NC ARML Practice Problems

NC ARML Practice Problems

The following topics are ones that seem to be favorites of the writers of many contests,

and are topics that are often overlooked in the normal curriculum. For each topic you

will find some brief introductory material followed by problems, some of which are from

previous NC State Math Contests or from ARML Competitions. I have tried to note the

source whenever possible. If you find a problem that looks like it came from a

previously published contest and I have not cited the source, please email me so that I can

give proper credit. Also, if you run across a really good problem that seems to fit into

one of the following categories, please forward me the problem, solution, and source and

I will add it to the topic.

Files that end in "_P" include practice problems. For topics for which we have solutions

to these problems, there will be a corresponding file ending in "_S".

____________________________________________________________________

AnalyticalGeometry_P AnalyticalGeometry_S

Bases_P Bases_S

Complex_P Complex_S

Distance_P Distance_S

Finite_P Finite_S

NumberTheory_P NumberTheory_S

Partial_P Partial_S

Perfect_P Perfect_S

Polynomials_P Polynomials_S

PowerofPoint_P PowerofPoint_S

Traingles_S Triangles_P

1 Written & Compiled by John Goebel, NCSSM Problem Solving Course, 2006

Analytical Geometry: Generally speaking, this topic refers to circles, parabolas, ellipses and hyperbolas. It is important to know the formulas of these and what the various parts represent. Circles: The definition of a circle is the set of all points in a plane that are equidistant from a given point, the center. The equal distance is, of course, the radius. Given the center points ( ),h k and radius r, the equation of the circle is

( ) ( )2 2 2x h y k r− + − = . If the equation is in standard form, 2 2 0x y Bx Cy D+ + + + = , you can complete the square to find the center and radius. Ellipses: An ellipse has to foci in addition to the center. The definition is the set of all points in a plane so that the sum of the distances from the two foci is constant. While the points can be anywhere in the plane, we almost always work with foci on the same vertical or horizontal line.

The center is (2,1) and the foci are (-1,1) and (5,1). The vertices (-3,1) and (7,1) are the endpoints of the major axis. The points (2,5) and (2,-3) are endpoints of the minor axis.

The equation of this ellipse is 2 2

2 2

( ) ( ) 1x h y ka b− −

+ = , where 2a is the length of the major

axis, 2b the length of the minor axis, (h,k) the center, and the foci are c units from the center along the major axis, and 2 2 2a b c− = . Of course, if b a> the major axis is vertical and 2 2 2b a c− = . The eccentricity of an ellipse is the ratio c

a (horizontal major

axis) or cb (vertical axis). The length of the major axis is the same as the sum of the

distances from any point to the two foci. In standard form, 2 2 0Ax By Cx Dy E+ + + + = , A and B will have the same sign, but are unequal. Complete the square to get the equation in the desired form. Hyperbolas: The definition of a hyperbola is similar to that of the ellipse, except that the absolute value of the distance from any point to the two foci is constant. We still have a center, foci, vertices, and major axis.


The center is (2,1) and the foci are (-3,1) and (7,1). The vertices (5,1) and (-1,1) are the

endpoints of the transverse axis. The equation of the hyperbola is 2 2

2 2

( ) ( ) 1x h y ka b− −

− = ,

if the transverse axis is horizontal. The center is still (h,k). The foci are c units from the center, along the line containing the transverse axis, where 2 2 2c a b= + . In addition,

hyperbolas have asymptotes. Their equations are ( )by k x ha

− = ± − . If the transverse

axis is vertical, the equation is 2 2

2 2

( ) ( ) 1y k x hb a− −

− = . The eccentricity is ce a= or cb ,

depending on the transverse axis. Parabolas: A parabola is the set of all points in a plane that are equidistant from a given point (focus) and a given line (directrix). The simple form requires that the directrix be either horizontal or vertical. For a horizontal directrix, the equation is

( )214

y k x hp

− = − , where (h,k) is the vertex and p is the distance from the vertex to the

focus. The line x h= is the axis of symmetry.


The standard form for this equation is 2y ax bx c= + + . In this form, the vertex is 2(4 ),2 4

ac bba a

⎛ ⎞−−⎜ ⎟⎝ ⎠

.

If the directrix is vertical, the x- and y-terms are interchanged. The equations becomes

( )214

x h y kp

− = − .

Problems: 1. The vertex of the graph of the parabola 23y x bx c= + + is at (-2,-7); determine

the value of b-c; NC ALGII 2002 33 2. The distance between the centers of the circles 2 28 28 112 0x x y y− + − + = and

6 26 20 12 0x x y y+ + + − = is: FURMAN 2003 JR11 3. Through which quadrants does the circle 2 24 6 1 0x x y y+ + − + = pass?

FURMAN 2003 JR13 4. If a parabola has vertex (0,0) and focus (1,0), the equation is: FURMAN 2003

JR 15 5. If Q is the point on the circle 2 210 6 29 0x x y y− + + + = which is furthest from

the point ( 1, 6)P − − , then the distance from P to Q is: FURMAN 2003 SR 2. 6. The parabola 2y ax bx c= + + intersects the y-axis in the point (0,8) and intersects

the x-axis in the single point (2,0). How many pairs of integers (m,n) with 2000 2000m− ≤ ≤ lie on the parabola? NC SMC 2000 INT 4

7. Let f(x) be a quadratic polynomial such that f(3) = 15 and f(-3) = -9. Find the coefficient of x in f(x). NC SMC 2002 MC 1

8. The center © and the vertices (V) of the ellipse 2 24 9 16 54 61 0x y x y+ − − + = are: NC ALGII 2000 9

9. The quadratic equation 2y ax bx c= + + is known to pass through the points (0,5), (2,11) and (-2,15). Find the sum of a and b. NC ALGII 2000 17

10. Write an equation for the hyperbola with horizontal transverse axis and asymptotes 4

3y x= ± . NC ALGII 2000 25

11. A parabola of the form 2y x bx c= + + contains the points (2,3) and (4,3), Find the value of c. AMC10 2006 .

Different Number Bases We are so familiar with base-ten numbers that we often forget want a number like 123.456 really means. Of course it means that we have

2 1 0 1 2 31 10 2 10 3 10 4 10 5 10 6 10− − −⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ .

In base 10 we must have ten symbols to represent the numbers from 0 to one less than the base (or 9). In a base smaller than 10, say base 5, we only need the digits 0 through 4, and a number like 432.12 means 2 1 0 1 24 5 3 5 2 5 1 5 2 5− −⋅ + ⋅ + ⋅ + ⋅ + ⋅ which would be

104(25) 3(5) 2 1(0.2) 2(0.04) 117.28+ + + + = as a base ten number. For bases greater than ten we need additional symbols to stand for all the numbers from 0 to the base minus 1. So in base 16 (hexadecimal) we have the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, A, B, C, D, E, F, and G. The number 3 2

16 103 4 14 16 3 16 9 16 4 58260F A = ⋅ + ⋅ + ⋅ + = in base 10. Of course base 2 (binary) has only the symbols 0 and 1. The base-two number 1101011 would mean (in base ten)

6 5 4 3 2101 2 1 2 0 2 1 2 0 2 1 2 1 64 32 8 2 1 107⋅ + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + = + + + + =

Converting from base ten to another base is fairly easy using the following algorithm. Suppose we want to convert the number 83 to base 2.

Now list the remainders in reverse order giving 1010011. A check reveals that

6 421010011 2 2 2 1 64 16 2 1 83= + + + = + + + = .

To go from base 2 to base 16, group the digits, from right to left in groups of 4. For this number we have [0101] [0011]. Convert each group of 4 now to base 16, giving 1 0

16 1053 5 16 3 16 83= ⋅ + ⋅ =

Problems:

1. The expression 32 in base b represents the same number as 21 in base c, while the expression 21 in base b represents the same number as 13 in base c. What is b?

2. In what base does 62 14 808× = ? 3. A certain 3-digit number exceeds the sum of its digits by 126. The three-digit number

obtained by reversing the order of the digits exceeds the sum of the digits by 225. What is the sum of the digits?

4. The base b numbers 13, 42, and 101 are in arithmetic progression. Find b. 5. A certain fraction r is represented in base b by 0.111111… while in base 2b it takes

the simpler form 0.2b. What is r? 6. How many positive integers less than 1000 have a 1 in their base 16 expansion?

FURMAN SR 2000 #26

83 2(41) 141 2(20) 120 2(10) 010 2(5) 05 2(2) 12 2(1) 01 0(2) 1

= += += += += += += +

7. A 6-digit number has its first digit a 9. If you move it to the last digit instead, you get a number which is only one fourth the size of the original number. What is the sum of the digits of the original number? FURMAN SR 2000 #32


Complex Numbers Everyone knows that a complex number is a number of the form x yi+ , where x and y are real numbers and 1i = − .

Let’s go on from there. It turns out that the polar form of complex numbers is more useful. To convert from the standard or rectangular form x yi+ to the polar form, we first find the norm or

modulus of the number, which is 2 2r x y= + . The argument or angle is found using the

inverse tangent,

1

1

tan 0

tan 0

3, 0, 0 , 0, 02 2

y xx

y xx

x y x y

θ π

π π

−

−

⎧ ⎛ ⎞ >⎜ ⎟⎪ ⎝ ⎠⎪⎪ ⎛ ⎞= + <⎨ ⎜ ⎟

⎝ ⎠⎪⎪

= > = <⎪⎩

.

Once found, the polar form is ( )cos sinr iθ θ+ , which is sometimes written r cisθ⋅ . In this form arithmetic of complex numbers becomes less complex. DeMoivres Theorem really makes things (multiplication, division, powers) simple:

Multiplication:

( ) ( )( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )( )( ) ( )( )

1 1 1 2 2 2

1 2 1 2 1 2 2 1 1 2

1 2 1 2 1 2 1 2 2 1

1 2 1 2 1 2

cos sin cos sin

cos cos sin cos sin cos sin sin

cos cos sin sin sin cos sin cos

cos sin

r i r i

r r i i

r r i

r r i

θ θ θ θ

θ θ θ θ θ θ θ θ


θ θ θ θ

+ ⋅ + =

+ + − =

− + + =

+ + +

Powers: ( ) ( ) ( ) ( )( )cos sin cos sin cos sin

n nn nr i r i r n i nθ θ θ θ θ θ+ = ⋅ + = +⎡ ⎤⎣ ⎦ . The proof of this, for positive integers at least, can be shown with induction and the multiplication theorem. Roots: The above theorem, when used with fractional exponents, yields the roots of a complex

number. ( ) ( )( ) 1 2 2cos sin cos sin , 0,1, , 1nnk kr i r i k n

n nθ π θ πθ θ ⎛ + + ⎞⎛ ⎞ ⎛ ⎞+ = + = −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠


Example:

( ) ( )( )

( ) ( )( )

13 330 2 0 28 8 cos 0 sin 0 8 cos sin , 0,1,2

3 3

2 2 4 42 cos 0 sin 0 ,2 cos sin , cos sin3 3 3 3

1 3 1 32, 2 1 3, 2 1 32 2 2 2

k ki i k

i i i

i i i i

π π

π π π π

⎛ + + ⎞⎛ ⎞ ⎛ ⎞= ⋅ + = + =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞− + = − + − − = − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Notice that the three complex cube roots of 8 are all 2 units from the pole, so they are one a

circle with radius 2, and they are evenly spaced around that circle at 0, 2 4, and 3 3π π . This will

always be the case. The fifth roots of 32, for example, will all have magnitude 2 and arguments

0, 2 4 6 8, , , and 5 5 5 5π π π π .

