + 2

3(k−l) 1

6(k−l) 1

6(k−l)

l

Second case. If k 6 l, Leo wins. Let Sue divide her segment intoparts of length k1, k2, k3, where k1 > k2 > k3. Then divides his segmentl so that his largest part is of the same length as Sue’s largest part, andthe remaining two of the same size: k1,

12(l − k1) and 1

2(l − k1). (See figureon the next page.) Since k1 > k2, it is possible to construct two isosceles


triangles from the resulting segments, of lengths

k1, k1, k2,l − k1

2,l − k1

2, k3.

Indeed, from segments of lengths a, a, b it is possible to construct an isoscelestriangle if and only if b < 2a. Obviously k2 < 2k1. On the other hand,

2 · l − k1

2= l − k1 > k3,

since k1 + k3 < k 6 l.

k

l

k1 k2 k3

k11

2(l−k1)

1

2(l−k1)

k2

k1 k1

k3

1

2(l−k1)

1

2(l−k1)

Problem 3. Here is an infinite set of solutions:

x = 2k2 + 1, y = k(2k2 + 1), z = −k(2k2 + 1), for k = 0, 1, 2, . . . .

Verification is straightforward.How can this answer be guessed? Since we only have to find infinitely

many solutions, not all of them, we can get rid of one variable and two cubesby taking z = −x. Then we get the equation 2x2 + y2 = y3, which we canrewrite as x2 = 1

2(y−1)y2 or

x = ±√

y − 1

2y.

So if (y−1)/2 is a perfect square, x is an integer. Thus we take (y−1)/2 = k2,which gives

y = 2k2 + 1, x = k(2k2 + 1).

Remark. There certainly exist other solutions, e.g.,

x = 12n(n2 − 1) + 1, y = − 1

2n(n2 − 1) + 1, z = −n2 + 1 for any n ∈ Z.

Problem 4. First solution. Let the circles be disposedas in the figure, that is, with the center of each outsidethe other. The complementary case is treated similarly;see the Remark after the second solution.

It suffices to prove that the triangles AQNand AMP are congruent. For this, we provethat they are similar and that the sides AQand AM are equal.

By the theorem on inscribed angles, \P =\N and \Q = \M , so △AQN is similar to△AMP .

A

B

M

N

P

Q

To prove that the segments AQ and AM are congruent, we prove that thecorresponding arcs, measured counterclockwise, are the same. The first of

68 SOLUTIONS

these arcs has measure twice the inscribed angle ABQ; see Fact 15. Similarly,the arc AM equals twice the angle MAN . Thus, it suffices to prove that\ABQ = \MAN .

Now, the angle ABQ is an exterior angle of the triangle ABN , and hence\ABQ = \ANB + \BAN . We see that \MAN = \BAM + \BAN ; but\ANB = \BAM , since each of them equals half the arc of the circle throughA, B and N . Hence

\MAN = \ANB + \BAN = \ABQ.

The problem is solved.

Second solution. By the tangent-secant theorem (power of a point), we have

MP =AM2

MB, NQ =

AN2

NB.

Thus it suffices to prove that

AM2

AN2=

MB

NB. (1)

Let’s show first that the triangles AMB and NAB aresimilar. We have \AMB = \BAN , since these are bothinscribed angles corresponding to the same arcin the circle ABM (see figure). By the sametoken, \MAB = \NAB. This proves the simi-larity of AMB and NAB, which in turn impliesthat

AM

AN=

AB

NB, and

AM

AN=

MB

AB.

Multiplying these equalities we get (1).

A

B

M

N

P

Q

Remark. If the center of one of the circles lies inside the other circle, the picturewill be different. If the center of each of the circles is inside the other circle, weget still another picture. In these cases, in the first solution, instead of provingthat \ABQ = \MAN , we have to prove that \ABQ + \MAN = 180◦. Thesecond solution remains valid in either case. The reader is encouraged to workout the details.

Problem 5. Let x be the dropped digit, a the part of the number to theleft of x, and c the part of the number to the right of x, containing n digits.Then the original number equals a · 10n+1 + x · 10n + c; see Fact 11 on page123. After x is struck out it becomes a · 10n + c.

(These formulas encompass the case n = 0, with the convention c = 0,but we can dispose of this case quickly: 10a is a multiple of a, so 10a + x I’ve added this case, since

otherwise it’s awkward to

make claims about c.cannot be one as well, unless a 6 x. That would mean the original numberhas only two digits, rendering this case irrelevant, since there are three-digitexamples, such as 105 = 15 · 7.)


Now consider the ratio between the two numbers:

r =a · 10n+1 + x · 10n + c

a · 10n + c, where c < 10n.

Subtracting 10 from both sides of this equation and simplifying we reach

r − 10 =x · 10n − 9c

a · 10n + c6

x

a6

9

a6 9. (1)

Set l = r − 10. Multiplying both sides of the equality in (1) by a · 10n + cand manipulating we get

(x − la) · 10n = (l + 9)c. (2)

If l 6 0, the left-hand side is positive, so l + 9 > 0. Thus,

−8 6 l 6 9.

Moreover, l 6= 0; otherwise the original number ends in 0.

Lemma. a < 10.

Proof. Consider the two cases: l > 0 and l < 0.If l > 0, equality (2) implies that x − la > 0, and hence

a <xl

69l

6 9.

If l < 0, then

x − la =(l + 9)c

10n<

9 · 10n

10n= 9,

implying −la < 9, so a < 9. ˜

The lemma implies that the original number has n + 2 digits. The nextstep is to find the greatest possible value of n.

By assumption c does not end in 0. Thus c cannot be divisible by both2 and 5 (see Fact 10 on page 123).

Suppose first that c is not divisible by 2. Consider the right-hand sideof (2). Since 1 6 l + 9 6 18, the number l + 9 can be divisible by 24 = 16,but it cannot be divisible by 25. Hence n 6 4. If n = 4, then l + 9 = 16,and (2) becomes

(x − 7a) · 54 = c.

Since x is a one-digit number, it follows that a = 1, and hence x = 8 orx = 9. If x = 9, then c ends with a 0 and hence does not fit. If x = 8, thenc = 625 and the final answer is

180625 = 10625 · 17.If, on the other hand, c is not a multiple of 5, then l + 9 can be divisible

by 5 but not by 52, so n 6 1, and the number will certainly not be thelargest possible.

Problem 6. For part (b), it is easy to give an example wherethe first few ships are arranged so that there is no room forthe last on; see figure on the right.

70 SOLUTIONS

In part (a), there is a hidden difficulty: many students assumed it’senough to prove that one can place the last 1× 1 ship, but in fact one mustshow that each ship in sequence can be placed.

The 1× 4 ship can be placed. To prove that each of the two 1× 3 shipscan be placed, too, we mark eight “phantom” 1 × 3 ships, all in the samedirection and leaving two cells in between them (see first diagram below).If a 1 × 4 and at most one 1 × 3 ship have been placed, each of them cantouch or overlap with at most two phantom ships; therefore some phantomship will be untouched and can be replaced by an actual 1 × 3 ship.

The reasoning for the 1 × 2 ships is similar; it involves twelve parallel1×2 phantom ships, again with two cells between them (see middle diagram).Suppose the 1 × 4 ship, two 1 × 3 ships and no more than two 1 × 2 shipshave been laid down. Each touches at most two phantom ships, so at leastone phantom ship will be untouched and can be replaced by an actual ship.

For the final step of the proof we consider sixteen 1×1 phantom ships (seerightmost diagram above). Each of the six long ships already placed touchesat most two phantom ships, while each of the (at most three) 1 × 1 shipsalready placed touches one phantom ship. Thus at most fifteen phantomships are affected, and one is left over, showing that the last 1 × 1 ship canbe placed.

Remark. It is interesting to find the maximal number of identical ships of, forexample, size 1 × 4, that one can place.

Level C

Problem 1. We offered two detailed solutions to the very similar problem57.8.2. Here we give the corresponding arguments in slightly abbreviated REF 57.8.2

form.First solution. Let x and y be the numbers to be found. Then 3xy =

107x + y, or 3y = 107 + y/x. Since 0 6 y/x 6 10, we have 107 < 3y <107 + 10,and therefore 33333331

3 < y < 333333623 . Now one can either

consider the three cases y = 3333334, y = 3333335, and y = 3333336, orobserve that

y

x6

3333336

1000000< 4,

and hence 107 + 1 6 3y < 107 + 4. Only one number in this interval isdivisible by 3, namely, 107 +2. So 3y = 107 +2, and hence y = 3333334 andx = 1666667.


Second solution. Since107x = (3x − 1)y (1)

and since x and 3x−1 have no common divisors, 107 is a multiple of 3x−1.But 3x − 1 > 3 · 106 − 1, so either 3x − 1 = 5 · 106 or 3x − 1 = 107. Only3x − 1 = 5 · 106 fits, so x = 1666667. The relation (1) implies y = 3333334.

Problem 2. We first show that all the xn lie in the interval [0, 1]. Indeed,

0 6 xn 6 1 =⇒ −1 6 1− 2xn 6 1 =⇒ 0 6 |1− 2xn| 6 1 =⇒ 0 6 xn+1 6 1.

Rationality implies periodicity. If xn is rational, then so is xn+1 and itsdenominator is no greater than that of xn (assuming both are irreducible,of course). Indeed, let xn = pn/qn be irreducible. Then

xn+1 = 1 −∣∣∣∣

qn−2pn

qn

∣∣∣∣=

qn − |qn−2pn|qn

.

If this fraction is irreducible, its denominator is the same as that of xn; if itis reducible, its denominator decreases after simplification.

Thus, all the terms of the sequence are rational numbers between 0 and1. But there are only finitely many such irreducible fractions with denomi-nators not greater than a given q. Therefore some term of the sequence willreappear and at that point the sequence will become periodic.

Periodicity implies rationality. First solution. Eliminating the absolutevalue sign in the equation xn+1 = 1−|1−2xn|, we see that either xn+1 = 2xn

or xn+1 = 2− 2xn. Therefore xn+1 = a1 +2b1xn, where a1 is an integer andb1 = ±1. Similarly, xn+2 = a2 + 4b2xn. Continuing this process we get

xn+k = ak + 2kbkxn,

where ak is an integer and bk = ±1.Suppose that xn+k = xn. Then xn is a solution of a linear equation with

integer coefficients:ak + 2kbkxn = xn.

This equation has a unique solution, since 2kbk = ±2k 6= 1. The solution isrational, so xn is rational. But then xn−1 is rational, xn−2 is rational, andso on. Finally, x1 is a rational number.

Second solution. Write x1 in binary notation (Fact 12 on page 123):

xn = 0.a1a2a3 . . . .

If a1 = 0, the binary representation of xn+1 is obtained from the binaryrepresentation of xn by a shift (verify this):

xn+1 = 0.a2a3a4 . . .

If a1 = 1, the binary representation of xn+1 is obtained from the binaryrepresentation of xn by a shift and the simultaneous interchange of 0 with 1,a process we will call inversion. Let us prove that if xn is periodic, then an

is also periodic. If xn+k = xn and an even number of inversions took placebetween them, then an+k = an. If an odd number of inversions occurred

72 SOLUTIONS

between xn and xn+k, we have an+2k = 1 − an+k = an. Either way thesequence an is periodic. But periodic binary representations, like decimalones, correspond to rational numbers (see Fact 13 on page 123).

Remarks. 1. The function y = f(x) = 1−|1−2x| is linear on each of the segments[0, 1

2

]and

[12 , 1

]:

y =

{2x for 0 6 x 6 1

2 ,

2 − 2x for 12 6 x 6 1.

Similarly, the function

y = fn(x) = f(f . . . f(x) . . . )︸︷︷︸

n times

sending x1 to xn is linear on each segment [k/2n, (k+1)/2n]. The graphs of thefirst three of these functions are as follows:

changed f1 and f2 to f2

and f3 to agree with thetext

y

xy = f(x)

y

xy = f1(x)

y

xy = f2(x)

Because of the shape of its graph, the function f is called the tent map. It is merged Remark 3 here

an important object in symbolic dynamics.

2. For every T = 2, 3, . . . , there exists at least one point with period T : for instance,the x-coordinate of the last intersection point of the segment {(x, y) | y = x,0 6 x 6 1} with the graph of the function y = fT (x). Explicitly, this value isx1 = 2T /(2T + 1).

Contemplate the question: How many periodic trajectories are there foreach period T? (Or, which is almost the same, how many points x are there forwhich x = fT (x) and x 6= fk(x) if k < T?)

Problem 3. First solution. Two members of Parliament will be said to beenemies if one of them has slapped the other. The statement of the problemis the particular case n = 665 of the following result:

Lemma. If there are M > 3n − 2 members of Parliament and each hasslapped exactly one colleague, it is possible to make up a parliamentary com-mittee with n members, none of whom is enemies with another.

Base of the induction. For n = 1, the statement is obvious.

Induction step. Since there are as many members as there were slaps,the pigeonhole principle (Fact 1) implies that there is a member who wasslapped at most once. This member, therefore, has at most two enemies (oneslapee and perhaps one slapper). We draft this member into the committee,and set aside his or her enemies, reducing the number of available membersby at most 3. The conditions of the lemma are then satisfied with n replacedby n−1, and by the inductive hypothesis one can select a committee of n−1


members of this reduced parliament containing no enemies. With the onepreviously chosen we get n committee members none of which are enemies.

Second solution. Consider a directed graph Γ (see fact 3 on page 120), whosevertices are members and whose edges correspond to slaps, the source beingthe slapper and the target the slapee.

Define a directed tree as any directed graph whose vertices can be par-titioned into levels satisfying the following properties: there is one vertex oflevel 1, called the root, with no edges emanating from it; and every vertexof level j > 1 has one outgoing edge, which goes into a vertex of level j − 1.

Lemma. Each connected component of Γ consists of disjoint directed treeswhose roots are linked together in an oriented cycle.