Division: In multiplication the norms are multiplied and the arguments added. In division the norms are divided and the arguments divided.

( )1 1 11 2

2 2 2

rcis r cisr cis r

θ θ θθ

= −

Problems: 1. Show that the power rule works for n = 2, and n = 3.

2. Show that ( )( ) ( )1 1 1

1 22 2 2

rcis r cisr cis r

θθ θ

θ= − .

3. Raise ( )81 3+ .

4. Find 4 i . 5. A certain complex number satisfies 2 1ω ω= − . What is 99ω ? FURMAN 2001 SR #13. 6. If ( )3 74x iy ki+ = − + , find the absolute value of k, given that 1 and 1x i= = − . NC

SMC 2002 INT1. 7. Find the sum of the cube of the roots of the equation 10 9 8 2 1 0x x x x x+ + + + + + =

DUKE 2003.

8. Find the sum of the squares of the roots of the equation 6

2

1 01

xx−

=−


Distance Formulas The Pythagorean Distance formula for the shortest distance between two points ( )1 1,x y

and ( )2 2,x y is ( ) ( )2 22 2 2 1d x x y y= − − − .

The comparable formula for distance in space between the points with coordinates

( )1 1 1, ,x y z and ( )2 2 2, ,x y z is ( ) ( ) ( )2 2 22 2 2 1 1 2d x x y y z z= − − − + − .

The distance from a point ( )1 1,x y to a line 0ax by c+ + = is one of those problems that seems easy, but is very time consuming, unless you know the following formula.

1 12 2

ax by cda b+ +

=+

The following proof illustrates why it pays to know the formula.

Drop perpendiculars to the line from the given point ( )1 1,x y . Now 2 1

a cy xb b

= − − and

3a cx xb b

= − − . The triangle in the picture is a right triangle, so the area can be found two

ways. The product of the legs will equal the altitude (d) times the hypotenuse.

2 2

1 1 1 1 1 1 1 1a c b c a c b cy x x y d y x x yb b a a b b a a

⎛ ⎞ ⎛ ⎞+ + × + + = × + + + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

, so

1 1 1 11 1

2 2 2 2

1 1 1 1

a c b cy x x y ax by cb b a ada ba c b cy x x y

b b a a

+ + × + ++ +

= =+⎛ ⎞ ⎛ ⎞+ + + + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠


Example: Find the distance between the point (2,-3) and the line 3 4 12 0x y− − = . The

distance is 2 2

63(2) 4( 3) 12 65 53 4

d − − −= = =

+.

The distance from a point in space to a plane can be found using an extension of this formula. The distance from the points ( )1 1 1, ,x y z to the plane 0ax by cz d+ + + = is

1 1 1

2 2 2

ax by cz d

a b c

+ + +

+ +

Problems. 1. What is the closest the line 4 5 20x y+ = comes to the origin? 2. What is the distance between the points ( )(2, (2)) and (2), 2f f if

( ) 4 1f x x= + . 3. How far from the origin are the points of intersection of the conics 2 27 47x y+ =

and 2 22 4 28x y− = ? 4. What is the shortest distance between the circle 2 2 25x y+ = and the line

3 4 48x y+ = ? 5. Find the distance between the parallel lines 2 3 12x y− = and 2 3 36x y− = . 6. A triangle has vertices (2,3), (6,-2), and (-1,-4). Find the area of the triangle. 7. Find the shortest distance between the parallel planes 3 4 5 12x y x+ − = and

3 4 5 20x y z+ − = .

8. What is the closest that the line 4 17 5

y x= + comes to a lattice point? DUKE

2003 Team10.

9. What is the closest the plane 1 3 43 5 15

z x y= + + comes to a lattice point?

10. Under what conditions would the line 0Ax By C+ + = be guaranteed to hit a lattice point?

11. The circles 2 2 10 6 18 0x y x y+ + − + = and 2 2 14 4 44 0x y x y+ − + + = have two

internal tangents. On each the distance between the points of tangency are equal. Find this distance.

Written & Compiled by John Goebel, NCSSM Problem Solving Course, 2006

1

Finite Differences and Recurrence Relations Finding a closed formula for a set of ordered pairs is a useful skill. There are two methods that work nicely. The first is finite differences and the second recurrence relations. In order to fit a function to a set of points, it is helpful to know the type of function you are dealing with. This is where finite difference can help. If the x-values are sequential and only one unit apart, the finite differences are simply the differences in the dependent values of a function. Let’s look at the following set of linear data first. It should be fairly obvious that the first

finite differences are constant, and that this is the slope of the linear function that would include these points. These finite differences are simply the change in y while the change in x each time is just 1. We know the function then is

( ) 4f x x b= + and it is each to substitute in any value to determine that b is -1. Now let’s look at some non-linear data.

The first finite differences are not constant, so this is clearly not a linear function. The second differences are constant, and this tells us that the function is quadratic. To see why, let’s look at a generic quadratic function of the form 2y ax bx c= + + .

Three points with successive x-coordinates might be

( ) ( ) ( )1, ( 1) , , ( ) , 1, ( 1)n f n n f n n f n− − + + . If we put these in a table, we get

1 34

2 74

3 114

4 15

x y

1 50

2 5 22

3 7 24

4 11 26

5 17


2

2 2

2 2

2 2

1 ( 1) ( 1) ( 2 ) ( )2 ( )

22 ( )

1 ( 1) ( 1) ( 2 ) ( )

n a n b n c an b a n a b can b a

n an bn c an bn c aan a b

n a n b n c an b a n a b c

− − + − + = + − + − ++ −

+ + = + ++ +

+ + + + + = + + + + + The first differences were not equal (as long as a is not zero), but the second differences are not only constant, but equal to 2a. Returning to our specific example, we now know that the function is quadratic and a = 1. Simple substitution can be used to determine the values of b and c as -3 and 7. If the function is cubic, the third difference will be constant and will be 6a. Look at the following example: X Y 1 2 3

1 -3 4

2 1 8 12 6

3 13 14 26 6

4 39 20 46 6

5 85 26 72

6 157

The 3rd differences are constant and equal to 6. This is actually 3!a or 6a, so a = 1. The second differences are 6ax + 2b, so when x = 2, (6)(1)(2) + 2b = 8, making b = -2. Simple substitution with two points can be used to find the last two coefficients. This function is 3 2( ) 2 3 5f x x x x= − + − ,

Exponential functions are a little more difficult. The general form for an exponential function is ( ) xf x a b c= ⋅ + . Let’s look at the following example.

x y 1 2 1 10 12 = 12(1) 2 22 24 = 24(1) 36 = 12(3) 3 58 72 = 24(3) 108 = 12(32) 4 166 216 = 24(32) 324 = 12(33) 5 490 648 = 24(33) 972 = 12(34) 6 1462

The first and second differences are not constant, and never will be. Notice, however, that there is a common factor of 12 in the first difference and a common factor of 24 in the second differences. From this we can conclude that the base of our function is 3. The function now is of the form 3xy a c= ⋅ + . Again, simple substitution can be used to find that a is 2 and c is 4, so the exponential function that fits these points is ( ) 2 3 4xf x = ⋅ + .


3

Recurrence Relations Often we are given a recursive formula to generate values of a sequence and we would like to find the closed form. You are familiar with arithmetic and geometric sequences and know how to find both the recursive and closed formulas for the terms of such sequences. If, however, the recurrence relation is a little more complicated, it is not as easy to find the closed formula. Take for example the simple case where you both multiply and add a number at each step in the sequence. So 0t a= and at each step we will multiply by r and then add d, making the recursive equation 1n nt t r d+ = ⋅ + . This is called a first order recurrence relation, and it is easy to solve algebraically. List the first several terms and you will see the closed formula: Term Value 0 A 1 Ar d+ 2 ( ) 2Ar d r d Ar dr d+ + = + + 3 ( )2 3 2Ar dr d r d Ar dr dr d+ + + = + + + 4 ( )

33 2 4

0

j

j

Ar dr dr d r d Ar dr=

+ + + + = +∑

So if we apply the closed formula for the ending geometric series, we get the final closed formula for the entire sequence:

11

nn

nrt A r dr−

= ⋅ +−

Recurrence equations of the form 1 2n n nt A t B t− −= ⋅ + ⋅ can be rewritten in the form

1 2 0n n nt A t B t− −− ⋅ − ⋅ = ,

and such equations are called second order homogeneous recurrence equations or relations. Clearly you must know two initial values and the values of A and B in order to recursively generate the terms of the sequence. Finding the closed form for the terms is another matter entirely. First, we will assume that there are solutions of the form n

nt c r= ⋅ , so that our equation becomes

1 2 0n n ncr Acr Bcr− −− − =


4

which is equivalent to

( )2 2 0ncr r Ar B− − − =

so we can solve the second factor of this last equation to find values for r. The two solutions for this equation are

2 42

A A B± + .

The closed form then for 1 1 2 2

n nnt c r c r= ⋅ + ⋅ , where

2 2

1 24 4,

2 2A A B A A Br r+ + − +

= = .

One very familiar recurrence relation is the Fibonacci sequence. Here 0 10, 1,t t= = and

1 2n n nt t t− −= + . The first few terms are 0, 1, 1, 2, 3, 5, 8, 13, 21, … If we use the solution

outlined above, you get 1 2 0n n ncr cr cr− −− − = , so that ( )2 2 1 0ncr r r− − − = . The solutions

to the second factor are 1 21 1 4 1 5 1 1 4 1 5,

2 2 2 2r r+ + + − + −= = = = . Now we have

1 1 2 2 1 21 5 1 5

2 2

n n

n nnt c r c r c c

⎛ ⎞ ⎛ ⎞+ −= ⋅ + ⋅ = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠. Now take the first two given values to

find the values for 1 2,c c . We get the system of equations:

( )

0 0

1 2 1 2 2 1

1 1

1 1 1 1 2

1 5 1 502 2

1 5 1 5 1 11 5 ,2 2 5 5

c c c c c c

c c c c c

⎛ ⎞ ⎛ ⎞+ −= + = + ⇒ = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞+ −= − = ⇒ = = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

.

This means then that 1 1 5 1 52 25

n n

nt⎛ ⎞⎛ ⎞ ⎛ ⎞+ −⎜ ⎟= −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

.

Austin’s Problem. The first term of a sequence is 0 and the second term 1. Each successive term is defined as follows: 1 24 , 2n n nt t t n− −= ⋅ − ≥ . Prove that for and

( )21 11, 1n n nn t t t− +≥ ⋅ = − .

Solution. First, notice that the first few terms of this sequence are 0, 1, 4, 15, 56, 208, … and it looks like the result is certainly true as 20 4 1 1 0⋅ = − = , 21 15 4 1 15⋅ = − = ,


5

24 56 15 1 224⋅ = − = , …. Certainly one way to solve this is to find a closed form for the terms of this sequence. As in the Fibonacci sequence, this is a homogeneous recurrence relation with 1 24 0n n ncr cr cr− −− + = , so ( )2 2 4 1 0ncr r r− − + = . The solutions to the second factor are

1 24 16 4 4 2 3 4 16 4 4 2 32 3 , 2 3

2 2 2 2r r+ − + − − −= = = + = = = = − . Now we have

( ) ( )1 1 2 2 1 22 3 2 3n n

n nnt c r c r c c= ⋅ + ⋅ = + + − . Now take the first two given values to

find the values for 1 2,c c . We get the system of equations:

( ) ( )( ) ( ) ( )

0 0

1 2 1 2 2 1

1 1

1 1 1 1 2

0 2 3 2 3

1 11 2 3 2 3 2 3 ,2 3 2 3

c c c c c c

c c c c c

= + + − = + ⇒ = −

= + − − = ⇒ = = −.

This means then that ( ) ( )( )1 2 3 2 32 3

n n

nt = + − − . Now to establish the desired

result, that ( )21 11, 1n n nn t t t− +≥ ⋅ = − .

First ( ) ( )( )1 1

11 2 3 2 3

2 3

n n

nt− −

− = + − − and ( ) ( )( )1 1

11 2 3 2 3

2 3

n n

nt+ +

+ = + − − . It

really helps here to realize that ( ) 112 3 2 32 3

−− = = +

+. So the product

( ) ( )( ) ( ) ( )( )( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( ) ( )( )

1 1 1 1

1 1

1 1 1 1

2 2 2 2

2 2 2 2

2 2

2 2

1 12 3 2 3 2 3 2 32 3 2 31 2 3 2 3 2 3 2 3

121 2 3 2 3 2 3 2 3

121 2 3 2 3 2 3 2 3

121 2 3 7 4 3 7 4 3 2 3

121 2 3 14 2 3

12

n n n n

n n

n n n n

n n

n n

n n

n n

t t− − + +

− +

− − + − +

− −

−

−

−

⋅ = + − − ⋅ + − −

⎛ ⎞= + − + + − +⎜ ⎟⎝ ⎠

= + − + − + + +

= + − − − + + +

= + − − − + + +

= + − + +

Now look at

Written and Compiled by John Goebel, NCSSM Problem Solving Course, 2006

1

Frequently Asked Number Theory Properties Prime Factorization: Every positive integer has a unique prime factorization. Suppose N factors as

follows: 1 21 2

nqq qnN p p p= ⋅ where each p is a prime number and each q is the

power of that prime. Since each prime can be used anywhere from 0 to nq times, the total number of different factors of N is ( )( ) ( )1 21 1 1nq q q+ + + . For

example, the total number of factors or 360 is 24 since 3 2 1360 2 3 5= ⋅ ⋅ , making ( )( )(3 1) 2 1 1 1 4 3 2 24+ + + = ⋅ ⋅ = .

Number of Subsets:

The number of subsets of a given set is equal to 2N , where N is the number of elements in the set. This includes the entire set and the null or empty set. A set with only 5 elements therefore has 52 32= distinct subsets. This is easy to understand since each element is either used or not. You might Related to #2 is the number of subsets containing a certain number of elements. This is just a combination problem. Note that the combination of N

object taken R at a time is ( )

!! !N R

N NCR R N R⎛ ⎞

= =⎜ ⎟ −⎝ ⎠. So the different number of

different sets of 5 elements taken from a set of 8 elements is

8 5

8 8! 8 7 6 565 5! 3! 3 2

C ⎛ ⎞ ⋅ ⋅= = = =⎜ ⎟ ⋅ ⋅⎝ ⎠

. Thus the number of subsets of any size, 0

through N (of a set of N elements) is 0 1 2 2 1N N N N N N N N NC C C C C C− −+ + + + + + but this is just 2N .

Factorial:

Factorial notation is simply ( )( )! 1 2 3 2 1n n n n= − − ⋅ ⋅ . 0! is defined to be 1. A favorite question on contests is to give the number of zeros at the end of N!. An example will illustrate the method used to compute this. Let’s work with 27!.

27! 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

Each factor of 10 will add a zero to the end of the number, and of course, factors of 10 result from factors of 5 and factors of 2. There are fewer factors of 5, so those are the ones we need to count. Each factor of 5 results in 1 zero, but each factor of 25 or 52 results in two zeros. Likewise any factor of 125 = 53 would result in 3 zeros. So, in 27 there are 5 factors of 5, one of which is also a factor of 25, so we have 5 zeros for the factors of 5 and one additional zero for the factor of 25. So there are 6 zeros at the end of 27!. If we wanted to know how many factors of 6 there were in 27!, we would look for powers of 3.


2

Binomial Coefficients: The combination formula in # 3 gives us the coefficients in Pascal’s Triangle.

Here is a part.

5 0 5 1 5 2 5 3 5 4 5 5

11 1

1 2 11 3 3 1

1 4 6 4 1

1 5 10 10 5 1C C C C C C

The sum of each row is

0

1

2

3

4

5

1 22 24 28 216 2

32 2

=

=

=

=

=

=

Floor and Ceiling:

In number theory where you work primarily with integers, 27 / 5 = 5 with no remainder. One symbol that we sometimes use to denote that we are going to drop the remainder is the floor or greatest integer symbol or function. This looks like this: greatest integer less than or equal to n n=⎢ ⎥⎣ ⎦ . So

272.7 2, 5, 3.6 45

⎢ ⎥= = − = −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦. Using this notation, we can find the

number of zeros at the end of 2003! As 2003 2003 2003 2003

5 25 125 625400 80 16 3 499

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ + + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦+ + + =

Modular Arithmetic

The remainders when dividing one integer by another always range from zero to one less than the divisor. A convenient way of denoting these remainders is with the mod notation. We write 12 2 mod 5≡ because the remainder when 12 is divided by 5 is 2. The integers can be divided into congruence classes mod m, sets of numbers which have the same remainder upon division by m. The congruence classes everyone is familiar with is the odd/even division of the integers. The odd numbers are congruent to 1 mod 2, while the evens are congruent to 0 mod 2.


3

Problems using these facts.

1. What is the highest power of 6 to divide 40!?

2. What is the smallest factorial to end in exactly 200 zeros? The largest?

3. What is the smallest positive integer that has exactly 30 factors?

4. Find the largest integer value of n so that 3n divides (25! + 26!).

5. How many different positive integer factors of 425 are there?

6. If we write 30! As 2n k where k is odd, find n.

7. How many distinct factors of 2004 are there? Be sure to do this with the current year as well.

8. If a factor of 4200 is chosen at random from all of its distinct factors, what is

the probability that it will be odd?

9. The number of divisors of 360 is?

10. Find the highest power of 12 that divides 100!

1 Written and Compiled by John Goebel, NCSSM Problem Solving Course, 2006

Partial Fractions Since elementary school you have known how to add fractions and since Algebra you have been able to add fractions with variables. The process of taking a single fraction and breaking it up into two or more terms is less obvious and often is not learned by students until they need this skill in a calculus class to do integration problems. It is, however, a useful skill.

Suppose we have the fraction 2( 1)x x −

. This is obviously equal to 2 1 1 2, or 1 x 1x x x

⋅ ⋅− −

, or any

other product where the product of the numerators is 2. This decomposition is obviously not unique. On the other hand, if we try to express the original fraction as a sum (or difference) of two fractions, this decomposition is unique.

Suppose 2( 1) 1

A Bx x x x

= +− −

, where A and B are real numbers. Then

( 1) ( )1 ( 1) ( 1) ( 1)

A B A x Bx A B x Ax x x x x x x x

− + −+ = + =

− − − −. Now for this last expression to equal the

original fraction, the numerators must be equal, so 2 ( )A B x A= + − , for all values of x. This means that the coefficient of x, which is A + B, must equal 0 and –A must = 2. Therefore

2, 2A B= − = , so 2 2 2( 1) 1x x x x

−= +

− −. In calculus, we use this decomposition to evaluate the

integral 2 2 2 12 ln( 1) 2 ln( ) 2 ln( 1) 1

xdx dx dx x x C Cx x x x x

−⎛ ⎞= − = ⋅ − − ⋅ + = ⋅ +⎜ ⎟− − ⎝ ⎠∫ ∫ ∫ .

In a problem like the following, this method can be used to greatly simplify the

solution.

Problem: Find the sum of the first 200 terms of the series 1 1 1 11 2 2 3 3 4 1n n

+ + + + +⋅ ⋅ ⋅ ⋅ +

.

Solution: Rewrite each term, using partial fractions as follows:

1 ( 1) ( ) 1 1, 1( 1) 1

A B A n B n A Bn n n n

= + ⇒ + + = ⇒ = = −+ +

, so the series can be rewritten as

1 1 1 1 1 1 1 1 1 11 2 2 3 3 4 1 1n n n n⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

. Now notice that the terms, upon re-

association, is 1 1 1 1 1 1 1 1 111 2 2 3 3 1 1n n n n

− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + + + + − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠. This makes the answer

to the problem simply 1 2001201 201

− = .

As long as the factors in the denominator are linear, the numerators will all be constants. In general, the numerator will always be in degree one less than the degree of the numerator.


Suppose we want to write the fraction 2

2 3( 1)

nn n

++

as the sum of two other fractions. To do this

write this fraction as 2 1A Bn Cn n

++

+ and then add these two to get

( )( )

( )22

2 2

( 1) ( )1 1

A B n Cn AA n Bn C nn n n n

+ + ++ + +=

+ +. The numerator of this fraction must equal 2 3n + ,

so 0, 2, 3A B C A+ = = = , so B must be -3. So 2 2

2 3 3 3 2( 1) 1

n nn n n n

+ − += +

+ +.

Problems:

1. Find the sum of the first 100 terms of the series 1 1 1 1 11 2 2 3 3 4 4 5 ( 1)n n

+ + + +⋅ ⋅ ⋅ ⋅ ⋅ +

.