1

1

1 1

1

2

3

34 3 2

Proof. Starting from any vertex, we follow directed edges until we return toa vertex already visited, which must happen since the graph is finite. Hence,each connected component has a cycle. We assign the vertices in these cycleslevel 1, and assign levels to other vertices according to the number of edgesthat separate them from a cycle (this number is well-defined since the pathleading from a vertex to a cycle is unique). ˜

Next we assign one of three colors to each vertex, starting with thecycles. In cycles of even length, we color the vertices alternatingly red (R)and green (G). In cycles of odd length, we paint one vertex blue (B), thenalternatingly paint the other vertices R and G.

R

G

R G

B

G

R

RRGR

We next color the trees according to their roots. If the root is R or B,even levels are colored G and odd levels starting from 3 are colored R. If theroot is G, even levels are colored R and odd levels are colored G.

Clearly, at least one of the colors is used in at least one third of vertices.The committee is then declared to consist of all members of that color.

Remarks. 1. If each cycle contains an even number of nodes, two colors suffice.

2. Give an example where it is impossible to select a committee with 668 members(out of 1994).

74 SOLUTIONS

3. The lemma in the first solution can be generalized: If there are M members ofParliament and each has slapped exactly k colleagues, it is possible to make upa parliamentary committee with M/(2k + 1) members containing no enemies.

Supply a proof; you can follow closely the first solution given above.

Problem 4. Let’s try to guess the length of the segment AK. Considerthe limit case as the point D tends to C. Then one of the circle shrinks toa point and the other becomes the circle inscribed in △ABC; the point Kmerges with P , the tangency point of this circle with side AC.

Now, there is a formula for the length of thesegment between a vertex A of a triangle ABC andthe tangency point of the incircle to a sidecontaining A: it reads

AB + AC − BC

2.

Let us prove that AK is given by this formula,not only in the limit case but in general.

A

B CDE

PB′

F

Q

KM

NC′

Denote the tangency points as in the figure. Observe that AB1 = AP ,AC1 = AQ, KM = KP , KN = KQ, since these are pairs of tangentsegments drawn from a single point. Moreover, MN = EF since these aresegments of common tangents to two circles (see Fact 17). We can now write

AB + AC − BC = AB′ + AC ′ − EF = AB′ + AC ′ − MN

= AP + AQ − MN = AP + AQ − (MK + KN)

= (AP − KP ) + (AQ − KQ) = 2AK.

Hence

AK =AB + AC − BC

2.

Problem 5. (a) The answer is negative. For example, take the polygonin the figure, consisting of three identical square roomsconnected by thin bent corridors. We will prove that nochord splits the polygon into two pieces of equal area.

Let the area of the polygon be S, and let the areaof each room be 0.3S, so the total area of the corri-dors is 0.1S. If the chord intersects only a corridor,then two rooms are on one side of it, and their areais greater than 0.5S.

0.3S

0.3S

0.3S

If the chord intersects one of the rooms, it cannot intersect the “hub” ofthe corridors, and so again there are two rooms on one side of the chord.

(b) First solution. The idea is to move the ends of a chord through thepolygon in such a way that the area of the smallest piece is minimized. Fornotational simplicity we assume the total area is 1.

Select the direction of the chord so it is parallel to any side or diagonalof the polygon; let this be the vertical direction. Then the chord cannotpass through more than one vertex of the polygon.


The chord divides the polygon into exactly two pieces if no vertex liesin the interior of the chord, and into exactly three pieces otherwise.

If at any time area of the smallest piece has area at least 13 , the problem

is solved.Assuming the smallest area is under 1

3 , we start moving the chord eitherright or left, so as to make the area of the smallest piece grow. As longas the chord does not touch a concave vertex, the area grows continuously(see Remark to the solution of Problem 63.11.4). However, when the chord REF 63.11.4

reaches a concave vertex, one of its endpoints may jump suddenly to anotherpart of the perimeter.

What happens when the chord such a troublesome vertex? (Note thatthe chord cannot meet two vertices simultaneously.)

Consider the situation at the exact moment when three regions exist. The original was a bit

laconic here. A case

analysis is helpful inshowing what it meant to

push the chord to the area

with the largest area.

Let a be the area of the region where the chord had been so far, and b, c,with b > c, the areas of the other two regions. Note that b > 1

3 , since a < 13

(otherwise we’d be done).There are three possible configurations, shown below, and in all of them

we push the chord slightly into the b region, thus combining the two regionsof smallest area. In one case, shown in the rightmost diagram, this involvesa reversal in the direction of movement of the chord.

b

c

a

c

a

b

b

a

c

If b < 23 we are done, because the combined piece has area just above

a + c = 1 − b > 13 , while the other has area just below b, which as we know

exceeds 13 .

If b > 23 , we continue moving the chord into the area b; the area of the

smallest piece has jumped from a to a + c, but remains less than 13 , and we

repeat the process.We eventually run out of troublesome concave vertices, meaning that

no obstacle remains for the smaller piece to attain area 13 . More precisely,

the total number of vertices is finite, and each troublesome vertex has awell defined associated value of a; but the value of a keeps increasing, sowe cannot encounter the same troublesome vertex twice, and the processterminates. (Compare Fact 2 on page 119.)

Second solution. Any polygon can be partitioned into triangles whose sidesare sides or internal diagonals of the polygon. (We will not prove this fact,so this solution is only a sketch.)

Consider such a triangulation. Each of the internal edges of the trian-gulation splits the polygon into two pieces. Let AB be an edge for which

76 SOLUTIONS

the area of the smaller piece is maximal. Let ABC be a triangle of thetriangulation adjacent to AB on the side of the larger piece.

Denote by SAB, SBC and SAC the areas of the pieces adjacent to thecorresponding sides of the triangle ABC and not containing it. (If a side isan edge of the polygon, the corresponding area is defined to be 0.) Then

SAB + SBC + SAC + SABC = 1,

where, for simplicity, we have defined the area of the polygon to be 1. Wemay assume that SBC > SAC . Further, from the maximality conditionimposed on the chord one can deduce that

SAB > SBC > SAC .

These inequalities imply that SAB+SABC > 13 . If SAB > 1

3 , then the problem

is solved; otherwise let X be a point on side AC such that SAB +SABX = 13 .

The chord passing through B and X satisfies the conditions of the problem.(Check it!)

Remarks. 1. A variation of part (a) is “Can a chord always be found that dividesthe polygon it into some number of pieces of equal area?” The answer is still no;it is not difficult to adapt the solution given above to this case. For example,let the areas of the rooms in the figure on page 74 be 0.3S, 0.33S and 0.36S.Prove that such a polygon cannot be cut by a single chord into any number ofpieces of equal area.

2. A polygonal annulus is a region whose boundary consists of two polygonal curves,instead of one. There are polygonal annuli that cannot be divided by a chordinto two polygons whose areas are more than a third the total area. Can youcome up with an example? Also, explain what breaks down in the proof givenin the first solution to (b) above.

3. The second solution of part (b) shows that it is always possible to select thedesired chord so that it contains through a vertex.

Problem 6. We call a polynomial anxn+an−1xn−1+· · ·+a1x+a0 positive if

all its coefficients are positive (ai > 0 for i = 0, . . . , n). Clearly, the productof positive polynomials is positive. Therefore if the square and cube of apolynomial are positive, so is any higher power, because any higher poweris a product of squares and cubes.

We first find a nonpositive polynomial whose square and cube are posi-tive. We start with 0 and 1 as coefficients to simplify the computations. Wenote that a “hole” — a single 0 between 1’s — gets “filled” when we multiplya polynomial by x + 1, because of the adjacent 1 coefficients of the latter.For example, we have (x2 + x)(x + 1) = x3 + x2 + x + 1, which can berepresented by the schematic calculation

1 0 11 0 1

1 1 1 1


Experimenting a bit with adjacent 1 coefficients we find that

f(x) = x4 + x3 + x + 1,

corresponding to the pattern 1 1 0 1 1, “fills it own hole”; that is, f2 ispositive even though f is not. Moreover, f3 is also positive. (Why?)

However, we need a polynomial with a negative coefficient. For this, we“jiggle” the polynomial f a bit (mathematicians say “perturb”). That is,we replace the 0 coefficient by a small negative coefficient, defining

g(x) = f(x) − εx2, (1)

where ε > 0 is very small. Then the coefficients of g2 and g3 are close tothose of f2 and f3, and so must still be positive.

So the answer to the question is affirmative, and an example is given byx4 + x3 − εx2 + x + 1 for any ε positive and small enough. One can checkexplicitly that ε = 1

10 will work.

Remarks. 1. To be rigorous, the closeness argument depends on the fact that thecoefficients of the square of a polynomial F (and also those of the cube) dependcontinuously on the coefficients of F , as they are themselves given by polynomialexpressions. That is, if F (x) = a4x

4 + a3x3 + a2x

2 + a1x + a0, the coefficientsof F 2 are a2

4, 2a3a4, a23 + 2a2a4, and so on; therefore a small change in the ai

causes only a small change in the coefficients of F 2.

2. Write the coefficients of f(x) = x4 + x3 + x + 1 compactly as a single integer,11011. Then the compact expressions of f2 and f3 coincide with the square andcube of that integer:

110112 = 121242121 and 110113 = 1334996994331,

Why? (Hint: Multiplication of polynomials is done in the same way as multi-plication of integers in decimal notation.)

How far can you push this observation? What happens if you square thepolynomial f(x) = x4 + 2x3 + 2x + 1, represented by 12021?

Level D

Problem 1. Let the polyhedron have n faces. Since no more than two faceshave the same number of edges, there are at least

[12(n+1)

]types of faces

(triangles, quadrilaterals, and so on). Since n > 4, this means there arefaces with 4 or more edges. But such a face has at least4 neighbors, so n > 5. Therefore there are faces with 5or more edges. But then n > 6. Let’s find an examplewith 6 faces: two triangles, two quadrilaterals, and twopentagons. Such a polyhedron can be obtained from atetrahedron by clipping off two corners (see figure).

Another solution, with more than the minimal number of faces, can beobtained from a cube by clipping off two neighboring corners.

78 SOLUTIONS

Remarks. 1. Solve the following related problem from the 1973 Moscow Mathe-matical Olympiad: Prove that any convex polyhedron has two faces with thesame number of edges. (Hint: Euler’s formula.)

2. Prove or disprove: Any polyhedron with 10n faces has n faces with the samenumber of edges.

Problem 3. We first solve an auxiliary problem: to find the smallest circlewith center on the y axis, tangent to the x axis, and intersecting the graph

of the function y = x4 somewhere beside the origin(see figure). In other words, what is the least r forwhich the system

x2 + (y − r)2 = r2, (1)

y = x4. (2)

has a nonzero solution? It is intuitively clear that thisproblem is equivalent to the initial one, and we shallprove this equivalence rigorously later.

y

x

y = x4

Substituting (2) into (1), simplifying and dividing by x2 (assuming x >0) we get

x6 − 2rx2 + 1 = 0

whence we deduce that

r(x) =12

(

x4 +1x2

)

.

The smallest value r0 is the minimum value of r(x) for x > 0. We have

r′(x) = 2x3 − 1x3 .

On the semiaxis x > 0 we see that r′(x) < 0 for x < x0 = 1/ 6√

2, r′(x0) = 0,and r′(x) > 0 for x > x0. Hence r′(x) decreases for x < x0 = 1/ 6

√2, it

attains its minimum at x0, and it increases for x > x0.Thus, the smallest value of r for which the circle has a common point

with the curve y = x4 (away from the origin) is

r0 = r(x0) =3 3√

2

4.

It remains to show that the same r0 solves the initial problem as well.First, let’s check that the cherry of this radius can indeed lie in the goblet.Indeed,

r(x) > r0 for any x 6= 0. (3)

Substituting the expression for r(x) in (3) we get

x6 − 2r0x2 + 1 > 0 for any x 6= 0.

Multiplying both sides of this inequality by x2 and replacing x4 by y we get

x2 + (y − r0)2 > r2

0 for any x and y = x4.

But this means that the cherry of this radius can lie in the goblet.


Finally we verify that if r > r0 the cherry will not touch the bottom ofthe goblet. Indeed, in this case,

x60 − 2rx2

0 + 1 < 0,

so for y0 = x40, we have

x20 + (y0 − r)2 < r2.

This means that the circle of radius r tangent to the x-axis at the originintersects the graph of the function y = x4; so the cherry will not touch thebottom.

Remarks. 1. It is not difficult to see that, in cross-section, the biggest cherry istangent to the goblet at the bottom and at two more points. In space, thatmeans the bottom and a circle of tangency.

2. A similar problem can be solved for the goblet whose vertical cross-section isof the form y = |x|a, for any real a > 0. For the x-coordinate of the tangencypoint of the thickest cherry with the goblet we have an equation

x2(a−1) =a − 2

a.

(a) For a > 2 there is a nonzero solution; the situation is similar to that ofa = 4.

(b) For a = 2, we get x = 0 and r = 12 , which is the curvature radius of the

graph at the origin; so the biggest cherry touches the goblet only once, atthe bottom.

(c) For 0 < a < 2 no cherry can touch the bottom of the goblet.

Problem 4. The homothety with center A and scale factor 2 sends theinitial polyhedron M into a polyhedron M ′ whose volume is 8 times greaterthan that of M . Let us prove that all 8 translated polyhedralie inside M ′.

Let B be the translate of the vertex A, let X be anarbitrary point of M , and let Y be the image of the pointX under the same translation (see figure). We must provethat Y belongs to M ′. A X

B Y

K

Since M is convex, the segment BX is entirely contained in M ; in par-ticular, its midpoint K belongs to M . The quadrilateral ABY X is a par-allelogram, so Y is obtained from K under a homothety with the center atpoint A and factor 2; hence Y belongs to M ′.