2. Find the sum of the first 50 terms of the series

1 1 1 1 11 3 3 5 5 7 7 9 (2 1) (2 1)n n

+ + + +⋅ ⋅ ⋅ ⋅ − ⋅ +

.

3. Find the sum of the series 1 1 1 11 2 3 2 3 4 3 4 5 98 99 100

+ + + +⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

.

FURMAN SR 2001 #23

4. How many terms of the series ( )( )

1 1 1 1 12 3 3 4 4 5 5 6 1 2n n

+ + + +⋅ ⋅ ⋅ ⋅ + +

must be added

to be greater than 0.4999?

5. Find the sum of the first 50 terms of the series ( )2

1 1 1 1 , 2.3 8 15 1

nn

+ + + + + ≥−

What is the sum of the infinite series?

6. Find the sum of the first 50 terms of the series 2

1 1 1 1 , 2.6 11 20

nn n

+ + + + + ≥+


7. The series 22

1n n

∞

=∑ is sandwiched in between the two series in 5 and 6. Can you find the

sum of this infinite series by the methods of this section? Why or why not?

Written & Compiled by John Goebel, NCSSM Problem Solving Course: 2006

Perfect Squares: There are several strategies for solving problems that involve perfect squares. Differences of two squares. On the 2000 State Math Contest question 3 asked: Let p be an odd prime number. How many ordered pairs of positive integers ( , ) with x y x y< satisfy the condition 2 2y x p− = ? Since the difference of two perfect squares always factors, this is what we will do. ( )( )y x y x p− + = . But since p is prime, its only factors are 1 and p, so

1 and y x y x p− = + = . Solving this system we find that 2 1y p= + which works for all odd primes, and 2 1x p= − , which works for all odd primes as well. Thus there are infinitely many solutions. 3 2, 1; 5 3, 2; etc.p y x p y x= ⇒ = = = ⇒ = = Completing the Square For how many positive integers n is 2 18n n+ a perfect square? Well

( ) ( )( )

2 2 2 2

2 2

18 18 81 81

9 81 ( 9) ( 9) 81

n n k n n k

n k n k n k

+ = ⇔ + + = +

⇔ + − = ⇒ + + + − =.

Look at the factors of 81 in pairs: (1,81), (3,27), and (9,9). Let ( 9) 1 and (n+9) 81n k k+ − = + = and subtract to find k. In this first case, 40k = . The other pairs yield 12 and 0k = , making 32,6,0n = , but we cannot use 0. Checking we see that 2 232 18 32 1600 40+ ⋅ = = 2 26 18 6 144 12+ ⋅ = = . Writing as a square: In this method take the desired quantity and write it as the square of some number, perhaps involving some or all of the original expression. Then see what makes sense. In this problem from the 2002 NC State Math Contest, we are asked to find the four values for which 2 109n n+ + is a perfect square.

( )( )

22 2 2

2 2

2

109 2

109 2 1 2 109

1092 1

n n n k n nk k

n nk k n k k

knk

+ + = + = + +

⇒ + = + ⇒ − = −

−⇒ =

−

Now check values of k from 1 through 10.

Written & Compiled by John Goebel, NCSSM Problem Solving Course: 2006

2

1 2 3 4 5 6 7 8 9 10109 108 35 20 13.2.. 9.3.. 6.6.. 4.6.. 3 1.6.. 0.47..

2 1

kkn

k−

=−

The integer quantities for n, paired with the integer values for k yield 109, 37, 23, and 11. Problems: 1. The number of solutions of the equation 2 2 63x y− = in positive integers is:

FURMAN 2003 Jr #29 2. How many integers between 500 and 1500 are perfect squares?

FURMAN 2001 JR #1 3. Find the sum of all positive integers n so that 22001 n+ will be a perfect square.

NC SMC 2001 INT8. 4. Find the sum of all positive integers n so that 22004 n+ will be a perfect square. 5. How many perfect squares are divisors of the product 1! 2! 3! 4! 5! 6! 7! 8! 9!⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ?

AMC12 2003 #23 6. For what positive integer values of n is 2 19 99n n− + a perfect square?

AIME 1999 7. For how many positive integers n is 2 2004n n− a perfect square?

Polynomials The general form for a polynomial is

1 2 21 2 2 1 0( ) n n n

n n np x a x a x a x a x a x a− −− −= + + + + + + .

Generally, when we work with polynomials, we are restricted to the real numbers. You are well aware that a quadratic polynomial can have two distinct real zeros, one double zero, or no real roots. Factoring or the quadratic formula can be used to find all zeros. A cubic polynomial can have one real and two complex roots, or three real roots. Factoring, if one zero is fairly obvious, along with long or synthetic division, can be used to find roots. There is a cubic formula, but too complicated to learn and use. A quartic polynomial can have four real roots (possible two distinct and one double, two double zeros, or one zero with multiplicity 4), or two real and two complex, or no real roots and 4 complex roots. Like the cubic, factoring works, but the quartic formula exists but is not useful. Roots and Coefficients. If the coefficients of a polynomial are rational, there can be individual rational roots, but irrational roots will always occur in conjugate pairs. For example, suppose 1, -2, and 2 3− are zero of P(x) and we are told that the coefficients are rational, then 2 3+ is also a zero. We know too that zeros and factors are related,

so ( )( ) ( )( )

( )( ) ( )2 2 4 2 2

( ) ( 1)( 2) 2 3 2 3

2 4 1 3 5 9 2

P x k x x x x

k x x x x k x x x x

= − + − − − +

= + − − + = − − + −.

Ignoring the constant k since it does not affect the zeros and notice the following. The sum of the zeros is ( ) ( )1 ( 2) 2 3 2 3 3+ − + − + − = , that the sum of all possible products

for zeros taken two at a time is

( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )( )1( 2) 1 2 3 1 2 3 2 2 3 2 2 3 2 3 2 3

2 2 3 2 3 4 2 3 4 2 3 4 3 5

− + − + + + − − + − + + − +

− + − + + − + − − + − = −

and the sum of the products of the roots taken three at a time is

( )( ) ( )( ) ( )( )( ) ( )( )( )(1)( 2) 2 3 (1)( 2) 2 3 (1) 2 3 2 3 ( 2) 2 3 2 3

4 2 3 4 2 3 4 3 8 6 9

− − + − + + − + + − − + =

− + − − + − − + = −

and finally the product of all 4 zeros is ( )( )( )( )1 2 2 3 2 3 2− − + = − . The polynomial

now is 4 3 2( )P x x Ax Bx Cx D= − + − + , where A is the sum of the zeros, B the sum of the products taken two at a time, C the sum of the products taken 3 at a time, and D the sum of the products taken 4 at a time. This pattern continues for higher degree polynomials.


2

You will see it most often in problems where you are told something about the polynomial and you will use or find the sum and or product of the zeros. Not that you do not have to find the roots to know their sum or product. A simple example might be as follows: Given 3 2( ) 5 2 12P x x x x= − + + , find the sum (or product) of the zeros. The slow way is to actually find the zeros by trial and error maybe (3, 1 5,1 5+ − ) and then the sum (or product). The fast answer is just 5 for the sum and -12 for the product. Rational Roots Theorem. If the coefficients of a polynomial are integers, then the

rational roots, if any, will have to be of the form pq

±± , where p is a integer factor of the

constant term and q is an integer factor of the leading term. Let’s use a simple example to see why this works. Suppose

3 2( ) ( )( )( ) ( ) ( ) ( )P x ax b cx d gx h acgx ach adg bcg x adh bch bdg x bdh= − − − = − + + + + + −

The rational zeros of the polynomial were , , and b d ha c g

. Notice that the numerators are

all factors of the constant term (bdh) and the numerators are all factors of the leading coefficient (acg). Synthetic Division and Remainder Theorem Synthetic division is based on long division and is useful in finding the value of a polynomial for a given x as well as the factors of the polynomial. (If the value of a polynomial for a given number a, then x – a is a factor.

Long Division

Synthetic Division In Synthetic Division, write the number being checked (as a possible zero perhaps) and then the coefficients as follows: ]2 1 1 1 14−

Now leave a row blank and draw a horizontal line below the coefficients. ]2 1 1 1 14−

Bring down the leading coefficient and then multiply it by the zero candidate, placing this product in the space below the next coefficient. ]2 1 1 1 14

21

−

Now add, write the total below, and repeat.

2

3 2

3 2

2

2

3 7

( 2) 1423 143 6

7 147 14

0

x x

x x x xx x

x xx x

xx

+ +

− + + −

−

+ + −

−−−


3

Notice that the final sum is zero, telling us that 2 is a zero. But the nice thing about synthetic division, is that long division, the other coefficients are the coefficients of the quotient polynomial.

The Remainder Theorem states that when dividing a polynomial by a factor of the form ( )x a− there is a quotient polynomial and a constant remainder. The remainder is P(a) since ( ) ( ) ( ) ( ) ( ) ( )P x x a Q x r P a a a Q a r r= − ⋅ + ⇒ = − + = . Descartes’ Rule of Sign Changes. In a polynomial with real coefficients, the number of positive real roots equals the number of sign changes in the coefficients, or that number less multiples of 2. The number of negative real roots is the number of sign changes (less multiples of 2) in the polynomial P(-x). For example: 5 4 2( ) 9 24 5 126P x x x x x= − − + + has only two sign changes, so there are either 2 or 0 positive real roots.

5 3 2( ) 9 24 5 126P x x x x x− = − + − − + has 3 sign changes so there are either 3 or 1 negative real roots. Turns out that there are two positive real roots, one rational and one irrational and only one negative (irrational) root. The other two roots are complex. Miscellaneous Facts; It is obvious in any polynomial that P(0) is the y-intercept and equally as obvious that P(1) is the sum of all the coefficients. Problems:

1. Given the cubic polynomial 3 2( ) 7 4 28P x x x x= − − + . Two of the zeros are additive inverses. Find the zeros.

2. If 2( )p x ax bx c= + + leaves a remainder of 4 when divided by x, a remainder of 3 when divided by x + 1, and a remainder of 1 when divided by x – 1, then p(2) is?

3. If P(x) is a polynomial with rational coefficients and roots at 0, 1, 2, and 1 3− , then the degree of p(x) is at least?

4. When Madison’s dog chewed up her mathematics assignment, one particular equation was ripped apart. I found a piece of the beginning of the equation and a piece at the end, but the middle was missing. The beginning piece was

5 49x x− + and the ending piece was 11 0+ = . Fortunately the teacher had promised that all of the roots would be integers. How many times is -1 a root? [Furman ????]

5. The following is a polynomial. Find the sum of the squares of its coefficients. 3 9 8 7 6 5 4 3 23 18 28 84 42 98 72 15 1x x x x x x x x x− + − + − + + + + . FURMAN

6. If a cubic polynomial p(x) has roots at -1, 2, and 3, and if p(0) = 1, then the remainder when p(x) is divided by x – 1 is:

7. If 2 is a solution of 3 10 0x hx+ + = , then h equals:

]2 1 1 1 14

2 6 14

1 3 7 0

−


4

8. The number of distinct real solutions of the equation 3 24 8 5 1 0x x x− + − = is: 9. What is the sum of the squares of the roots of 4 25 6 0x x− + = 10. For how many integers N is 4 3 26 6N N N N+ < + ? 11. How any times does the graph of 3 2( ) 2 4f x x x x= − + + cross the x axis? 12. Madison’s dog chewed on her homework before she could finish it. The

fragment saved from the horrible canine’s mouth reveal only the two terms of highest degree of the polynomial p(x). It looked like 18 17( ) 3p x x x= − + . Madison found roots 1,2,3,…17. What was the missing root? FURMAN 2000 SR#27

13. Find the largest x so that 12

x < and 4 3 23 0.x x x x+ − + = FURMAN 2000

SR #8 14. The sum of the solutions of ( ) ( ) ( )3 3 32 3 6 4 4x x x+ + − = − is k, Find k.

ARML 1980 Relay 2-3. 15. The real value of x which satisfies

( ) ( ) ( ) ( ) ( )3 3 3 3 33 31 2 3 4 5 3x x x x x x+ − + − + − + − + − = is k. Find k. ARML 1980 Relay 2-4.

16. Find the four values of x for which ( ) ( )4 43 5 8x x− + − = − . ARML 1978 Team 8.

17. The equations ( ) ( )42 2 0x x− − − = and 2 0x kx k− + = have two roots in common. Find the value of k. ARML 1978 Ind 5

18. Find the remainder that results when ( ) ( ) ( ) ( ) ( )5 4 3 21 2 3 4 5x x x x x+ + + + + + + + + is divided by 2x + . ARML 1977 Team 2

19. If , 3 , 5 , , 3, 5a a a b b b+ + are all roots of a 4th degree polynomial equation, where 0 a b< < , compute all possible values of a.

20. Let ( )P x be a polynomial whose degree is 1996. If 1( )P nn

= for

1,2,3, ,1997n = , compute the value of (1998)P . ARML 1997 Team 9 21. Let r and s denote the two real roots of 2 5 1 0x x− + = . Then 8 8r s+ is an

integer. Determine this integer. ARML 1981 Team 4 22. There are two values for k for which the cubic polynomial 3 22 9 12x x x k− + −

has a double root. What are these two values? ARML 1981 Team 5 23. Determine the sum of the y-coordinates of the four points of intersection of

4 25 4y x x x= − − + and 2 3y x x= − . ARML 2006 Ind 6. 24. Determine the sum of the y-coordinates of the three points of intersection of

2 5y x x= − − and 1 .yx

=


Geometry Facts – Circles & Cyclic Quadrilaterals Circles, chords, secants and tangents combine to give us many relationships that are useful in solving problems. Power of a Point Theorem: The simplest of these theorems pertains to two chords of a circle that intersect in the interior of the circle.

The theorem tells us that AP PB CP PD⋅ = ⋅ . This is easy to prove since APC BPD≈ .

We know that AP PDCP PB

= .

Before going on to the other Power of a Point Theorems, it might be worth noting something about the angles formed by secants, chords, and tangents to circles. In the above figure, APC BPD∠ ≅ ∠ , since they are vertical angles, but they are both equal to

( )12

mAC mBD+ .

Likewise, CDB CAB∠ ≅ ∠ , and both have measure 12

mBC .

Now let’s look at two secant lines.


Is it easy to show that PAD PCB≈ , so PB PDPC PA

= . When this is written as

PB PA PD PC⋅ = ⋅ , we have our second “Power of a Point” theorem. The angle

( )12

m APC mAC mBD∠ = − .

Now let’s look at a secant and a tangent,

We need to know that 1

2m CAB mCB∠ = and 1

2m PCB mCB∠ = , so we have

PCA PBC≈ , and PC PBPA PC

= , or 2PC PA PB= ⋅ . The angle at P, has measure

( )12

m CPB mAC mBC∠ = − .

Closely related to circles are the Cyclic Quadrilaterals. These are quadrilaterals that are inscribed in a circle, that is, their vertices are on a circle.

First, it should be obvious that and A C∠ ∠ are supplementary, as are and B D∠ ∠ , since both pairs cut off opposite halves of the circle.


Ptolemy’s Theorem: If ABCD is a cyclic quadrilateral, then the sum of the products of opposite sides is equal to the product of the diagonals.

Proof: Drop segment BE so that ABE CBD∠ ≅ ∠ . Since BAE CDB∠ ≅ ∠ , we know that

ABE DBC∼ . Thus AB BD AB DC AE BDAE DC

= ⇒ ⋅ = ⋅ .

Also since ADB ACB∠ ≅ ∠ and ABD CBE∠ ≅ ∠ we have ABD EBC∼ . Thus AD EC AD BC BD ECBD BC

= ⇒ ⋅ = ⋅ . Therefore

( )AB DC AD BC AE BD BD EC

AE EC BD AC BD⋅ + ⋅ = ⋅ + ⋅

= + = ⋅

Bretschneider’s Formula: In any quadrilaterial, ABCD, with sides a, b, c, and d, the

area is 2( )( )( )( ) cos2

B DK s a s b s c s d abcd ∠ +∠⎛ ⎞= − − − − − ⎜ ⎟⎝ ⎠

, where 2

a b c ds + + +=

Proof: First draw diagonal AC and calculate the area of the quadrilateral as the sum of the areas of the two triangles formed.

1 1( ) ( )2 2

K ab sin B cd sin D= ⋅ + ⋅

Square this equation:

(1) 2 2 2 2 2 2 21 1 1sin ( ) sin( )sin( ) sin ( )4 2 4

K a b B abcd B D c d D= + + .


Now use the law of cosines to twice as follows:

2 2 2 2 22 cos( ) 2 cos( )p a b ab B c d cd D= + − = + − . Rewrite this last equation as 2 2 2 2 2 cos( ) 2 cos( )a b c d ab B cd D+ − − = − and then square this equation, set equal to

zero, and write as follows: ( ) ( )22 2 2 2 22 cos( ) 2 cos( ) 0ab B cd D a b c d− − + − − = . Square the first term and divide through by 16 to get the following equation:

(2) ( )22 2 2 2 2 2 2 2 2 24 cos ( ) 8 cos( )cos( ) 4 cos ( )

016

a b B abcd B D c d D a b c d− + − + − −=

Now add equations (1) and (2) :

( )

( ) ( )

( )( ) ( )

2 2 2 2 2 22

22 2 2 22 2 2 2 2 2

22 2 2 22 2 2 2

22 2 2 22 2 2 2

sin ( ) sin( )sin( ) sin ( )4 2 4

cos ( ) cos( ) cos( ) cos ( )4 2 4 16

sin( )sin( ) cos( )cos( )4 2 4 16

1 2 cos4

a b B abcd B D c d DK

a b c da b B abcd B D c d D

a b c da b abcd c dB D B D

a b c da b c d abcd B D

= + +

+ − −+ − + −

+ − −= + − + −

+ − −= + − + −

( ) ( )22 2 2 2 2 2 2 2

16

4 4 cos16 2

a b c d a b c d abcd B D+ − + − − += −

Square the term on the left.

( )2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 42 cos2 2 2 2 2 2

16 2abcd B Da b a c a d b c b d c d a b c dK

++ + + + + − − − −= −

Now by adding and subtracting 8abcd to the numerator on the left, and some pretty fancy factoring, we have

( )2 cos( )( )( )( )16 2 2

abcd B Da b c d a b c d a b c d a b c d abcdK+− + + + − + + + − + + + −

= − − .

Now, if 2

a b c ds + + += , we have

( )2 ( )( )( )( ) 1 cos( )2

abcdK s a s b s c s d B D= − − − − − + + . Now, using the half angle cosine

formula, we have


2 2( )( )( )( ) cos2

B DK s a s b s c s d abcd +⎛ ⎞= − − − − − ⎜ ⎟⎝ ⎠

, and finally,

2( )( )( )( ) cos2

B DK s a s b s c s d abcd +⎛ ⎞= − − − − − ⎜ ⎟⎝ ⎠

Brahmagupta’s Theorem: In a cyclic quadrilateral the area

( )( )( )( )K s a s b s c s d= − − − − , where 2

a b c ds + + += .

Proof. Once Bretschneider’s theorem has been proved, Brahmagupta’s theorem follows almost immediately, since 180B D∠ +∠ = , the last term in Bretschneider’s becomes zero.

Problem Set 1. Two circles or radius 4 and 1 are externally tangent. Compute the sine of the

angle formed by their common external tangents. ARML 1986, Team 1 2. The sides of a quadrilateral are 3, 3, 4 and 8 (in some order). Two of its angles

have equal sines but unequal cosines, yet the quadrilateral cannot be inscribed in a circle. Compute the area of the quadrilateral. ARML 1986, Team 3

3. Show that if a quadrilateral is cyclic, [that is, it is inscribable in a circle], and its

consecutive sides are a,b,c, and d, and its diagonals are p and q, then

( )( )2 2 2 2pq a b c d≤ + + . ARML 1987, Power I(c)

4. A convex n-gon will be called “Pythagorean” if it has integer sides, it is cyclic,

and its longest side is a diameter for its circumscribing circle. It shall be denoted by Pn, or Pn:( a,b,…), where a, b, … are the lengths of its sides. We shall always use the letter d for its longest side. [Thus P3 is a Pythagorean triangle. Note that it would be a right triangle.]

I. [There is a theorem which states (in part) that: If a prime d is the hypotenuse of a Pythagorean triangle, then 2d is the hypotenuse of two Pythagorean triangles,

3d is the hypotenuse of three Pythagorean triangles, etc. I. A. Find two P3’s for which d = 25. B. Find three P3’s for which d = 125. II. Ptolemy’s Theorem says: A convex quadrilateral is cyclic if and only if the

product of its diagonals equals the sum of the products of the two pairs of opposite sides.


A. If the P3:(3,4,5) is reflected as shown,

A quadrilateral EFGH can be formed (it will not be a P4, as FG is not an integer). Multiplying each side by 5 produces a P4. Find the sides of this P4. B. Find a P4 with two equals sides and with d = 25 that is different from the

answer to part IIA. [Note: Two Pn’s are not considered different if their sides are equal, but in a different order.]