Next we observe that points close to A do not belong to any of thetranslated polyhedra. Indeed, let us turn M so that A lies above all theother vertices of M . Then there exists a plane that passes below A but aboveall the other vertices of M . This plane cuts off of M a small polyhedronN containing A. Clearly, N does not intersect any of the 8 translatedpolyhedra.

If all the translated polyhedra had disjoint interiors, the volume of M ′

would be at least the sum of volumes of the translated polyhedra plus thevolume of N . But the sum of volumes of the translated polyhedra is equalto the volume of M ′. This is a contradiction.

80 SOLUTIONS

Remarks. 1. It is easy to generalize the result to n-dimensional space.

2. At the Thirteenth International Mathematical Olympiad, held in the town ofZilina, Czechoslovakia, in 1971, a weaker formulation was offered: Prove that atleast two of the nine (not eight) polyhedra intersect.

Problem 5. It is difficult to illustrate the problem (the intersection pointslie far from each other), so let’s think. First, recall that the points of abisector are equidistant from the sides of the angle it bisects.

For each of the four lines containing a side of the quadrilateral ABCD,define a function fi, where i = 1, 2, 3, 4, equal to the directed distance to thisline: if the point lies on the same side of the line as the quadrilateral, we takethe usual distance, otherwise we take minus the distance. The functions fi

depend linearly on the coordinates of the points (see Fact 25), i.e.,

fi(x, y) = aix + biy + ci.

A point lies on the bisector of an outer angle of the quadrilateral if andonly if the sum of the values of the two functions corresponding to the legsof the angle vanishes.

For each of the two intersection points of the bisectors spoken about inthe formulation of the problem, we see that the sum of all four functionsvanishes.

But the sum of linear functions is a linear function, and the set of pointswhere a nonconstant linear function vanishes is a straight line. The sum ofthe fi does not vanish identically since it is positive inside the quadrilateral.Hence the intersection points of the bisectors lie on one line.

Remarks. 1. A spatial generalization of the result of the problem can be obtainedusing the same idea.

2. Here is another problem that can be solved using a similar idea: in a triangle,find the locus of the points the sum of whose distances to two sides equalsthe distance to the third. The intersection of a bisector with the opposite sidesatisfies this condition; therefore so does any point on a line segment connectingtwo of these intersections. (Why? Be careful.)

3. In school, a linear function is any polynomial of degree at most 1. In linearalgebra, a linear function is any polynomial of degree at most 1 that does nothave a constant term. Functions with a constant term are said to be affine.

Problem 6. What causes the appearance of many 9s in a row in powers of2? The fact that the power of 2 is just a bit less than a number divisibleby a high power of 10. For example, 212 + 4 is divisible by 100, 253 + 8 isdivisible by 1000.

Let’s first try to find numbers of the form 2n+1 divisible by a high powerof 5. Then we will multiply those numbers by the corresponding power of2 and obtain numbers of the form 2k(2n + 1) divisible by a high power of10. Simplifying and subtracting the smaller summand we get the requiredpower of 2.

Lemma 1. For all k > 1, the number 22·5k−1

+ 1 is divisible by 5k.


Proof. We use induction on k (see Fact 24 on page 24). The base of induc-tion (k = 1) is obvious. To prove the induction step, we write

22·5k−1

+ 1 = 45k−1

+ 1.

Let a = 45k−1

. By the induction hypothesis a + 1 is divisible by 5k. Then

45k

+ 1 = a5 + 1 = (a + 1)(a4 − a3 + a2 − a + 1).

Since a + 1 is divisible by 5k, it suffices to prove that the second factor isdivisible by 5. Indeed, a is of the form 5m − 1, so all the summands of thesecond factor give residue 1 after division by 5 (see Fact 7), and their sumis divisible by 5. Lemma is proved. ˜

Thus, the number 2k(22·5k−1

+ 1) ends with at least with k zeros. It iseasy to see that 2k has at most k/2 digits, if k > 1. Hence, among the lastk digits of the number

22·5k−1+k,

no more than k/2 digits can differ from 9.

Remarks. 1. Similar arguments prove that the numbers

40, 41, 42, . . . , 45k−1

have different residues after division by 5k. Try to prove a more general factoften encountered in number theory, called Hensel’s lemma:

If x − 1 is divisible by pk, where p > 2 is a prime and k > 0, but is notdivisible by pk+1, then xn − 1 is divisible by pk+r if and only if n is divisibleby pr.

2. At the Leningrad Mathematical Olympiad in 1981, a similar problem was posed:Is there a positive integer power of 5, whose last 100 digits contain at least 30zeros in a row?

Year 1995 Olympiad

Level A

Problem 1. Let a and b be the initial prices (in pennies) of ale and bread,respectively. Then

1 = b + a.

After the first rise, the prices satisfy

1 =(

b2

+ a)

· 1.2.Subtracting one equation from the other and simplifying we get 2b = a, soa = 2

3 . The price of ale after the second rise is a · 1.2 · 1.2 = 1.44 · a = 0.96.Thus, Ben can still buy his ale with a penny.

Problem 2. We work by induction (see Fact 24 on page 128).

Base of the induction. Since 10017 = 53 · 189, the claim is true forthe first element of the sequence.

Induction step. Suppose the number 100

k−1 times︷︸︸︷

1 . . . 17 is divisible by 53.Then so is the next number, 1001 . . . 1

︸︷︷︸

k times

7. Indeed, the difference between thetwo numbers,

1001 . . . 1︸︷︷︸

k times

7 − 1001 . . . 1︸︷︷︸

k−1 times

7 = (1001 − 100) · 10k = 901 · 10k,

is divisible by 53 since 901 = 53 · 17; but the sum of two numbers divisibleby 53 enjoys the same property (Fact 5 on page 121).

Remarks. 1. One can show that the quotients upon division by 53 are of the form18 . . . 89.

2. See problem 58.9.1 for a similar question. REF 58.9.1

Problem 3. (a) The segment KL is a midline ofthe triangle ABC, so its length is half that of AB.Similarly, LM has half the length of BD. ButAC is taken to BD under a 120◦ rotation aboutthe point O (clockwise in the case shown in thefigure). Hence they have the same length.

A

B

C

D

O

K

L

M

120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦

84 SOLUTIONS

(b) Again by the midline theorem, we have AC ‖ KL and LM ‖ BD, as In the original there was aminor gap: the proof

would have “shown”

equally well that △NKL

is equilateral, whereN = midpoint of AD.

oriented segments. The angle between the oriented segments AC and BDis 120◦ since a 120◦ rotation takes one to the other. Therefore the anglebetween the oriented segments KL and LM is 120◦, and the angle betweenLK and LM is the complement of that, or 60◦. Using part (a), we deducethat the triangle KLM is equilateral.

Problem 4. Among parallelipipeds of fixed volume, the one with the leastarea is the cube; see Fact 26 on page 130. Therefore the minimum amount I see no indication that

the problem is restricted

to integer lengths; doesn’t

“kvadratnaffl edinica

materiala” mean the

same as “square unit of

material”, rather than

“square tile of material ofunit size”?

of material needed is

6 · ( 3√

1995)2. (1)

This is a messy number, so let’s try to find a solution in integers. The nearestcubes to 1995 are 1728 = 123 and 2197 = 133; moreover, a 12× 12× 13 boxdoesn’t have enough volume (1872). A 12× 13× 13 box does (2028), but itsarea, 2 · (12 · 13 + 12 · 13 + 13 · 13) = 962, only gives an answer to part (a).

To do better, we can try adding 1 to one dimension while subtracting 1from another. Because the factors are roughly the same, this should changethe volume and area very little (recall that (x− 1)(x + 1) is “almost equal”to x2). We have a 1.5% excess margin in the volume— we can go down from2028 to 1995 — while the area only needs to go down by about 0.5%, from962 to 958. So the strategy might work. And indeed, the 11 × 13 × 14 boxhas volume 2002, and area 958, so it works for (a), (b) and (c).

Remarks. 1. One can check that this is the best possible answer in integers: 957units are not enough. Allowing real numbers, the minimum area is 950.853, thevalue of (1).

2. The two-dimensional counterpart of the result we used is that the rectangle ofleast perimenter for a given area is a square. Derive this from the inequalitybetween the geometric and arithmetic means.

Problem 5. First solution. It suffices to prove that the fuel cost does notchange under an interchange of two consecutive deliveries. Indeed, by ap-plying such transpositions we can reorder the deliveries in any way we want.(This statement, which is easily proved by induction on the number of vil-lages, is an important theorem in permutation theory; see Chapter 2 inTranspositions.) CITE Transpositions

Suppose that at some point the bus visits village M and then visits Nnext, delivering goods of respective weights m and n. By assumption, theone-way distances to M and N are m and n as well. If we change the routejust by interchanging N and M , this obviously does not affect the amountof fuel spent on other legs of the trip. More subtly, it does not affect theportion of the fuel bill due to the goods delivered to any other villages. Thisis clear for the villages that come earlier in the route, but it’s also true forthose that come later; it’s a consequence of the fact that the cost per mile isproportional to the total weight, and so can be written as a sum of costs foreach package separately. Apart from the packages delivered to M and N ,


all packages on the truck travel the same miles under either route (beforeor after the interchange).

Thus we just have to compare the cost of carrying goods of weights mand n to the villages M and N . In the first case, a load of m + n will becarried over a distance m (to village M), then a load n will carried over adistance m + n (back to the city and out to N). Hence, the cost is

(m + n)m + n(m + n) = (m + n)2.

In the second case the cost is (m + n)n + m(m + n), which is the same.

Second solution. Let the weights of the goods be a1, a2, . . . , an in the orderthey are delivered. The fuel cost is proportional to

a1(a1 + 2a2 + 2a3 + · · · + 2an) + a2(a2 + 2a3 + · · · + 2an)

+ · · · + an−1(an−1 + 2an) + a2n.

Expanding the products and reordering we get

a21 + a2

2 + · · · + a2n + 2a1a2 + 2a1a3 + · · · + 2a1an + 2a2a3 + · · · + 2an−1an,

which equals (a1 + a2 + a3 + · · ·+ an)2. This expression does not depend onthe order of enumeration.

Problem 6. Let N lie on AB and K on AF , as in the figure. Observethat FK = AN . Select points P on BC, R on CD, S on DE and T onEF so that FK = AN = BP = CR = DS = ET .Then \KBN = \TAK, \KCN = \SAT , \KDN =\RAS, \KEN = \PAR, \KFN = \NAP . Thus

\KAN+\KBN+\KCN+\KDN+\KEN+\KFN

= \KAN+\TAK+\SAT+\RAS+\PAR+\NAP

= \KAN+\KAN = 120◦+120◦ = 240◦.A

B

C

D

E

F

K

N

PR

S

T

Level B

Problem 1. We work by induction (see Fact 24 on page 128).

Base of the induction. Since 12008 = 19 · 632, the claim is true forthe first element of the sequence.

Induction step. We need to show that if one of our numbers is divisibleby 19, so is the next. To this end it is enough to show that the differencebetween two consecutive numbers is divisible by 19. But

120 3 . . . 3︸︷︷︸

n times

08 − 120 3 . . . 3︸︷︷︸

n−1 times

08 = 1083 · 10n = 19 · 57 · 10n.

Remarks. 1. It is not difficult to show that the quotients upon division by 19 areof the form 63 . . . 32.

2. See problem 58.8.2 for a similar question. REF 58.8.2

86 SOLUTIONS

Problem 2. At first let A′ be any point on BC, and drawthe segment AA′. We claim that, among the segmentsfrom C to points on the side AB, there are at most twowhose length equals AA′: namely, the segments CC1 Unless I’m missing

something, the original

made this sound morecomplicated than it really

is.

and CC2 such that

\C1CA = \A′AC and \C2CB = \A′AC.A C

B

A′C1

C2

P1P2

(The two can coincide.) Indeed, CC1 = AA′ because△ACC1 and △CAA′ are congruent (case ASA). Similarly, CC2 = CC1 bythe congruence of △ACC1 and △BCC2. That these are the only two pointson AB whose distance to C equals AA′ follows from the fact that a circleintersects a line in at most two points.

Now we observe that the desired locus is made up of all intersections ofAA′ with CC1 and CC2, as A′ varies.

Let P1 be the intersection of AA′ with CC1. It lies on the altitude ofthe triangle ABC dropped from B: indeed, the triangle AP1C is isosceles,since the angles at its base are equal. Therefore AP1 = CP1, that is, P1 lieson the perpendicular bisector of AC, which is also the altitude in question.Conversely, all points on this altitude satisfy the condition of the problem,by symmetry.

The locus of P2, the intersection of AA′ and CC2, is less obvious. Weknow that the triangles A′AC and C2CB are congruent, and rotating thelatter by 120◦ about C (clockwise in the figure above) aligns it with theformer. Thus CC2 rotated by 120◦ becomes parallel to AA′; that is, theangle AP2C is 120◦. This can also be seen as follows:

\AP2C = 180◦ − \A′AC − \C2CA

= 180◦ − \A′AC − (60◦ − \A′AC) = 120◦.

The locus of points (inside the triangle ABC) that see thesegment AC under an angle of 120◦ is an arc with endpointsA and C and measure 2 · (180◦ − 120◦) = 120◦. (The casewhere C1 = C2 means this arc intersects the other part ofthe locus —the altitude — at the center of the triangle.)Conversely, all points on this arc satisfy the condition ofthe problem. (Why?) A C

B

Problem 3. Suppose first that n 6 1995, so no strip can be longer than1995. Since all strips have different, integer lengths, the most area they can Original had: “It is clear

that the sides of the strips

should be parallel to the

sides of the givenrectangle.” But why?

cover is 1+2+· · ·+1995 = 12 ·1995·(1995+1), or 1995·998 (see Remark after

solution). So the maximum possible value of n (not exceeding 1995) is 998.Now, one can show that n can be 998, by exhibiting a

decomposition of the 1995 × 998 rectangle into strips of dif-ferent lengths; we follow the pattern of the figure on the right(which was drawn with longest side 9 instead of 1995). Thatis, one row of the rectangle is occupied by a single strip of length m = 1995,


the next by strips of lengths 1 and m − 1, and so on until the 12(m + 1)-th

strip. (Note that m is odd.)It is also obvious that any value of n less than 998 will work; starting

from the 1995 × 998 rectangle, delete some rows until only n are left.Inspired by this particular case, we prove:

Lemma. An M × N rectangle, with M > N , can be divided into strips allof different lengths if and only if M > 2N − 1.

Proof. We follow the technique used above for M = 1995. The maximumarea that can be covered is 1

2 M(M+1) (the sum of integers from 1 to M),

so MN 6 12 M(M+1), or N 6 1

2(M+1). There remains to show that forall such values of N the rectangle can be covered. The case of M odd isverbatim the same as for M = 1995. When M is even, because N is aninteger, the largest possible N is actually M/2, and again it’s clear that wecan achieve coverage for such a rectangle (by omitting the strip of lengthM/2), as well as any rectangle with fewer rows. ˜

We are now solve the problem in the case where 1995 is the shorterdimension (n > 1995). We simply take N = 1995 in the lemma and obtainM > 2 · 1995 − 1 = 3989. Thus the final answer is “n 6 998 or n > 3989”.

Remark. We used the equality 1 + 2 + · · ·+ n = 12n(n+1). This formula can easily

be proved by induction. We take this occasion to give another elegant proof,whose underlying idea is often used to find the sum of any arithmetic sequence.

Set X = 1 + 2 + · · · + n and compute the sum of all numbers in the table

1 2 3 . . . n−2 n−1 n

n n−1 n−2 . . . 3 2 1

Since the sum of the elements in each column is equal to n + 1 and there are ncolumns, the sum is equal to n(n+1). On the other hand, the sum of elementsin each row is X, so 2X = n(n + 1), implying our statement.

Problem 4. First solution. The hypothesis implies that the number

a + b + c + d = a + b + c +ab

c=

(a + c)(b + c)

cis an integer. Hence the fraction is reducible. Since both factors in thenumerator are greater than in the denominator, after division by c each ofthe factors turns into a number greater than 1. Thus, a + b + c + d is theproduct of two factors, each greater than 1, so it cannot be prime.

Remark. We used a nontrivial statement: If the fraction (xy)/z is an integer, thenz can be represented as a product of integers ts so that x is divisible by t and yis divisible by s.

If you master the second solution, you will see how to prove this statementrigorously.

Second solution. We know that a, b, c, d are positive integers and thatab = cd. We claim that in this situation there exist positive integers u, v,w, z such that a = uv, b = wz, c = uw, d = vz.

88 SOLUTIONS

Consider the prime factorizations of a, b and c. Since c divides ab, allits prime factors are present among the prime factors of a and b combined;more precisely, a factor appearing m times in the prime factorization of cmust be accounted for at least m times in the prime factorizations of a andb taken together.

Let p1, . . . , pn be all the prime factors of a and b, and write

a = pk1

1 pk2

2 . . . pkn

n , b = pl11 pl2

2 . . . plnn , c = pm1

1 pm2

2 . . . pmn

n ,

where some of the numbers ki, li, mi can vanish. As we observed above,mi 6 ki + li for i = 1, . . . , n. Therefore we can write mi = ti + si, whereti 6 ki and si 6 li. Set

u = pt11 pt2

2 . . . ptnn and w = ps1

1 ps2

2 . . . psn

n .

Then c = uw. Besides, u divides a and w divides b. Hence a = uv andb = wz, where v and z are integer. Finally,

d =ab

c=

uvwz

uw= vz,

proving our claim.Now note that

a + b + c + d = uv + wz + uw + vz = (u + z)(v + w).

Both factors are greater than 1, so a + b + c + d is not a prime.

Problem 5. First solution. The general idea is this: At some point, three atleast of the starting triangles must be cut. After that there will be two groupsof three identical triangles each. Again, from each of these two trios, two “The order of cuttings is

irrelevant” was a bit too

glib, since some cutsdepend on others.

triangles must be cut. Two children coming from one group willbe identical with one another and with two children fromthe other group (see shaded triangles in the figure).We’re back to having four identical triangles!

But we must be careful about the argument: what if the cuts are madefirst in one triangle and its children, leaving the other starting triangles forlater? Perhaps then we might not get the trios of the previous paragraph?

To organize our thoughts, let’s devise a scheme for naming the trianglesand hence the cuts: A, B, C, D for the initial triangles; AL and AS for thechildren of A, where L and S stand for large and small (if the triangles areisosceles the problem is trivial); ALL, ALS for the children of AL, and soon. As mentioned, ALS and ASL are congruent. A cut made in a trianglewhose name has k letters will be called a level-k cut.

Suppose we have a recipe —a sequence of cuts—that takes us from fourequal triangles to all different triangles. The key observation is that theorder of the cuts doesn’t matter, so long as the recipe still makes sense. Inparticular, the end result of a recipe can be obtained via a different recipeconsisting of a reordering of the same cuts: all level-one cuts made first,then all level-two cuts, etc.


Now start with a recipe having as few steps (cuts) as possible. Afterthe reordering, which leaves the number of steps unchanged, there must bethree level-one cuts right at the beginning; otherwise two starting triangleswill never be touched. After all level-one cuts are made, there must befour level-two cuts; otherwise there will be two first-generation children leftuntouched at the end.

These four level-two cuts produce four identical triangles —say ALS,ASL, BLS, BSL— plus other stuff. We discard all other triangles apartfrom these four.

Now, if we apply to these triangles all the cuts in the recipe that stillmake sense (skipping steps that involve discarded triangles or their descen-dents), we end up with a subset of the triangles obtained from the full recipe,and of course these must be all different. But then we’ve found a shorterrecipe to go from four identical triangles to all different triangles! (At least7 steps shorter.) But we had assumed our starting recipe was as short aspossible, so we have reached a contradiction.

Second solution. We will look for an invariant of the problem (see Fact 2 on I rewrote to try to give the

reader an idea of how one

might come up with this

solution.

page 119) that we can use to our advantage. An obvious choice is the area.Let each starting triangle have area 1 and let p and q = 1−p be the areas

of the triangles resulting from the first cut. By induction, the area of anytriangle is pmqn, where m and n indicate respectively how many times thesmaller or the larger of the two children was chosen after a cut. Triangleswith the same pair (m,n) have the same area and, since they also have thesame shape, must be congruent.

Obviously, the total area of the triangles is 4 throughout the process.Now suppose that after some sequence of cuts, no congruent triangles are

left; we will try to obtain a contradiction. Each final triangle corresponds toa different pair (m,n). Let K be larger than all the values of m,n that occur.If we sum the areas of all (m,n)-triangles for 0 6 m < K and 0 6 n < K,whether or not they occur, we should obtain at least 4, since, by assumption,the final triangles are among this set.

Using the formula for the sum of a geometric series, we have

K−1∑

n=0

K−1∑

m=0

pmqn =

( K−1∑

n=0

pm

)

·( K−1∑

m=0

qn

)

=1−pK

1−p

1−qK

1−q<

1

1−p

1

1−q. (1)

It would be nice if p and q were equal to 12 , because then the expression

after the “<” would equal 4, the sum of areas would be less than 4, andwe’d have our contradiction. Too bad we don’t control p and q. . .

But wait! We’re free to devise any invariant we want. We can choose anew invariant to imitate the case p = q = 1

2 , whatever the actual values ofthese numbers. Let the “wealth” of a triangle be divided equally between itstwo children whenever a cut is made (and let it be 1 for the initial triangles).

The wealth of a triangle corresponding to the pair (m,n) is then 2−(n+m).Again, the total wealth is invariant and equal to 4. Now the partial sum

90 SOLUTIONS

that replaces (1) is

K−1∑

n=0

K−1∑

m=0

(12

)m+n=

( K−1∑

n=0

(12

)m)

·( K−1∑

m=0

(12

)n)

=1 − 2−K

1/2

1 − 2−K

1/2< 4,

and we are done.

Remark. The following problem posed by Fields Medal winner Maxim Kontsevichis solved similarly: A bacterium sits at the (0, 0) square of an infinite squaregrid. After one minute it divides into two bacteria, which go on living in thesquares immediately above and to the right of the initial square. One of thesesplits after one minute, again occupying the squares immediately above and tothe right of where the parent was. The splits continue, always one bacterium ata time; moreover each square can fit only one bacterium at any given moment.Prove that there will always be at least one bacterium in some square (x, y)with x + y 6 5, x > 0, y > 0.

Problem 6. (a) The cook introduced, for her convenience, an imaginaryweightless can, so the number of cans became divisible by 3. She thennumbered the cans 1 through 81 in order of increasing weight.

To prove that she numbered the cans correctly, she first divides theminto three groups: cans 1 to 27, cans 28 to 54, and cans 55 to 81. She placesthe first group on the left pan and the third group on the right pan. Thebalance shows the greatest possible difference in weight between groups of27 cans, something that can be confirmed from the inventory. No othergrouping of 27 cans on each pan can give that readout.

Next the cook divides each group into three piles of 9 cans each. Sheplaces the lightest piles from each group (cans 1–9, 28–36 and 55–63) onthe left pan and the heaviest piles from each group (cans 19–27, 46–54 and73–81) on the right pan. This yields the greatest difference in weight betweensets of 27 cans of which 9 are taken from each group. This again is a numberthat can be calculated from the inventory.

Thus, after the second weighing there are 9 piles ordered according totheir weight; each pile has 9 cans and the geologists believe that each pilehas the composition claimed by cook.

In the third weighing, the cook divides each pile into 3 trios. She placesthe lightest trio from each pile on the left pan and the heaviest three canson the right pan. Again the readout is the maximum difference that can beachieved between sets of 27 cans of which 3 are taken from each pile. Thusthe makeup of each trio has been proved.

In the fourth weighing the cook places the lightest can from each trio onthe left pan and the heaviest on the right pan. Again because the readoutis the maximum possible, the identity of each can has been established.

(b) Three weighings are not enough, whatever method the cook uses. Indeed,the first weighing defines three sets of cans: those on the left pan, those onthe right pan and those that didn’t take part in the weighing. At least oneof these sets contains 27 or more cans (since 26 ·3 < 80); let’s call it X. The


second weighing defines three subsets of X: one containing those elementsof X that were placed on the left pan, and so on. One of these subsetscontains nine or more cans; call it Y . The third and last weighing definesthree subsets of Y accordingly, at least one of which has three or more cans.The cans in this last set simply cannot be distinguished on the basis of theweighings.

Level C

Problem 1. (a) Up to two points on the unit circle existγ

(π−γ)/2

γ/2

π−γ

with the given sine, corresponding to γ and π − γ, for Here I included a fuller

analysis, which doesn’t

take much more space andgives some geometric

intuition.

some γ ∈ [−π, π] (heavy lines in the figure). The pont γstands for all arguments of the form α = γ +2πk, wherek is an integer; so it gives rise to two possibilities for α/2 on the unit circle,opposite one another and having opposite sines (thin solid lines). That givestwo values for sinα/2 (or one if γ = 0, since 0 = −0). Likewise π − γ givestwo possibilities for sinα/2 (thin dashed lines). These last two are differentfrom the first so long as γ 6= π−γ, for the following reason: the distinct half-angles γ/2 and (π − γ)/2 lie in the same quadrant (first or fourth) becauseγ ∈ [−π, π]. Inside any quadrant the sine function is one-to-one, so we gettwo values for the sine of the half-angle, with different absolute values.

Thus we get four values for sinα/2, with a few exceptions that are clearfrom the argument: sinα = ±1 (which implies γ = π−γ, so we get only twovalues for sinα/2) and sinα = 0 (which implies γ = 0, so there are threevalues of sinα/2, namely ±1 and 0).

(b) We can argue as in part (a) but we will use another method, based onthe formula sin 3θ = −4 sin3 θ + 3 sin θ. Let sin α/3 = t. We get:

sin α = sin(

3 · α

3

)

= −4t3 + 3t.

But a polynomial of degree three can have at most three roots (Fact 20 onpage 127). So the maximum number of values of t is three. When α = 0(for instance), three values do occur:

−4t3 + 3t = 0 =⇒ t(3 − 4t2) = 0 =⇒ t = 0,±√

32

,

corresponding to α/3 equal to various multiples of π/3 (not surprisingly).

Remarks. 1. The algebraic approach is available for part (a) too, but it’s not asneat as part (b) because cosines must come in too. Recall the identity cos 2θ =

cos2 θ − sin2 θ = 1 − 2 sin2 θ; it gives sin α/2 = ±√

(1− cos α)/2. Now, givensin α, there are two opposite choices for cos α (why?). So there are, in general,four choices for sinα/2,

2. More generally, if n is any odd integer, sinnφ is a polynomial of degree n insin φ containing only odd-power terms, while cos nφ is a polynomial of degreen in cos φ containing only even-power terms (so it can also be written as apolynomial in sinφ if desired). See Remark to Problem on page 114.

92 SOLUTIONS

3. If sinnφ is given, the highest number of values that sinφ can take is n for nodd, and 2n for n even. Prove this yourself.

4. If cos nφ is given, then cos φ can assume no more than n different values. Provethis too.

Problem 3. Let the lateral sides ofthe trapezoid be AB and CD. Letthe line supporting the diagonal BD I modified the proof

slightly to eliminate the

cosine, and the drawing to

make it clear that the

intersection may occuroutside the diagonal

proper.

cross the circle constructed on AB ata second point M ; in the case of atangency the “second intersection” is M = B itself. Similarly, let N be thesecond intersection of the line BD with the other circle.