C. Show that a Pn must exist for all integers 3n ≥ . [This may be done by

describing how to create such a Pn.] ARML 1989 Power Question

5. The bisectors of the angles of quadrilateral ABCD are

drawn. The form quadrilateral EFGH, as shown, in which 193oE F+ = . If A C> , compute the numerical

value of A C− . ARML 1989 Individual 2 6. A convex hexagon is inscribed in a circle. If its successive sides are 2, 2, 7, 7, 11,

and 11, compute the diameter of the circumscribed circle. ARML 1989 Individual 8.

7. “Nice” Angles and Polygons: Definition 1: We call an angle A “nice” if both sin(A) and cos(A) are rational. Definition 2: We call a convex polygon “nice” if all of its interior angles are “nice.” III. A convex quadrilateral has the following properties: 1. Its sides are integers whose product is a square. 2. It can be inscribed in a circle, and can be circumscribed about (another) circle. Prove that this quadrilateral is “nice.” ARML 1991 Power Question


6

( ) ( ) ( )( )( ) ( )( )

( ) ( )( )( ) ( )( )

22

2

2 2

2 2

11 2 3 2 3 12 3

1 2 3 2 3 12 3

1 122 3 2 2 312 121 2 3 14 2 3

12

n n

n

n n

n n

n n

t

−

−

−

⎛ ⎞− = + − − −⎜ ⎟⎝ ⎠

⎛ ⎞= + − + −⎜ ⎟⎝ ⎠

= + − + + −

= + − + +

These two quantities are equal, so that proves the given result. Problems. 1. Make an argument to show that the third differences for cubic data is actually 6a. 2. Find the sum of the squares of the first 100 positive integers. 3. Find the sum of the cubes of the first 50 positive integers. 4. Find the sum 21 4 9 16 100− + − + − . 5. The Lucas sequence begins with 2, then 1, and then the remaining terms are found

as in the Fibonacci sequence. Write the recursive and closed forms for the terms of this sequence.

6. The first term of a sequence is 0 and the second is 1. For all

1 22, 3 2n n nn t t t− −≥ = ⋅ − ⋅ . Find the first 5 terms of the sequence, the recursive, and the closed form for the terms of this sequence.

7. Sequence ( )1 2 3, , ,...a a a is defined recursively by 1 20, 100a a= = and

1 22 3n n na a a− −= − − . Find the greatest term in the sequence ( )1 2 3, , ,...a a a . DUKE 1998 Team #9

8. Find k given that (0)

and ( 50) 4000.( ) ( 1) 3 2

f kf

f n f n n=⎧

− =⎨ = + − −⎩

NC SMC 2003 Integer 11.


Triangles

Some of the following information is well known, but other bits are less known but useful, either in and of themselves (as theorems or formulas you might want to remember) or for the useful techniques and strategies used in their development. Area Formulas: Everyone knows the formula for the area of a triangle given a base and the corresponding

height. It is, of course, 2

b hK ⋅= . But if you are given 2 sides and the included angle

(SAS), or all three sides (SSS), the following area formulas are useful.

In the above diagram, A, b, c are given, so we know that the altitude from C to side c has

length ( )bSin A , so the area is ( )( ) ( )2 2

base height c bSin AK ⋅= = .

If, in the same diagram, only the sides are given, then we will use the Law of Cosines to find something about angle A. Note that we do not need to find the sin(A) from the cos(A), but will work with the square of the most recent formula.

2 2 22 2 2

2 22 2 2 2 2 22 2

2 cos( ) cos( )2

cos ( ) sin ( ) 12 2

b c aa b c bd A Abc

b c a b c aA Abc bc

+ −= + − ⇒ = ⇒

⎛ ⎞ ⎛ ⎞+ − + −= ⇒ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


( ) ( )( ) ( )( )

( )( ) ( )( )

( )( ) ( )( ) ( )( )( )( )

22 2 22 2

2 2 22

22 2 2 22 2

222 2 2 2 2

2 2 2 2 2 2

2 22 2

12sin ( )

4 4

24 2

4 162 2

16

( ) ( ) ( ) ( )16 16

b c ab cbcb c AK

bc b c ab c

b c bc b c a

bc b c a bc b c a

a b c b c a a b c a b c b c a b c a

⎛ ⎞⎛ ⎞+ −⎜ ⎟− ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= =

⎛ ⎞− + −⎜ ⎟⎜ ⎟ − + −⎝ ⎠= =

− + − + + −=

− − + − − − + − + − + += =

Now if we let P a b c= + + and 2PS = , we get

( )( )( )( )

( )( )( )

2

2

162 2 2

2 2 2 2( )( )( ) ( )( )( )

a c b a b c b c a a b cK

P b P c P a P s b s c s a s

K s s a s b s c K s s a s b s c

+ − + − + − + +=

− − −= ⋅ ⋅ ⋅ = − − −

∴ = − − − ⇒ = − − −

This last formula is known as Heron’s Formula. Special Points and Segments: The angle-bisectors of the angles of a triangle are concurrent and meet at the point called the incenter. Recall from your geometry that an angle-bisector is equidistant from the sides of the angle being bisected, so the incenter is equidistant from all three sides, making it the center of the circle that can be inscribed inside the triangle, the inscribed circle.


Since the area of the entire triangle is equal to the sum of the areas of the three smaller triangles , ,ABQ BCQ and ACQ , we have

2 2 2 2cr ar br a b c KK r rs r

s+ +⎛ ⎞= + + = = ⇒ =⎜ ⎟

⎝ ⎠

The perpendicular-bisectors of the sides meet at a point that is equidistant from the endpoints of each of the sides (the vertices), so this point is equidistant from all three vertices, making it the center of the circumscribed circle. This point is called the circumcenter.

The area of the triangle, using the formula 1 sin( )2

K bc A= leads us to a formula that

relates the area K to the radius R of the circumscribed circle. We know that 2m CPB m CAB∠ = ∠ (a central angle is measured by the arc and an inscribed angle is

one-half the measure of the arc), so m FPB m A∠ = ∠ , but in 2,sin( )a

FPB FPBR

∠ = , so

sin( )2aAR

= , making the area 1 1sin( )2 2 2 4 4

a abc abcK bc A bc RR R K

⎛ ⎞= = = ⇒ =⎜ ⎟⎝ ⎠

.

Cevians: A cevian is a line segment that joins a vertex of a triangle with a point on the opposite side. Medians and angle bisectors, and in some cases altitudes, are examples of cevians. There are some nice formulas for the lengths of these segments. But first, let’s review some of our basic geometry.


Theorem: An angle-bisector divides the opposite side into segments that are proportional

to their adjacent sides. c db e= , or d e

c b= .

This means that if the length of segment BC has length a, then acdb c

=+

and bceb c

=+

.

Stewart’s Theorem.

Let’s look at the proof of the general theorem for finding the length of a cevian. The Law of Cosines does the trick.

In ABC 2 2 2

cos( )2

a b cCab

+ −= and in ABN

2 2 2

( )2

b e xCos Cbe

+ −= , so

( ) ( )( )

( )

2 2 2 2 2 22 2 2 2 2 2

2 2 2 2 2

2 2 2

2 2 2 2 2 2

2 2

( )

a b c b e x e a b c a b e xab be

ax a e ae b a e c e

ax ae a e b d c e

ax aed b d c e a x de b d c e

+ − + −= ⇒ + − = + −

∴ + − = − +

⇔ + − = +

⇔ + = + ⇔ + = +

Now let’s apply this to the median.


Using Stewart’s Theorem we have

( )2

2 2 2 2 2

2 2 2 2 2 22 2

2 2 2

2 2 2 2

2 2 2 4

2 4

a a a aa m b c b c

a b c b c am m

b c am

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = + = +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

+ +⎛ ⎞∴ + = ⇒ = −⎜ ⎟⎝ ⎠

+∴ = −

If we apply Stewart’s Theorem to the angle-bisector, we have

( )

( )( )

2 22 2

2 2 2 2

2 22 2 2 22

2 2 21( ) ( ) ( )

ac abb cdb ec ac abb c b ca x de db ec x dea a b c b c

bc b c ab c bc a bc ax x bcb c b c b c b c

++ ⎛ ⎞⎛ ⎞+ ++ = + ⇒ = − = − ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠

+ − ⎛ ⎞+⇒ = − = ⇒ = −⎜ ⎟+ + + +⎝ ⎠

Ceva’s Theorem: Any time that the cevians of a triangle are concurrent, there is a special relationship that exists for the lengths of the segments on each of the sides. This relationship is known as


Ceva’s Theorem and states that if cevians AD, BE and CF are concurrent at point P if and

only if 1AF BD CEFB DC EA

⋅ ⋅ =

First extend segments AD and BE and construct line the line L through C that is parallel to AB. Label the points where AD and BE extended meet L as G and H.

From the following pairs of similar triangles we have the following proportions:

AEB CEH∼ AE AB

CE CH= (1)

ADB GDC∼ CD CGBD AB

= (2)

FPB CPH∼ FB FPCH CP

= (3)

APF GPC∼ AF FPCG CP

= (4)


From the last two proportions it follows that FB FACH CG

= or FB CHFA CG

= (5). Now multiply

(a), (2) and (5) together to get 1AE CD BF AB CG CHEC DB FA CH AB CG

⋅ ⋅ = ⋅ ⋅ = . This proves that if the

lines are concurrent, the identity is true. Now we must prove that if the identity is true then the cevians must be concurrent.

Let P’ be ther point where FC and BE intersect and draw segment AP’ and extend to point D’ on side BC. Since these three cevians are concurrent, the first part of the theorem holds so

1AE CD BFEC D B FA

′⋅ ⋅ =

′,

but from the hypothesis we know that

1AE CD BFEC DB FA

⋅ ⋅ = ,

so

AE CD BF AE CD BF CD CDEC D B FA EC DB FA D B DB

′ ′⋅ ⋅ = ⋅ ⋅ ⇒ =

′ ′

The last part of this proportion, CD CDD B DB

′=

′ implies that

'1 1'

CD CD CD D B CD DB CB CBD B DB D B DB D B DB

′ ′ + ++ = + ⇒ = ⇒ =

′ ′.

This last proportion implies that D and D’ are the same point and the second part of the theorem is proved.


Problems: 1. Two sides of a triangle are 7 and 9 while the median to the third side has length 5.

Find the length of the third side. 2. The sides of a triangle are consecutive integers and the area is an integer. Find

the triangle with the smallest perimeter that is not a right triangle. Are there others?

3. Triangle ABC is reflected in (or about) its median AM (extended) as shown. If

6, 12, 10AE EC BD= = = and 3AB k= , compute k. ARML 1978 Ind 8

4. Right triangle ABC (hypotenuse AB ) is inscribed in equilateral triangle PQR, as

shown. If 3PC = , and 2,BP CQ= = compute AQ. ARML 1991 Ind 7

5. In triangle ABC, the perpendicular bisector of AC intersects AC at M and AB

at T. If the area of triangle AMT is 14

the area of triangle ABC, and o128A C+ = , compute the number of degrees in angle A.