A

B

K

C

D

M

N

By the tangent-secant theorem, the square of the length of the tangentsfrom K to the circle on AB equals the product lengths of the secant segmentsKM and KB. (This area is called the power of the point K with respect tothe circle.) So to prove that the tangents to both circles have equal lengths,it is enough to prove that

KM · KB = KN · KD. (1)

Next we observe that AM intersects the line BD at right angles, since theangle AMB is subtended by a diameter (or, in the case of tangency, sincethe tangent is perpendicular to the radius whose end point is the tangencypoint). Likewise, CN is perpendicular to BD. But then the triangles AMKand CNK are similar, so

KA

KM=

KC

KN,

Multiplying this equation by (1) we reduce what we have to show to

KA · KB = KC · KD.

But this is a known property of any trapezoid, also provable using similarity(of the triangles AKD and CKB).

Remark. Instead of intersecting one diagonalwith both circles, we can intersect each I reworked the “variation

on the solution” to give

the reader something to

chew on.

diagonal with a different circle, say ACwith the circle on AB, determining a pointM, and BD with the circle on CD, deter-mining N (see figure). The proof of the statement remains the same, exceptthat the roles of KA and KB are interchanged.

A

B

K

C

D

M N

Further, in this situation, M and N lie on the circle with diameter BC.(Why?) Show that A,M,N,D also lie on one circle, either by comparing anglesor by using the equality KM ·KA = KN ·KD and the tangent-secant theorem.

Problem 5. First solution. Assume to the contrary that not all three num-bers have the same absolute value. Also assume that gcd(a, b, c) = 1; if thatis not so, substitute for the three original numbers their quotients by theirgcd, something that does not disturb either the hypotheses or the conclusion.


Now one of the three numbers is distinct from ±1; let it be divisible bythe prime p. At least one of the other two numbers is not divisible by p;otherwise p would divide the gcd. Assume that p does not divide c.

Given an integer x, denote by k(x) the power of p in the prime factoriza-tion of x; see Fact 10 on page 123. Thus k(c) = 0. Without loss of generalitywe assume that k(a) > k(b).

We havea

b+

b

c+

c

a=

a2c + b2a + c2b

abc. (1)

In the denominator of this fraction, p appears with power k(a)+k(b). Hence

the numerator should be divisible by pk(a)+k(b). We will show that this isimpossible.

The summand a2c contains p to the power 2k(a) > k(a) + k(b), and b2acontains p to the power k(a)+2k(b) > k(a)+k(b); hence a2c+b2a is divisible

by pk(a)+k(b). But c2b only contains p to the power k(b) < k(a) + k(b), so

c2b is not divisible by pk(a)+k(b). The sum (a2c+ b2a)+ c2b, therefore, is not

divisible by pk(a)+k(b) (see Fact 5). This is a contradiction.

Second solution. Set: x = a/b, y = b/c, z = c/a. Then xyz = 1 and

xy + yz + zx =xy + yz + zx

xyz=

1

x+

1

y+

1

z=

a

c+

c

b+

b

a

is an integer. Consider the polynomial

P (t) = t3 − (x+y+z)t2 + (xy+yz+zx)t − xyz.

We have shown that all the coefficients of this polynomial in t are integers.Since the highest coefficient is equal to 1, all rational roots of this polynomialare actually integers (see Remark). But as we recognize from Viete’s formulafor cubic polynomials (Fact 20 on page 127), or can check by expandingP (t) = (t−x)(t−y)(t−z), the roots are precisely x, y and z. Thus, x, y,and z are integers; since xyz = 1, we have x = ±1, y = ±1 and z = ±1.Equivalently, |a| = |b| = |c|.Remark. In the second solution we used a particular case (an = 1) of the following

fact:

Theorem. Let f(x) = a0+a1x+· · ·+anxn be a polynomial with integer coefficients.Any irreducible fraction r/q that is a root of f(x) satisfies r |a0 and q |an.

Proof. The condition f(r/q) = 0 is equivalent to

a0qn + a1rq

n−1 + · · · + anrn = 0. (2)

Suppose q does not divide an. Then there is a prime p and an integer t > 0 suchthat pt divides q but not an. Since r and q are relatively prime, p does not divide r.So the term anrn is not divisible by pt, whereas every other term on the left-handside of (2) is so divisible. But this is impossible (see Fact 5 on page 121).

The proof that r divides a0 is analogous, interchanging the roles of q and r. ˜

94 SOLUTIONS

Problem 6. First solution. Observe first that the result does not dependon the order in which the buttons are pushed. We will work by inductionon the number n of light bulbs. For n = 1, the statement is obvious.

Let the statement be true for n−1 lights; we must prove it for n lights.Consider the i-th light. By the induction hypothesis we can switch off

all lights, except perhaps this one. Denote the set of buttons we have topush for this by Si.

If, for some set Si, all the lights go out, we are done. Otherwise, aftereach Si only the i-th light is glowing.

What happens if we first push all the buttons from Si and then pushall the buttons from Sj? We invite the reader to check that this will flipthe state of exactly two lights, the i-th and j-th ones, leaving the remaininglights in the same state as at the beginning.

This means we can switch off any two lights, and hence any even numberof lights. Thus if initially an even number of lights is on, we are done.

Assume that initially an odd number of lights is on. By assumptionthere is a button that flips an odd number of lights. Push it; now then thenumber of lit bulbs is even, and we can switch them off in pairs as in theprevious case.

Second solution. This solution uses linear algebra over the field F2 with twoelements, and so falls outside the high-school curriculum. But the readermay be familiar with the usual linear algebra over real numbers, and thesame notions apply with minor changes, which we point out along the way.

Arithmetic in F2 = {0, 1} is modulo 2: the addition and multiplicationtables are the obvious ones, except that 1+1 = 0. (See Fact 25 and the firsttwo chapters of Field2elem.) CITE Field2elem

Let the lights be numbered from 1 through n. To every state of thepanel we assign a row vector x = (x1, . . . , xn), where xi = 1 if the i-th lightis on and xi = 0 otherwise. Such vectors are elements of the n-dimensionalspace F

n2 , analogous to R

n.We also assign a vector to each button: b = (b1, . . . , bn), where bi = 1

if and only if that button flips the i-th light. Pushing the button changesthe state from x to x + b. Pushing several buttons in sequence amounts toadding the sum of the corresponding b’s to the initial vector x. If that sumequals the initial vector, the final vector x +

∑b equals 0, meaning that

all the lights are off. Our problem, therefore, is to prove that the vectorscoming from buttons span the whole n-dimensional space.

To a set of lights we assign a linear functional (I.e., a linear functionF

n2 → F2), given by

(x1, . . . , xn) 7→∑

xi,

where the sum runs over the indices i of the lights in the given set. BecauseF2 has only one nonzero element, any linear functional is obtained is thisway, for some set of lights.


The functional defined by a certain set of lights vanishes at the vectorcoming from a button if and only if this button flips an even number oflights in the set. Therefore the assumption that for any (nonempty) set oflights there is a button flipping an odd number of lights from this set canbe recast in the language of linear algebra as follows: No nonzero functional the word “nonzero” is

missing in the original.vanishes on all vectors coming from buttons. (A nonzero functional is onethat takes a nonzero value somewhere.)

But this implies what we wish to show, for the following reason. Itis a fact of linear algebra that given a vector subspace and a point in thecomplement of this subspace, there is a linear functional on the whole spacethat vanishes on the given subspace but not at the chosen point. So if thevectors coming from buttons don’t span the whole space, there is a linearfunctional that vanishes on all vectors coming from buttons, but not on somevector outside the span; and this is contrary to the assumption.

Remark. Call a set of lights sticky if any button changes the state of an evennumber of lights in the set. The problem stipulates that there are no stickysets. Consider the more general situation where sticky sets are allowed. If theset of initially glowing lights has an odd number of lights in common with asticky set, it’s impossible to switch off all the lights. (Why?)

Prove, by induction or using linear algebra, that if the set of initially glow-ing lights has an even number of lights in common with any sticky set, then itis possible to switch off all the lights.

Level D

Problem 1. Since the absolute value of a sum does not exceed the sum ofabsolute values (see Remark), we have:

|x+y−z|+ |x−y+z| > |(x+y−z)+(x−y+z)| = 2|x|.We similarly get

|x−y+z|+ |−x+y+z| > 2|z| and |−x+y+z|+ |x+y−z| > 2|y|.Adding up the three inequalities and dividing the sum by 2, we obtain thedesired result.

Remark. Although the inequality

|x + y| 6 |x| + |y| (1)

can easily be proved case by case, we give an elegant proof. Since both sidesof the inequality are nonnegative, we can square them and get an equivalentinequality. In other words, it suffices to prove that

|x + y|2 6 (|x| + |y|)2.Expanding squares and using the fact that |a|2 = a2 for every a, we see thatthis is equivalent to x2 + 2xy + y2 6 x2 + 2|x| |y| + y2, which is obvious.

Inequality (1) is true for vectors x and y in n-dimensional space as well.The same proof works, with slight modifications. On the plane, (1) is equivalentto the triangle inequality.

96 SOLUTIONS

Problem 2. Notice first that the vertex condition is equivalent to havingthe three colors of the edge meeting at a vertex all different.

(a) Since 1995 is divisible by 3, we can color the bottomedges red, green and blue in turn. We then color the match-ing top edges with the same colors. Now each vertical edgeis adjacent to edges of two colors; we use the remainingcolor for the vertical edge.

br g b r

r

g

gbb

rg b

r

Clearly, the colors of the lateral edges repeat with period three; hence,on each lateral face, the lateral edges are of different colors, and the top andbottom edges have the third color, as required.

(b) Suppose for a contradiction that the prism is colored as required. Theremust be three consecutive edges of different colors on the top face; other-wise the perimeter of that face is colored alternately with only two colors,contrary to assumption.

Take such a run of three edges; let their colors be red, green, blue,proceeding counterclockwise as seen from above. The vertical edges goingdown from the two intermediate vertices clearly have their colors determined(blue and red). Therefore the bottom edges between and adjacent to thesevertical edges also have their colors determined, matching the upper ones.(Why?) Further, the next vertical edges, at both ends of the run, mustbe green, and so the next edges on the top face in both directions are alsodetermined. We now have a run of five edges: blue, red, green, blue, red, asin the figure above.

We conclude by induction that the top and bottom edges are coloredwith period 3, But 1996 is not divisible by 3, so this is not possible.

Problem 3. First solution. We extend the medianAA1 by its length and complete the triangle to aparallelogram ABDC, as in the figure.

A B

K

C D

A1

A2

Recall that a bisector divides the opposite sideof a triangle in a ratio proportional to the adjacentsides:

A2B

A2C=

AB

AC.

Because A2K is parallel to AC and BD, it divides the transversals AD andBC into proportional segments (this is sometimes called the “generalizedintercept theorem”: see Remark 1 below), Thus the point K divides AD inthe ratio

KD

KA=

A2B

A2C=

AB

AC=

CD

AC.

Hence, K divides AD in the same ratio as the bisector of \ACD does, bythe same property of the bisector already mentioned. Therefor CK is thebisector of \ACD! Thus, AA2 and CK are perpendicular, since

\KAC + \KCA = 12(\BAC + \DCA) = 1

2 · 180◦ = 90◦.


(Here we just proved that the bisectors of facing interior angles of a transver-sal to two parallel lines are perpendicular.)

Second solution. We use a notable property of trapezoids (see Remark 2below): the midpoints of the bases, the intersection of the diagonals and theintersection of the lines containing the nonparallel sides are all collinear.

A B

K

C

A1

A2

B1

C1

R

We work as follows (see figure). We first extend the segment CK to theintersection with line AB at point C1. In the trapezoid AKA2C, the

midpoint B1 of the base AC, the intersection of the diagonalsR and the intersection A1 of the extensions of the nonpar-

allel sides are collinear; therefore R lies on the midlineA1B1 of the triangle ABC, and we get CR = RC1.

Thus, in the triangle CAC1, the segment ARis simultaneously a median and a bisector, andhence it is also an altitude.

Remarks. 1. The generalized intercept theorem. Let A1, A2, A3 be pointson a straight line and B1, B2, B3 points on another line. If the segments A1B1,A2B2 and A3B3 are parallel (or exactly one is a point while the other two areparallel), we have

A1A2

A2A3=

B1B2

B2B3.

The proof is easy. We may assume that A2B2 is not a point. Through A2

draw a line parallel to the line containing B1, B2, B3. Let this line intersect thelines A1B1 and A3B3 at points C1 and C3, respectively. Then A2C1B1B2 is aparallelogram, so A2C1 = B2B1. Analogously, A2C3 = B2B3. It remains to usethe similarity of the triangles A1A2C1 and A3A2C3 (which is the usual intercepttheorem).

2. A remarkable property of the trapezoid. In a trapezoid, the midpointsof the bases, the intersection of the diagonals and the intersection of the linescontaining the nonparallel sides are all collinear.

The simplest proof uses homothety. There exists a homothety with centerP sending one base of the trapezoid onto the other base, and therefore sendingM to N . Therefore P lies on the line MN .

Similarly, there exists a homothety with center Q sending one base of thetrapezoid onto the other. So Q also lies on the line MN .

Problem 4. We prove by the induction on n that it is possible to divideany interval into white and black intervals so that the sum of the integralsof any polynomial of degree 6 n over white intervals is equal to that overblack intervals.

Base of the induction. For n = 0, the polynomial is a constant. Theintegral is the product of this constant by the length of the interval. So wedivide the interval in half.

Induction step. Denote the midpoint of the interval [a, b] by c. Bythe induction hypothesis, we can divide [a, c] into black and white intervalsso that the sum of integrals of any polynomial of degree 6 n− 1 over white

98 SOLUTIONS

intervals is equal to that over black intervals. Let B1, . . . , Br be the blackintervals and W1, . . . ,Ws be the white ones, numbered left to right. Wedenote the integral of the function f(x) over Bi by

∫

Bi

f(x) dx.

We similarly denote the integral of the function f(x) over Wi. Now trans-late each interval Bi to the right by c, color the result white and denoteit by Ws+i. Similarly, translate Wi to the right by c, color it black anddenote it by Br+i. Clearly, the intervals B1, . . . , Br+s together with theintervals W1, . . . ,Wr+s form a division of the interval [a, b] into black andwhite intervals. The figure illustrates the cases n = 1 and n = 2.

n = 1

B1 B2W1 W2

a c b

n = 2

B1 B2 B3 B4W1 W2 W3 W4

a c b

We claim that the division of [a, b] thus obtained has the desired propertyfor all polynomials of degree 6 n. To prove this, we first use the change ofvariables y = x − c, dy = dx to check that

∫

Wr+i

f(x) dx =

∫

Bi

f(y+c) dy,

(see 28 on page 131). Then, since the integral of a difference equals thedifference of the integrals, we have

∫

Wr+i

f(x) dx −∫

Bi

f(x) dx =

∫

Bi

(f(x+c) − f(x)) dx.

Analogously,∫

Wi

f(x) dx −∫

Bs+i

f(x) dx = −∫

Wi

(f(x+c) − f(x)) dx.

Adding up these equalities for all i, we get

r+s∑

i=1

∫

Wi

f(x) dx −r+s∑

i=1

∫

Bi

f(x) dx

=

r∑

i=1

∫

Bi

(f(x+c) − f(x)) dx −s∑

i=1

∫

Wi

(f(x+c) − f(x)) dx. (1)

Now let f(x) be a polynomial of degree 6 n. Then f(x + c) − f(x) isa polynomial of degree at most n − 1; see Fact 22 on page 128. By theinduction hypothesis the right-hand side of (1) vanishes, so the left-handside also vanishes.


Remarks. 1. Consider the subdivision of the unit interval into subintervals oflength 2−n according to the procedure above. Letting “+” stand for a blackinterval and “−” for a white one, the resulting sequence of signs is the n-th termof the Morse sequence

+, +−, +−−+, +−−+−++−, +−−+−++−−++−+−−+, . . .

where from each term A we obtain the next by writing A twice and then inter-changing + and − everywhere in the second copy.

The n-th term of the Morse sequence can also be constructed as follows:For every k = 0, . . . , 2n − 1, the k-the sign in the sequence is the parity of thenumber of 1’s in the binary representation of k (see Fact 12 on page 123).

One can show that in the strings of the Morse sequence no substring appearsthree times in a row.

2. The expression

Dcf(x) = f(x + c) − f(x)

is called the first difference of the function f with step c. The second differenceis the first difference of the first difference, and so on.

Before computers, first and higher differences were used for tabulating func-tions. For example, the function sinx can be approximated by a polynomial ofdegree n, and this can be used to compute a table whose k-th column containsthe values of the k-th differences. One starts filling out the n-th column, whichis constant for a degree-n polynomial. Then, given the k-th column, it’s easy tofill out the (k−1)-st, and so work backward to the function itself. Thus, one cantabulate any function using summation only, so long as it can be approximatedby polynomials.

Problem 5. We first show that n > 1994. Let A have period 1000 . . . 00(1994 zeros), and let B be periodic as well, with period 1000 . . . 000 (1995zeros). Any piece of B of length at most 1994 occurs in A, and the otherconditions are also satified.

We next show that n < 1995 by proving that, for any A of period 1995,if all substrings of B of length 1995 are found in A, then B = A, apart fromreindexing.

Indeed, assume that A and B satisfy the hypotheses. We first show thatB repeats with periodicity 1995 (that is, 1995 is a period of B). Otherwise,B would contain a substring of length 1996 with different first and lastsymbols:

B = . . . x . . . . . . . . .︸︷︷︸

1994 symbols

y . . . ,

where x 6= y. Let Z be the substring between x and y, and assume x occursin Z exactly k times. Consider the two strings xZ and Zy. Both occur inA, since they have length 1995. But their length coincides with the periodof A, so xZ must be a cyclic permutation of Zy. In particular, x shouldappear the same number of times in xZ as in Zy. But this is clearly not thecase. Therefore B repeats with periodicity 1995.

Now take any substring of B of length 1995. By assumption, this sub-string occurs somewhere in A. But we already know that B and A are

100 SOLUTIONS

completely determined by any substring of length equal to 1995; thereforeB and A coincide as claimed, and 1995 is the minimal period of B.

Remark. The proof also implies that if a two-sided infinite string has minimal periodn, any two identical segments of length n − 1 are offset by an integer numberof periods.

Problem 6. The difference of powers suggests using the fact that, if r is amultiple of s, then ar−br divides as−bs, where a, b are variables or numbers.(This is easily proved:

ar − br = (as − bs) · (a polynomial in as and bs).

See also Fact 8 on page 122.)Thus, if n−1 is a composite number and s is one of its factors, the

difference 3n−1 − 2n−1 is divisible by 3s − 2s. We want 3n−1 − 2n−1 to bedivisible by n, so we look for n in the form 3s − 2s and such that n−1is divisible by s. (We also need n itself to be composite, but this comes freewith our choice of n, so long as s is not prime.)

Let s = 2t. Let’s prove that n− 1 is divisible by s, for all integers t > 0.This amounts to showing that 32t−1 is divisible by 2t, since 22t

is certainlydivisible by 2t.

For t = 1 this is clear: 9 − 1 divides 2. Let it be true for some t > 1.Then for t + 1, we have

32t+1− 1 = (32t

+ 1)(32t− 1).

The first factor is even, and the second is divisible by 2t by the inductionassumption; hence the product is divisible by 2t+1 and we are done.

Remarks. 1. Why does the problem specify that n must be composite? If not,Fermat’s little theorem (see [Chapter 3, § 6]MTF, for example) makes the prob- CITE [

lem easy. The theorem says: If p is prime and a is not divisible by p, thenap−1 − 1 is divisible by p. Use it to show that 3n−1 − 2n−1 is divisible by n forn prime and distinct from 2 and 3.

2. There are infinitely may composite numbers n such that 2n−1 − 1 is divisibleby n.

Problem 7. Let’s arrange six sticks (long thin parallelepipeds)in the pattern shown in the figure, resting on the faces ofan imaginary central cube. The sticks almost touch theirneighbors, but not quite.

From the center of the construction, no vertices arevisible, because either end of any stick is hidden by a faceof another stick, which lies much closer. Looking in anydirection, we see only portions of faces and edges, as in the figure to the left.

Next we make as least five bridges between the sticks, to con-nect them all into a polyhedron. This creates new vertices, but wecan contrive to hide them from the center as well. For instance,a bridge might be made by welding a thin, straight wire (with a


square cross section) from one end of a stick to one end of another. The newvertices are very close to old ones, so they remain invisible.

Remark. In two dimensions such an example is impossible: For any polygon anda point outside it, at least one of the polygon’s vertices is seen from the point(although, perhaps, no side is seen completely).

Back in three dimensions, one can guarantee that at least one vertex isvisible if there is a plane that separates the eye from the polyhedron.

Year 1996 Olympiad

Level A

Problem 1. Reducing each side of the equation to a common denomina-tor — for instance, a + b2/a = (a2 + b2)/a — we obtain

a2 + b2

a=

a2 + b2

b.

From the initial equation it is clear that a 6= 0 and b 6= 0. Therefore,a2 + b2 > 0 (see the remark below). So we can eliminate the factor a2 + b2

from both sides to obtain 1/a = 1/b. Hence a = b.

Remark. The following obvious statements are often used in olympiad problemsolving (and in mathematics generally): (1) the square of a nonzero number ispositive; (2) the sum of the squares of several numbers is nonnegative. If sucha sum is zero, then each of the numbers is zero.

Problem 2. Denote by mi the masses of the weights and by xi the massesof the balls. The sum (m1−m2)+(m2−m3)+ · · ·+(m9−m10)+(m10−m1)equals 0, because each mi occurs exactly twice, with opposite signs, leadingto complete cancellation.

The mass xi of the i-th ball is the absolute value of the differencemi − mi+1, so the preceding condition can be rewritten as

±x1 ± x2 ± · · · ± x9 ± x10 = 0,

where some signs of the xi are pluses and the others are minuses. Choosethe balls whose masses enter into this sum with a plus sign and put themon the left pan of the balance. Put the other balls on the right pan. Thebalance will then be in equilibrium.

Problem 3. Consider a grid square with a flower init. Divide it into four equal quadrants (small squares).We can assume without loss of generality that thesides of the quadrants have unit length. Suppose ourflower is in the upper left quadrant, as in the diagram;the other cases are all similar. Denote the gardenersat the vertices of the square by A, B, C, D.

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1

A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2 B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1

B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2

104 SOLUTIONS

We will prove that the flower will be looked after by gardeners A, B, andC. We divide all the gardeners into four groups, each group being locatedin one of the four shaded sectors in the figure.

Gardener A is closer to the flower than B. This follows from thePythagorean Theorem, because the vertical separation from the flower isequal for both gardeners and the horizontal separation is less for A than forB (being less than 1 for the former and greater than 1 for the latter).

Similarly, gardener A is closer to the flower than C, and gardeners B andC are closer than D. Therefore, D does not look after the flower. Neitherdoes any other gardener from the D sector: they are all even farther fromthe flower than D (consider the vertical and horizontal distances again).

Now consider the B sector. Clearly, all its gardeners (apart from B) arefurther from the flower than B, and so even further than A. We will showthey are also further than C, and so do not look after the flower. Take,for instance, gardener B1. The vertical separation from B1 to the floweris greater than 2, and the horizontal separation is greater than 1. So thisgardener is farther from the flower than C.

It is exactly analogous to show that no gardener in the C sector, apartfrom C, takes care of the flower.

Finally, consider the A sector. We must show that A1 and A2, thenearest gardeners apart from A, are further from the flower than B and Care. Consider A1 (the proof for A2 is similar). To see that A1 is furtherthan B, just look at horizontal separations. To prove that A1 is further thanC, consider the perpendicular bisector to the segment A1C. Its points areequidistant from A1 and C. Points below the bisector are closer to C thanto A1. Clearly, the flower is below the bisector; therefore, it is closer to C.

Thus, the gardeners that look after the flower are A, B, and C.Now consider the gardener at the origin, say X. We already know that

X does not look after flowers unless they’re in one of the four squares (of theoriginal grid) touching the origin. And we need only find the intersection ofX’s area of responsibility with one of these squares: the pieces in the otherthree squares are found by symmetry.

For instance, consider the square XY ZT in the figure. Again, we divideit into quadrants. As we have seen, gardener X is one of those who looksafter the three shaded squares. Flowers in the fourth quadrantare looked after by Y , Z, and T . Hence, gardener X looks afterthe flowers growing in these three small squares and in the nineother squares obtained from them by reflection. X T

Y Z

Problem 4. Without loss of generality, assume K is closerto B than L, as in the figure. Then MKC is an equilateraltriangle (since MC = KC and \MCK = 60◦). Hence,AB is parallel to MK (since \MKC = \ABC = 60◦)and the angles AKM and BAK are equal (being alter-nate interior angles).

A

B CK L

M


Notice that the angles BAK and CAL are equal by symmetry (or by thecongruence of △BAK and △CAL, which follows from the SAS property).Hence \AKM = \CAL and

\AKM + \ALM = \CAL + \ALM = \LMC.

The last equation follows from the fact that \LMC, being an exterior angleof triangle AML, equals the sum of its two opposite interior angles.

Thus, it remains to prove that \LMC = 30◦. But ML is a median inthe equilateral triangle MKC; hence ML is an angle bisector of this triangle.

Remark. Compare with the solution to Problem 3 on page 111.

Problem 5. We can assume that the rook startsfrom the upper left corner. A solution for even n isillustrated on the right. The rook visits first all thesquares on the top two rows, then the squares on thethird and fourth rows, and so on.

We now show that the task is impossible for oddn. Consider any row apart from the top one. Whenthe rook hits this row, its next move must be to an-other square of the same row, and then to another row. Thus, the squaresof the chosen row are paired up by the moves within that row. (Note thatthe rook visits each square exactly once, since the number of moves is n2.)Therefore any row contains an even number of squares.

In other words, after its first visit to a row, the rook will have visitedtwo of its squares. After the second visit, there will be four such squares. Ifn is odd, a moment would come when the rook must visit the last unvisitedsquare on a row, and then it would be unable to move. (See also Fact 23 onpage 128.)

Problem 6. (a) Let N be the greatest number of problems solved by asingle student. So, for instance, if N = 8 one student solved all problemsand we’re done; if N = 7 we take a student who solved 7 problems and onewho solved the remaining problem, and again we’re done.

Consider the student/problem pairs that resulted in solutions. Thereare at least 40 such pairs (exactly 40 if we interpret “solved by five stu-dents” more strictly; either interpretation works). By the pigeonhole prin-ciple (page 119), at least one student solved 5 or more problems. So we’releft with two cases for N :

N = 6, so there is a student who solved exactly six problems. Each of theremaining two problems was solved by 5 among the remaining 7 stu-dents, and since 7 is less than 5 × 2, there must be some student whosolved both problems (pigeonhole principle again).

N = 5, that is, each student solved at most five problems. Since there were atleast 40 solutions, each student must have solved exactly five problems,and each problem must have been solved by exactly five students.

106 SOLUTIONS

So suppose student 1 solved problems 1–5. We claim there is astudent among the other seven who solved problems 6, 7, and 8. Ifthis were not true, each of the seven would have solved at least threeof the first five problems. Together with student 1, this would give5 + 7 × 3 = 26 solutions of problems 1–5. But this cannot be, becauseeach problem was solved by exactly five students. Therefore our claimis true.