ARML 1988 Ind 4


6. An isosceles triangle has a median equal to 15 and an altitude equal to 24. This information determines exactly two triangles. Compute the area of each of these triangles. ARML 1994 Team 4

7. Point P is inside ABC . Line segments , , and APD BPE CPE are drawn with D

on BC , E on CA , and F on AB . (See figure). Given that 6,AP = 9,BP = 6,PD = 3PE = , and 20CF = , find the area of ABC . AIME 1989 #15

8. The points (0,0), (a,11), and (b,37) are the vertices of an equilateral triangle. Find the value of ab. AIME 1994 # 8

9. Two medians in a triangle are congruent if and only if the sides which they bisect

are congruent.

SOLUTIONS


Analytical Geometry: Solutions: 1. The vertex of the graph of the parabola 23y x bx c= + + is at (-2,-7); determine

the value of b-c; NCALGII2002:33 2. The distance between the centers of the circles 2 28 28 112 0x x y y− + − + = and

6 26 20 12 0x x y y+ + + − = is: FUR2003JR11 3. Through which quadrants does the circle 2 24 6 1 0x x y y+ + − + = pass?

FUR2003JR13 4. If a parabola has vertex (0,0) and focus (1,0), the equation is: FUR2003JR!5 5. If Q is the point on the circle 2 210 6 29 0x x y y− + + + = which is furthest from

the point ( 1, 6)P − − , then the distance from P to Q is: FUR2003SR2.

( ) ( )

2 2 2 2

2 2

10 6 29 0 10 25 6 9 29 25 9 5

5 3 5

x x y y x x y y

x y

− + + + = ⇔ − + + + + = − + + =

⇔ − + + =

So this is a circle with center (5,-3) and radius 5 . The distance from P to Q includes the distance from P to the center and then an additional radius. So this

distance is ( )( ) ( )( )2 2 2 25 1 3 6 5 6 3 5 45 5− − + − − − + = + + = + 6. The parabola 2y ax bx c= + + intersects the y-axis in the point (0,8) and intersects

the x-axis in the single point (2,0). How many pairs of integers (m,n) with 2000 2000m− ≤ ≤ lie on the parabola? NCSMC2000INT4

The parabola must be of the form

( ) ( ) ( )2 2 22 8 0 2 2 2 2y a x a a y x= − ⇒ = − ⇒ = ⇒ = − . Since the y-values grow much more quickly than the x-values, we need to find all possible points whose y-values are less than or equal to 2000. So ( ) ( )2 22 2 2000 2 1000 1000 2 1000x x x− ≤ ⇒ − ≤ ⇒ − ≤ − ≤ . But since we

want integer coordinates, this means that 31 2 31 29 33x x− ≤ − ≤ ⇒ − ≤ ≤ , so there are 63 such lattice points.

7. Let f(x) be a quadratic polynomial such that f(3) = 15 and f(-3) = -9. Find the

coefficient of x in f(x). SMC2002MC1 We know that ( )215 3 (3)a b c= + + and ( )29 3 ( 3)a b c− = − + − + , and since we are

looking for b, subtract these to get 24 6 4.b b= − ⇒ = − 8. The center C and the vertices (V) of the ellipse 2 24 9 16 54 61 0x y x y+ − − + = are:

NCALGII2000#9

( ) ( )

( ) ( ) ( ) ( )

2 2 2 2

2 22 2

4 9 16 54 61 0 4 4 4 9 6 9 61 16 81 36

2 34 2 9 3 36 1

9 4

x y x y x x y y

x yx y

+ − − + = ⇔ − + + − + = − + + =

− −⇔ − + − = ⇔ + =

so the center of this ellipse is (2,3) and vertices (end points of the major axis) are (5,3) and (-1,3).

9. The quadratic equation 2y ax bx c= + + is known to pass through the points (0,5),

(2,11) and (-2,15). Find the sum of a and b. NCALGII2000#17 We know that 11 4 2 ,5 ,15 4 2a b c c a b c= + + = = − + , so

4 2 6,4 2 10 8 16 2a b a b a a+ = − = ⇒ = ⇒ = , so 1b = − and the sum 1a b+ = . 10. Write an equation for the hyperbola with horizontal transverse axis and

asymptotes 43y x= ± . NCALGII2000#25

2 2

19 16x y

− =

11. A parabola of the form 2y x bx c= + + contains the points (2,3) and (4,3), Find

the value of c. AMC102006. We know that 3 4 2 ,3 16 4b c b c= + + = + + , so

2 1,4 13 2 12, 6b c b c b b+ = − + = − ⇒ = − = − . This makes 11c = .


1

Different Number Bases Solutions:

1. 3b + 2 = 2c + 1 and 2b + 1 = c + 3 thus c = 2b – 2, so 3b + 2 = 2(2b - 2) + 1 , or 3b + 2 = 4b – 3, so b = 5.

2. 2 2 2(6 2)( 4) 8 8 6 26 8 8 8b b b b b b+ + = + ⇔ + + = + , so 22 26 0 0,13b b b− = ⇒ = , but since base 0 is meaningless, b = 13.

3. 100 10 ( ) 126a b c a b c+ + − + + = and 100 10 ( ) 225c b a a b c+ + − + + = . Simplify the first equation to get 99 9 126 11 14a b a b+ = ⇔ + = , so a must be 1 and b must be 3. Similarly, the second equation simplifies to 99 9 225 11 25c b c b+ = ⇔ + = , but since b = 3, c must be 2.

4. Written in base b we have 23,4 2, 1b b b+ + + as our numbers, so 2(4 2) ( 3) ( 1) (4 2)b b b b+ − + = + − + . Simply this to 2 7 0 0,7b b b− = ⇒ = , so b must

be 7.

5. We have 2 3

1 1 1rb b b

= + + + and 2

1 1 1 1 522 4 4 4

r bb b b b b

⎛ ⎞ ⎛ ⎞= ⋅ + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

. The first

sum is a geometric series with ratio 1 , 1bb

> , so it converges to 1 1

1 11br

bb= =

−−, so

we have 1 5 4 5 5 51 4

b b bb b

= ⇔ = − ⇒ =−

.


Complex Numbers Solutions: 1. Show that the power rule works for n = 2, and n = 3.

( ) ( )( ) ( )( )( ) ( )( )( )( )

2 22

2 2 2 2

2 2 2

2

cos sin cos sin

cos 2 cos( )sin( ) sin

cos sin 2 sin( ) cos( )

cos 2 sin(2 )

r i r i

r i i

r i

r i

θ θ θ θ

θ θ θ θ

θ θ θ θ

θ θ

+ = ⋅ + =⎡ ⎤⎣ ⎦

+ + =

− + =

+

( ) ( )( ) ( )( )( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )( ) ( )( ) ( )( )( ) ( ) ( )( ) ( )( )( )

3 2

3

3 2 2 2 2

3 2 2 2 2

3

cos sin cos 2 sin(2 ) cos sin( )

cos 2 cos sin sin 2 sin cos 2 sin 2 cos

cos 2cos 1 2sin cos sin 1 2cos 2sin cos

cos 2cos 1 2 1 cos sin 1 2sin 2cos

r i r i r i

r i

r i

r i

r

θ θ θ θ θ θ



θ θ θ θ θ θ

+ = + + =⎡ ⎤⎣ ⎦

⎡ ⎤− + + =⎣ ⎦⎡ ⎤− − + − + =⎣ ⎦⎡ ⎤− − − + − + =⎣ ⎦

( ) ( )( ) ( ) ( )( ) ( )( )( )( )( ) ( )( ) ( ) ( )( )( )

( ) ( )

2 2 2

3 3 2

3

cos 4cos 3 sin 1 2sin 2 1 sin

4cos 3cos sin 3 4sin

cos 3 sin 3

i

r i

r i

θ θ θ θ θ

θ θ θ θ

θ θ

⎡ ⎤− + − + − =⎣ ⎦⎡ ⎤− + − =⎣ ⎦

+⎡ ⎤⎣ ⎦

2.

( ) ( )( )( ) ( )( )

( ) ( )( )( ) ( )( )

( ) ( )( )( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )( )( ) ( )( )

( ) ( )( ) ( )( )

1 1 1 1 1 1 2 2

2 2 2 2 2 2 2 2

1 2 1 2 1 2 1 212 2

2 2 2

1 11 2 1 2 1 2

2 2

cos sin cos sin cos sincos sin cos sin cos sin

cos cos sin sin sin cos cos sin

cos sin

cos sin

r i r i ir i r i i

irr

r ri cisr r

θ θ θ θ θ θθ θ θ θ θ θ


θ θ

θ θ θ θ θ θ

+ + −= ⋅ =

+ + −

+ + −=

+

− + − = −

3. Raise ( )81 3+ .


( )

( )

88

8

8 7

8 81 3 2 cos sin 2 cos sin3 3 3 3

1 32 2 1 32 2

i i

i i

π π π π⎡ ⎤⎛ ⎞ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = + = +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎣ ⎦⎣ ⎦⎛ ⎞

= − + = − +⎜ ⎟⎜ ⎟⎝ ⎠

4. Find 4 i .

( ) ( ) ( )( ) ( ) ( )( )1

1 44 40 1 cos sin 1 cos sin2 2 8 8

2 2 2 22 2

i i i i

i

π π π π⎡ ⎤= + = + = +⎢ ⎥⎣ ⎦⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

5. A certain complex number satisfies 2 1ω ω= − . What is 99ω ? FURMAN 2001 SR #13.

2 2

9999

1 31 1 0 cos sin2 3 3

99 99cos sin cos sin cos( ) sin( ) 13 3 3 3

i i

i i i

π πω ω ω ω ω

π π π πω π π

± ⎛ ⎞ ⎛ ⎞= − ⇒ − + = ⇒ = = ±⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞∴ = ± = ± = ± = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

6. If ( )3 74x iy ki+ = − + , find the absolute value of k, given that 1 and 1x i= = − . NC

SMC 2002 INT1.

( )3 3 2 2 3

3 2 2 3

74 3 ( ) 3 ( ) ( )

3 (3 ) 74

x iy ki x x iy x iy iy

x xy i x y y ki

+ = − + ⇒ + + + =

− + − = − +

If 1x = , then 3 2 2 23 1 3 74 3 75 5x xy y y y− = − = − ⇒ = ⇒ = ± , and

( ) ( )323 1 5 5 15 125 110 110or⋅ ± − ± = ± = −∓ , so the absolute value of y is 110. 7. Find the sum of the cube of the roots of the equation 10 9 8 2 1 0x x x x x+ + + + + + =

Duke 2003.