(Another way to think of this last paragraph is as follows. Eachproblem was left unsolved by three students; in the case of problems 6,7, and 8, these are student 1 and two more. Therefore, all in all, thereare at most seven students— including student 1— who failed to solveone or more problems among the last three. That leaves at least onestudent who solved all three.)

(b) We need to fill an 8 × 8 table (rows representing students, columnsrepresenting problems) so that each column has four or more entries and notwo rows complete each other to form the full set {1, 2, . . . , 8}. We may aswell assume that each column has exactly four entries.

Any solution is likely to involve some trial and error, but we use what welearned in part (a) to direct the search. In particular, we keep the definitionof N . The cases N = 8, N = 7 and N = 6 won’t help us, because our earlierarguments show that in these cases, there are two students who solved allproblems between themselves. (In the case N = 6 the relevant inequality isnow 7 < 4 × 2.)

Next we try for a counterexample assuming N = 5. Naturally, we startby filling row 1 in positions 1–5. By assumption, in the subtable consistingof the last three columns and seven rows, there are 3× 4 = 12 entries. Eachof these rows has at most two entries in columns 6–8; otherwise it wouldcomplete row 1. It is reasonable to place exactly two entries on each of rows3–8, leaving row 2 with nothing in columns 6–8; thisway we can make row 2 a duplicate of row 1 withoutlosing any freedom.

The simplest way to choose two of the numbers 6,7, 8 for each of the last six rows so that each numberis chosen four times is to group the rows in pairs, as inthe figure: rows 3 and 4 skip column 6 and get entriesin columns 7 and 8; the next two rows skip column 7instead; and the last two rows skip column 8.

1

2

1 2 3 4 5

3

4

5

6

7

8

6 7 8

students

pro

ble

ms

We have now filled everything in, except for the subtable consiting ofcolumns 1–5, rows 3–7. This is similar to the original task, but with twoentries per column (out of six), so it’s easy to solve. Also, we must take carethat no two rows that complete each other in the first five columns shouldalso complete each other in the last three. This is best achieved by keepingthe pairing of rows already introduced; we divide the first five columns intothree chunks (of lengths 2,2,1 or 3,1,1) and fill one chunk per pair of rows,


like this:

or

Level B

Problem 1. First solution. Suppose the statement is not true. Then acertain n-gon has at least 36 angles less than 170◦, and the remaining n−36angles less than 180◦ (convexity). Therefore, the sum of all angles of thispolygon is less than 36 ·170◦ +(n− 36) ·180◦. But it is well known that thissum is equal to (n − 2)180◦. This yields the inequality

(n − 2) · 180 < 36 · 170 + (n − 36) · 180.

After removing the parentheses and an obvious simplification, we obtain theinequality 180 · 34 < 170 · 36, which is false.

Second solution. If an interior angle of a polygon is less than 170◦, thecorresponding exterior angle is greater than 10◦. If we had 36 or more suchangles, their sum would be greater than 360◦. But the sum of exterior anglesof a convex polygon is 360◦. This contradiction completes the proof.

Problem 2. First solution. First assume that one of the numbers is zero.For instance, let a = 0 (the other cases are similar). Then we have theinequalities |b| > |c| and |c| > |b|, whence |b| = |c|, that is, b = c or b = −c.In the first case, we have b = a + c; in the second case, a = b + c, and we’redone.

Now suppose that none of the numbers a, b, and c is zero. Withoutloss of generality we can assume that a is the greatest of the three numbersin absolute value (that is, |a| > |b| and |a| > |c|). We can also assumethat a > 0 (if it isn’t, we replace all three variables by their negatives, anoperation that affects neither the problem’s hypotheses nor the conclusion).Then |a| = a, |a−b| = a−b, |a−c| = a−c.

Under these assumptions, the inequality |b − c| > |a| implies that b andc have different signs. (Why?)

Consider the two possible cases.(1) b > 0, c < 0. Then |b| = b, |c| = −c, |b − c| = b − c, and the

given inequalities take the form a − b > −c, b − c > a, a − c > b. The firstinequality implies that b 6 a + c; the second one, that b > a + c; therefore,b = a + c.

(2) b < 0, c > 0. Then, as in the previous case, we obtain the inequalitiesa−b > c, c−b > a, a−c > −b. Therefore, in this case we have simultaneous

108 SOLUTIONS

inequalities c > a + b and c 6 a + b, that is, c = a + b. Thus, the statementis proved in both cases.

Second solution. Square both sides of the inequality |a − b| > |c| and carryall terms to the left-hand side. We get (a− b)2 − c2 > 0. The left-hand sidecan be factored as a difference of squares: (a − b − c)(a − b + c) > 0, or,which is the same,

(a − b − c)(b − c − a) 6 0.

By a similar argument, the products (b−c−a)(c−a−b) and (c−a−b)(a−b−c)are also nonpositive.

Multiplying the three products, we find that

(a − b − c)2(b − c − a)2(c − a − b)2 6 0.

But if a product of nonnegative numbers is not positive, it must be zero.Hence at least one of the numbers is zero and we’re done.

Problem 3. The angles BNM and NCA

A

B

C

M Nare equal, because the lines MN and ACare parallel. The inscribed angle NCA isequal to half the measure of its interceptedarc AB

⌢. The angle BAM between a tangent

and a chord is also equal to half the measureof the arc AB

⌢which it intercepts (Fact 15 on

page 124), so \NCA = \BAM . Hence the anglesBNM and BAM are equal, and the quadrilateralAMBN is cyclic. We have the following chain of equations for angles:

\NCA = \BAM = \MBA = \MNA = \NAC.

The second equation follows from the fact that AMB is an isosceles triangle(the tangent segments drawn to a circle from an external point are equal),the third one from the fact that AMBN is a cyclic quadrilateral, and thelast one from the equality of alternate interior angles of parallel lines.

It follows that \NCA = \NAC; hence ANC is an isosceles triangle,i.e., AN = NC.

Remark. The attentive reader may have noticed that we used the fact that \BCAis acute. Check that otherwise there is no point N on the side BC such thatMN ‖ AC.

Problem 4. (a) The sum of the number 9 and the number beneath it liesbetween 10 and 18. Since there is only one perfect square in this interval,the number under 9 must be 7. Similarly, the number 7 must be writtenabove the number 9 on the second row. In the same manner, it can be shownthat the numbers under 4, 5, and 6 must be 5, 4, and 3, respectively. Nowit is not difficult to complete the answer:

1 2 3 4 5 6 7 8 9

8 2 6 5 4 3 9 1 7


(b) Using the same reasoning we see that the only number that can bewritten under 11 is 5. But the number under 4 must also be 5, which isimpossible.

(c) The idea is to reduce the problem to a similar one for smaller n. Thefirst square past 1996 is 2025 = 452, which equals 1996 + 29. So if undereach number k from 29 through 1996 we write the number 2025 − k, thesum-is-a-square condition is satisfied for all columns after the first 28. Thisreduces the problem to the case n = 28.

Repeating the same strategy we write under each of the numbers k =21, 22, . . . , 28 its counterpart 49−k, reducing the problem to the case n = 20.Then we write 36−k under each k = 16, 17, 18, 19, 20, reducing the problemto n = 15. Finally, under each k = 1, 2, . . . , 15 we write 16 − k.

Remark. When do you run into trouble with this strategy? Try to prove that theproblem can be solved whenever n > 11.

Problem 5. We begin with two simple statements:

(1) The segment joining the midpoint of a chord to the circle’s center isperpendicular to the chord.

(2) The condition that the endpoints of a chord lie on different arcs ABis equivalent to the chord intersecting the segment AB (in an interior point).

Thus the problem can be reworded as follows: Given three points A,B, and O with AO = BO, find the locus of all points M such that theperpendicular to MO going through M crosses the open segment AB.

We now prove that the perpendicular to MO at M intersects the opensegment AB if and only if exactly one of the two angles OMA and OMB isobtuse. Indeed, the perpendicular intersects AB if and only ifA and B lie in opposite half-planes with respect to thisperpendicular; this condition can also be expressed bysaying that exactly one of the two points (A or B) liesin the same half-plane as the point O, and the other liesin the half-plane opposite O. But a point P lies in thehalf-plane opposite to O if and only if the angle PMOis obtuse (as illustrated by point A in the figure on theright), and P lies in same half-plane as O if it is acute(point B in the figure).

A

B

OM

The locus of points M such that \AMO is the interior of the circle withdiameter AO, and the locus of points M such that \BMOexceeds 90◦ is the interior of the circle with diameter BO(see Fact 14 on page 124). This means that the locus inquestion consists of the points lying inside one, but notboth, circles with respective diameters AO and BO. Inother words, it is the union of the interior of these circlesminus their intersection —the shaded region in the figure.

A B

O

110 SOLUTIONS

Problem 6. Let us show that Ali Baba can always arrange for seven pilesto end up with at most four coins, and the thief can ensure that no pile evercontains less than four coins. That will mean that Ali Baba can walk awaywith 100 − 7 · 4 = 72 coins.

We start by proving that the thief can ensure there are at least four coinsin each pile after each exchange. This is certainly true at the beginning.Suppose it is still true at a certain point and some coins are then placed inthe cups. If there are two cups with the same number of coins, the thief cansimply swap them, leaving the situation unchanged. If the number of coinsis different for each cup, then the two cups with most coins will contain atleast three and four coins, respectively, and the thief can swap these cups.Then the new piles will again consist of at least four coins each.

Now we show that Ali Baba can always reduce seven piles to four coinsor less.

Suppose there are four piles with more than four coins each. Let x1 >

x2 > x3 > x4 > 5 be the numbers of coins in these piles. We’ll show thatAli Baba can work with these piles alone until one of them has four coins orless. He chooses the partitions

x1 = y1 + 1, x2 = y2 + 2, x3 = y3 + 3, x4 = y4 + 4,

and puts 1, 2, 3, and 4 coins from these piles into the respective cups. Afterthe thief rearranges the cups, the new piles will consist of

y1 + z1, y2 + z2, y3 + z3, y4 + z4 (1)

coins, where z1, z2, z3, z4 is a certain permutation of the numbers 1, 2, 3, 4.Let x′

1, . . . , x′4 be the four numbers in (1), reordered if necessary to preserve

their nonincreasing order. There are three possibilities:

(1) If the first cup was moved, the tallest pile got even taller (sincez1 > 1). In other words, x′

1 > x1.(2) If the first cup stayed put and the second was moved, the second

tallest pile grew; it either surpassed the first and we have x′1 > x1 (since

we renumbered the piles), or this did not happen and we have x′1 = x1 and

x′2 > x2.

(3) If the first two cups stayed put, the last two were interchanged; inthis case x′

1 = x1, x′2 = x2, and x′

3 > x3.

Ali Baba then repeats this process so long as these same four piles stillhave more than four coins each. We must show that this cannot go onforever. Conclusions (1)–(3) can be rephrased as follows: on each applicationof the process, x1 can only go up, never down; if x1 doesn’t go up, x2 canonly go up, never down; if neither x1 and x2 go up, x3 must go up. Thismeans that no triple (x1, x2, x3) can ever occur more than once. (To convinceyourself of this, consider the parallel situation where x1, x2, x3 represent thedigits of a decimal number and obey the same rules: the number as a wholemust increase at each step.) But there are only finitely many triples, so


eventually it must be impossible to apply the process again, indicating thatsome pile now has four coins or less.

Considering now all the piles, the net effect so far is that the number ofpiles with four coins or less has increased. As long as Ali Baba can repeatthe whole procedure, this number will keep increasing. It only stops whenthere are at least seven piles with four coins or less.

Level C

Problem 1. First solution. We can assume that a > b (the case a 6 b istreated similarly). Then b2 6 ab, a2 > ab, and therefore,

a2 > a2 + b2 − ab > b2,

whence a > c > b. It follows that the first factor in the product (a−c)(b−c)is nonnegative and the second one is nonpositive. But then the product isnonpositive, completing the proof.

Second solution. Consider a triangle with a 60◦ anglebetween two sides of lengths a and b. By the law ofcosines, the length of the third side is

b a

c

60 ◦

√

a2 + b2 − 2ab cos 60◦ = c.

Since the greatest angle of any triangle is at least 60◦ and the smallestangle is at most 60◦, the angle opposite side c is of intermediate value inour triangle. Since a greater angle in a triangle lies opposite a longer side,either a 6 c 6 b or b 6 c 6 a. Hence one of the two factors a − c andb − c is nonnegative and the other is nonpositive. Therefore, their productis nonpositive.

Problem 2. Draw all 17 lines parallel to a diagonal ofthe square and passing through at least two of the markedpoints (see figure). This takes care of all points except twocorners of the array, which require an 18th line.

To see that fewer lines will not suffice, consider thecenters of the 36 unit squares around the edges of thearray. Clearly, any line not parallel to the edges of the array crosses at mosttwo of these points, so we need at least 18 lines to account for all of them(pigeonhole principle, see page 119).

Problem 3. Since an exterior angle of a triangle is equal to the sum ofits two opposite interior angles, the triangle APkM yields the equation\PkMC = \PkAC + \APkM , or \APkM = \PkMC − \PkAC. Addingtogether all these equations, we see that the sum in question is equal to

(\P1MC + · · · + \Pn−1MC) − (\P1AC + · · · + \Pn−1AC).

Suppose that n is odd; the complementary case is similar. Reflection aboutthe altitude of the equilateral triangle P1MC shows that the angles PkMCand Pn+1−kMC add up to 60◦ (see figure on the next page).

112 SOLUTIONS

Then the terms in the first sum fall into (n − 3)/2 pairs,each of which adds up to 60◦, and two unpaired terms:\P1MC = 60◦ and \P(n+1)/2MC = 30◦. Hence the firstsum is equal to 30◦ ·n. The terms of the second sum fallinto (n− 1)/2 pairs with a total of 60◦ each. Hence thesecond sum equals 30◦ · (n− 1). The statement of theproblem follows immediately.

A

P1 Pk Pn+1−kB C

M

Remark. Compare with Problem 4 on page 15.

Problem 4. The case m = n = 1 is obvious: there’s no room for even asingle move. For definiteness, assume that the board consists of n rows andm columns, with m > n, and that the rook starts from the upper left corner.

It turns out that the winning strategy for the first player is to make eachmove the longest possible. To prove this, we’ll assume that there are boardson which this strategy does not win and refute this conjecture by derivinga contradiction. Among all such boards, we can take one with smallestpossible area. Since on an 1 × n board the first player obviously wins, wecan assume that both dimensions are greater than 1: m > n > 2.

The first player moves the rook along the entire longer side, the top row.The second one is forced to move down. Consider three cases:

(a) The second player makes a one-square move. We’re nowin the starting situation of the same game played on anm × (n−1) board (see figure). Since m > 2, this is notthe 1 × 1 board.

(b) The second player moves all the way down. Then the firstmoves all the way across, per the strategy. If m = n = 2,the first player wins immediately. If not, we’re in thesituation we’d be in after the initial move if the gamewere being played on an (m−1) × (n−1) board.

(c) The second player moves by k squares, where 1 < k < n−1. Thefirst player moves all the way across, per the strategy. At this point,whether the next move by the second player is up or down, we’re in the in the original, the

black-edged rectangle istoo long.

situation we’d be in after the first player’s initialmove if the game were being played on a smallerboard: either an (m−1) × (n−1) board (thick solidline in the figure; note that this cannot be the 1× 1board since n > 4) or an m × (n−k) board (doublethin line).

In every case, we have reduced the situation to onethat is encountered, if the strategy is followed, during a game played ona board of lesser area (and distinct from 1 × 1). But, by our least-areaassumption, on such a board the first-player strategy wins. Hence it winson the m × n board as well, and we arrive at a contradiction.


Problem 5. We answer the question with an example. Suppose there are 10inhabitants in the country, and that their houses are placed along a straightline in ascending order of their owners’ heights. Suppose the intervals be-tween the houses, in kilometers, are 1, 2, 3, 4, 5, 4, 3, 2, and 1:

Then everyone except the tallest person can travel free of charge. Indeed,the five shortest persons can choose circles of enormous radius, so each willbe shorter than 5 of their 9 neighbors. The rest of them must choose a circleenclosing just one neighbor.

On the other hand, everyone except the shortest person will be allowedto play basketball, if the five tallest choose a large circle and all the restchoose a circle enclosing just one neighbor.

Problem 6. It is readily seen that P (N) > P (M) for N > M > 0, becausethe coefficients are positive. We also have P (N) > 1 for N > 0.

Next we claim that if k divides x− y, then k divides P (x)− P (y). Thisfollows by induction on the number of monomials in P . For one monomialit is obvious since

xn − yn = (x − y)(xn−1 + xn−2y + · · · + yn−1).

But if the statement is true for two polynomials, it’s also true for their sum.Now let’s set

A = P (1)P (2) . . . P (1996).

Since P (k) divides A, it also divides P (A + k) − P (k), for k = 1, . . . , 1996.Hence P (k) divides P (A + k). But P (k) > 1 and P (A + k) > P (k). Itfollows that P (A+k) is a composite number for k = 1, . . . , 1996, completingthe proof.

Level D

Problem 2. Introduce the symbols a =5√

2 +√

3 and b =5√

2 −√

3, sothe desired root is x = a + b. By direct substitution we see that a5 + b5 = 4and ab = 1, so

a +1a

= x and a5 +1a5 = 4.

114 SOLUTIONS

Our goal is to relate the expression on the right to powers of x. By expandingthe binomial powers we see that

x5 =(

a +1a

)5= a5 +

1a5 + 5

(

a3 +1a3

)

+ 10(

a +1a

)

,

x3 =(

a +1a

)3= a3 +

1a3 + 3

(

a +1a

)

.

Substituting the value of a3 +1a3 from the second equation into the first, we

obtain(

a +1a

)5=

(

a5 +1a5

)

+ 5

((

a +1a

)3− 3

(

a +1a

))

+ 10(

a +1a

)

.

Hencex5 = 4 + 5(x3 − 3x) + 10x

orx5 − 5x3 + 5x − 4 = 0.

Remark. It is possible to express an +1

ancan be expressed in terms of a +

1

afor

any n:

an +1

an= Pn

(

a +1

a

)

.

The polynomials Pn are related to the so-called Chebyshev polynomials Cn bythe relation Cn(x) = 1

2Pn(2x). The Chebyshev polynomials are defined by theformula

cos(nx) = Cn(cos x).

The link between these formulas follows from the relation cosx = 12 (eix +e−ix),

where i =√−1; see Zorich. See also Remark 2 to Problem 2 (page 91). CITE Zorich

Problem 3. Any set of evenly spaced parallel planes is equivalent to anyother under a similarity, and any cube is also similar to any other. So thequestion can be rephrased thus: Can the vertices of the unit cube [0, 1]3 lieon evenly spaced parallel planes?

Now, a family of parallel planes is made up of the constant sets of alinear function ax+by+cz; and the spacing between two such planes isproportional to the difference between the values of the function on each.We wish to find a, b, c so that the values of the function on the vertices ofthe unit cube are evenly spaced. These values are 0, a, b, c, a+b, b+c,a+c, and a+b+c. A moment’s thought will show that our condition will besatisfied if, for example, a = 1, b = 2 and c = 4. In other words, the planesdefined by x+2y+4z = k, for k = 0, . . . , 7, contain each one vertex of theunit cube.

Problem 4. This problem and many similar ones can be solved using mod-ular arithmetic, which is the consideration of remainders upon division byvarious integers. Choosing the best integer by which to divide in order todraw useful conclusions (the modulus) usually involves some trial and error,but where squares are involved, it’s often useful to work modulo 4. The re-mainder of a perfect square upon division by 4 is either 0 or 1 (see Remark


1 after the solution). So sums of two squares can only give 0 + 0, 0 + 1 or1 + 1 modulo 4; numbers that give a remainder of 3 modulo 4 cannot besums of two squares.

Similarly, the remainders of a square modulo 9 are 0, 1, 4, or 7 (seeRemark 1); considering the possible sums of these remainders we find thatsums of two squares can only give 0, 1, 2, 4, 5, 7, 8 modulo 9. (Note that4 + 7 ≡ 2 modulo 9, since 11 = 9 + 2.)

Here are two of many possible solutions based on these facts:

First solution. The sum n = (36k + 2)2 + 42, where k is an integer, givesthe same as 22 + 42 modulo 4 and also modulo 9, since 4 and 9 both divide36. Therefore:

• n =≡ 0 modulo 4, which means that n−1 ≡ 3 modulo 4. So n − 1cannot be a sum of two squares.

• n =≡ 0 modulo 9, which means that n+1 ≡ 3 modulo 9. So n + 1cannot be a sum of two squares.

Since k can take any value and n grows with k if k > 0, we have foundinfinitely many values of n as desired.

Second solution. The sum n = 9k + 1, where k > 0 is an integer, gives 2modulo 4 and gives 1 modulo 9. Therefore:

• n + 1 cannot be the sum of two squares since n+1 ≡ 3 modulo 4.• n−1 gives 1 modulo 4 and 0 modulo 9, which are both allowed. So this

path doesn’t look promising, but now we look at arithmetic modulo 3.If n − 1 = 9k can be written in the form 9k = a2 + b2, there are twopossibilities: either a and b are both non-multiples of 3, so a2 and b2

each have remainder 1 modulo 3, which disagrees with their sum beinga multiple of 3; or a and b are both multiples of 3, and we can divide by9 to obtain a similar equation, 9k−1 = (a/3)2+(b/3)2. In the latter casewe just apply the argument again to the new equation, and concludethe proof by reverse induction (see Fact 24). Note that if k − 1 = 0 wecannot apply the argument, but then there is obviously no solution.

Remarks. 1. It is easy to find what remainders are allowed for a square, for anymodulus q. For suppose that n gives r modulo q; that is, we have an equationn = kp + r, with k an integer. Then n2 = k2q2 + 2kqr + r2; that is, n2 has thesame remainder as r2, modulo q. So all we need to do is look at all values or rand take their squares modulo q.

• q = 3: We have 02 = 0, 12 = 1, 22 = 4 ≡ 1 modulo 3. So the possibilitiesare 0 and 1. The square of any non-multiple of 3 leaves a remainder of 1modulo 3.

• q = 4: We have 02 = 0, 12 = 1, 22 = 4 ≡ 0, 32 = 9 ≡ 1 modulo 4. So thepossibilities are again 0 and 1.

• q = 9: We can save time by noting that the remainders 5 to 8 can berelated to their “complements”; that is, 9− t is the same (modulo 9) as −t,so (9−t)2 is also the same (modulo 9) as (−t)2 = t2. Thus we only need

116 SOLUTIONS

to consider the squares of 0, 1, 2, 3, 4. They give 02 = 0, 12 = 1, 22 = 4,32 ≡ 0 and 42 ≡ 7 modulo 9.

• Also useful: the square of an odd number always yields a remainder of 1when divided by 8. (Check it!)

2. A famous criterion for a positive integer n to be representable as the sum of twosquares requires that any prime factor of n of the form 4k + 3 must appear inthe prime factorization of n with an even power [§ 4–5]NumbRepresSumSquar. CITE [

See also Remark 2 on page 41.

Problem 5. Denote by O1 and O2 the centers of the circles and by r1 andr2 their radii, respectively.

Consider the point D at which a common interior tangent to the circlesmeets the segment O1O2. Then (see Fact 17)

DO1

DO2=

r1

r2.

Let AC be a diagonal of the quadrilateral in question and let S be thepoint at which it meets O1O2 (see figure).

A

B

C

D=S

X

O1 O2

By the law of sines, we have

SO1

sin \O1AS=

r1

sin \O1SA,

SO2

sin \O2CS=

r2

sin \O2SC. (1)

Consider the circle with center X passing through A and C. It is not hardto see that the lines AO1 and CO2 are tangent to this circle. Expressingthe angles between these tangents and the chord AC in terms of interceptedarcs (see Fact 15), we obtain the equation

\O1AS + \O2CS = 12(

⌢AC +

⌢CA) = π

(where the second arc contains the point B in the figure and the firstdoesn’t); hence sin \O1AS = sin \O2CS. The angles O1SA and O2SC,being vertical angles, are equal. Therefore, by (1), we have

SO1

SO2=

r1

r2.

Since there is only one point dividing a line segment in a given ratio, thepoints D and S coincide. A similar reasoning is true for the second diagonal.


This means that the diagonals of the quadrilateral and the common internaltangents cross the line of centers at the same point, and we’re done.

Remark. This problem implies the following remarkable statement: If a quadrilat-eral ABCD is circumscribed around a circle, the intersection point of the linesjoining the opposite points of contact coincides with the intersection point ofits diagonals. Indeed, take as X the incenter of the quadrilateral. Let ω1 be thecircle centered at A and passing through the points at which the incircle of thequadrilateral touches AB and AD. Denote by ω2 a similar circle with centerC. Applying the statement of the problem to the configuration thus obtained,we see that the lines joining the opposite points of contact of the incircle meeton AC and, for similar reasons, on BD at the same time.

Problem 6. First solution. We will choose certain rows of the modifiedarray one by one, and keep track of the sum S = S(m) of the m rows chosenso far. The first row we take is the modified version of the row (1, 1, . . . , 1)of the original array. The string S(1) simply coincides with this row and,obviously, consists of zeros and ones. If it has only zeros, this row aloneyields the desired set. Otherwise, we find in the original array the rowobtained from S(1) by replacing 1 by −1 and 0 by 1. The correspondingrwo in the modified array row is chosen to be our second string. Then S(2),the sum of S(1) and the chosen row, consists of 0s and 1s again.

Suppose that we have already chosen k rows. If the sum S(k) coincideswith one of the previous sums S(m), where m < k, then the sum of thek−m rows indexed from m+1 to k is the zero string and we’re done. If thesum S(k) does not coincide with any of the previous sums, then we take upnext the row in the modified array corresponding to the row in the originalarray obtained from S(k) by replacing 1 by −1 and 0 by 1. This row couldnot have been chosen before, because different sums S(m) specify differentrows to be chosen in the array, and the sum S(k) has never occurred before.

If at a certain step we obtain a sum that has occurred before, then, aswe have already seen, the problem will be solved; if not, we will eventuallyhave exhausted all the rows of the array. That is, we’ll have 2n differentsums S(k). Since the number of different strings of 0s and 1s of length nis also equal to 2n, each of these strings will coincide with one of the sumsS(k). In particular, for a certain k, the string S(k) will consist of zeros, andthe desired set is the set of the first k chosen rows.

Second solution. Denote by ai the rows of the initial array and by bi the rowsof the modified array, where i = 1, 2, . . . , 2n. We will construct yet a thirdarray, with the rows ci = ai − 2bi. In other words, any two correspondingelements of ci and ai either coincide (if the element of ai was replaced byzero) or are opposite (in all the other places). In particular, the new arrayconsists of ±1. Therefore, for any i there exists a j(i) such that ci = aj(i).Now let us consider the sequence ik specified by the recurrence relationik+1 = j(ik). (The first term of the sequence is chosen at will.) Since thissequence can take only finitely many values, it must contain equal terms.

118 SOLUTIONS

Suppose that ik = il for certain k and l, k < l, and all the terms of thesequence with numbers less than l are different. Then

bik +bik+1+ · · ·+bil−1

= 12(aik−cik)+ 1

2(aik+1−cik+1

)+ · · ·+ 12(ail−1

−cil−1)

= 12(aik−aik+1

+aik+1−aik+2

+ · · ·+ail−1−ail)

= 12(aik−ail) = 0,

which completes the proof.

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Transcript of Problems - MSRImentioned in allusions to times long forgotten. Problem 5. Prove or disprove: There...