11

10 9 8 2 111

xx x x x xx−

+ + + + + + =−

, so the roots of the first are exactly the same as the

roots of the second, except for x = 1. In the rational expression, we are looking for the eleven 11th roots of unity. If the problem asked for the sum of the roots, this would be easy, since the sum of the roots of 11 1 0x − = is zero, but we can’t use 1, so the sum of roots is -1, but that is not the question. We know the roots are evenly spaced around the unit circle at the following angles. Notice that the cubes (hence 3 times the angles) are written underneath.


We see that the cubes of the complex numbers, as indicated by taking three times the

angle, are the same as the original roots, except for the order in which they occur. So the sum of the cubes of the roots is the same as the sum of the roots, which is -1.

8. Find the sum of the squares of the roots of the equation 6

2

1 01

xx−

=−

The roots of the numerator are as follows:

2 2 4 4cos 0 sin 0, cos sin , cos sin ,6 6 6 6

6 6 8 8 10 10cos sin , cos sin , cos sin6 6 6 6 6 6

i i i

i i i

π π π π

π π π π π π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

The first zero is simply 1, and the fourth is -1, which we cannot use (because of the denominator), so we are looking at the squares of the other four. These squares are

4 4 8 8cos sin , cos sin ,6 6 6 616 16 20 20cos sin , cos sin

6 6 6 6

i i

i i

π π π π

π π π π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

and these simplify to

2 2 1 3 4 4 1 3cos sin , cos sin ,3 3 2 2 3 3 2 2

2 2 1 3 4 4 1 3cos sin , cos sin3 6 2 2 3 3 2 2

i i i i

i i i i

π π π π

π π π π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = − + + = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = − + + = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

, so the sum of

these squares is 14 22

⎛ ⎞− = −⎜ ⎟⎝ ⎠

. If we simply wanted to know the sum of the roots, we

could notice that the sum of the roots of the numerator is zero, the sum of the roots of the denominator is also zero, and includes two of the roots of the numerator, so the sum of the roots (not the squares of the roots) of the entire equation is still zero.

0 211π 4

11π 6

11π 8

11π 10

11π 12

11π 14

11π 16

11π 18

11π 20

11π

0 611π 12

11π 18

11π 2

11π 8

11π 14

11π 20

11π 4

11π 10

11π 16

11π


Distance Formulas Solutions: 1. What is the closest the line 4 5 20x y+ = comes to the origin?

Since 4 5 20 0x y+ − = , we have 2 2

4(0) 5(0) 20 20 20 4141414 5

d+ −

= = =+

.

2. What is the distance between the points ( )(2, (2)) and (2), 2f f if

( ) 4 1f x x= + . The two points are (2,3) and (3,2) , so the distance is 2 2(2 3) (3 2) 2− + − = . 3. How far from the origin are the points of intersection of the conics 2 27 47x y+ =

and 2 22 4 28x y− = ? Since the distance we want is simply 2 2x y+ , notice that the linear

combination using 1 times each equation is simply 2 2 2 23 3 75 25x y x y+ = ⇔ + = , so the points of intersection, whatever they are,

are only 5 units from the origin. 4. What is the shortest distance between the circle 2 2 25x y+ = and the line

3 4 48x y+ = ?

First notice that the closest the line gets to the origin is 2 2

3(0) 4(0) 48 4853 4

+ −=

+, so

it is more than 5 units from the origin. Now subtract the radius of the circle,

yielding the distance 48 2355 5− = .

5. Find the distance between the parallel lines 2 3 12x y− = and 2 3 36x y− = . Pick any point on one line and plug it into the distance formula for a point and a

line. So let’s pick (6,0) from the first line. Now

2 2

2(6) 3(0) 36 24 24 1313132 3

d− −

= = =+

.

6. A triangle has vertices (2,3), (6,-2), and (-1,-4). Find the area of the triangle.


You could use the determinant-based formula to get the area (if you know it), or find the length of one segment and the altitude to it (using distance from a point to a line) and be done. If A = (2,3) and B = (6,-2), then 2 24 5 41AB = + = . The equation for the line through A and B is 5 4 22 0x y+ − = , so the altitude to AB

has length 2 2

5( 1) 4( 4) 22 43415 4

− + − −=

+, making the area ( )1 43 4341

2 241⎛ ⎞ =⎜ ⎟⎝ ⎠

.

7. Find the shortest distance between the parallel planes 3 4 5 12x y x+ − = and

3 4 5 20x y z+ − = . Pick any point on the first plane, say (4,0,0) and plug this into

2 2 2

3(4) 4(0) 5(0) 20 8 8 2 4 210 5503 4 5

+ − −= = =

+ +.

8. What is the closest that the line 4 17 5

y x= + comes to a lattice point? Duke03-

Team10. The line, in standard form, is 20 35 7 0x y− + = , so the distance to any lattice

point (x,y) is 2 2

20 35 7

20 35

x y− +

+, so the closest point will occur when the numerator

is minimized. Since it is an absolute value, and must be an integer, we need to see if this quantity can be zero, or if not 0, what is the smallest possible value. This is a Diophantine problem. Well, clearly 20 35x y− will always be a multiple of 5, so we could make it -5, making the entire numerator 2. This is the smallest possible value for the numerator, so the shortest distance is

2 2

2 2 2 653255 6520 35

= =+

.

9. What is the closest the plane 1 3 43 5 15

z x y= + + comes to a lattice point?

Write the equation for the plane in standard form: 15 5 9 4 0x y z− − + = . Now

since 5 and 9 are relatively prime, the quantity 5 9y z+ can be made to equal any number we would like it to be (Chinese Remainder Theorem). When 0x = , we can make 5 9 4y z+ = − , when 1, 1y z= = − and the quantity 15 5 9 4 0x y z− − + = at the lattice point (0,1,-1), making the minimum distance zero.

10. Under what conditions would the line 0Ax By C+ + = be guaranteed to hit a

lattice point?


If A and B are relatively prime, then the linear combination 1Ax By+ = will always have a solution, so we can make ( )( )Ax By C C+ − = − , which would make

the numerator of the distance formula 2 2

Ax By C

A B

+ +

+ equal to zero, hence the line

hits some lattice point. 11. The circles 2 2 10 6 18 0x y x y+ + − + = and 2 2 14 4 44 0x y x y+ − + + = have two

internal tangents. On each the distance between the points of tangency are equal. Find this distance.

11. and

( ) ( )

2 2 2 2

2 2

14 4 44 0 14 49 4 4 44 4 49

7 2 9

x y x y x x y y

x y

+ − + + = ⇔ − + + + + = − + +

⇔ − + + =

The first circle has center (-5,3) and radius 4 while the second has center (7,-2) and radius 3. The distance between the centers P and Q is

( )2 2 2 27 ( 5) ( 2 3) 12 5 13PQ = − − + − − = + =

Since APC BQC∼ , we know that 4

3AC PCBC QC

= = . Thus ( )4 137

PC = and

( )3 137

QC = . This makes 2

2 2 24 13 4 4 84 13 7 120 307 7 7 7

AC ⋅⎛ ⎞= − = − = =⎜ ⎟⎝ ⎠

.

Likewise 2

2 2 23 13 3 3 63 13 7 120 307 7 7 7

BC ⋅⎛ ⎞= − = − = =⎜ ⎟⎝ ⎠

. Thus the length

of the internal tangent is 8 630 30 2 307 7

AB = + = .

( ) ( )

2 2 2 2

2 2

10 6 18 0 10 25 6 9 18 25 9

5 3 16

x y x y x x y y

x y

+ + − + = ⇔ + + + − + = − + +

⇔ + + − =


1

Finite Differences and Recurrence Relations Solutions. 1. Make an argument to show that the third differences for cubic data is actually 6a. X Y 1 2 3 x - 2 ( ) ( )3 22 2

( 1)a x b x

c x d− + −

+ − +

( )( )

23 9 7

2 3

a x x

b x c

+ + +

+ +

x - 1 ( ) ( )3 21 1( 1)

a x b xc x d

− + −

+ − +

( )6 6 2a x b+ +

( )( )

23 3 1

2 1

a x x

b x c

+ + +

+ +

6a

x ( ) ( )3 2a x b xcx d

+

+ +

( )6 2a x b+

( )( )

23 3 1

2 1

a x x

b x c

− + +

− +

6a

x + 1 ( ) ( )3 21 1( 1)

a x b xc x d

+ + +

+ + +

( )6 6 2a x b− +

( )( )

23 9 7

2 3

a x x

b x c

− + +

− +

x + 2 ( ) ( )3 22 2( 2)

a x b xc x d

+ + +

+ + +

2. Find the sum of the squares of the first 100 positive integers.

Of course this is arithmetic, so we all know the quick formula for this, that is

( )1001001 100 (101)(50) 5050

2S = + = = , but let’s do it with the finite differences.

X 1 2 3 4 5 Y 1 3 6 10 15

1 2 3 4 5 2 1 1 1


2

It looks like the second finite differences are all one, so the function for the partial

sums is quadratic with leading coefficient ½. So we get 21( )2

S n n bn c= + + and

see that ( )21(1) 1 12

S b c= + + = and ( )21(2) 2 2 32

S b c= + + = , so 1 ,2

b c+ = and

2 1b c+ = . Solving this system of equations gives us 1 , 02

b c= = , so

( )21 1 1( ) 12 2 2

S n n n n n= + = ⋅ + .

3. Find the sum of the cubes of the first 50 positive integers.

Since the fourth differrences are constant, we know that this is a fourth-degree

polynomial of the form 4 3 26( )4!

S n n bn cn dn f= + + + + . There are several

approaches to finding the remaining coefficients. The most efficient way may be

to notice that the desired sums are all perfect squares, so 2

21( )2

S n n bn c⎛ ⎞= + +⎜ ⎟⎝ ⎠

,

and 2

21 1(1) 1 12 2

S b c b c⎛ ⎞= ⋅ + + = ⇒ + =⎜ ⎟⎝ ⎠

and

221(2) 2 2 9 2 2 3

2S b c b c⎛ ⎞= ⋅ + + = ⇒ + + =⎜ ⎟

⎝ ⎠. This system yields 1 , 0

2b c= = , so

our desired sum is ( )2

22 21 1 1( ) 12 2 4

S n n n n n⎛ ⎞= + = +⎜ ⎟⎝ ⎠

,

4. Find the sum 21 4 9 16 100− + − + − .

Notice that the partial sums for the triangular numbers with the added fact that every other one is negative. So if we find the formula for the triangular numbers, we are nearly done.

n 1 2 3 4 5 6 S(n) 1 -3 6 -10 15 -21

X 1 2 3 4 5 6 7 Y 1 9 36 100 225 441 784

1 8 27 64 125 216 343 2 19 37 61 91 127 3 18 24 30 36 4 6 6 6


3

n 1 2 3 4 5

S(n) 1 3 6 10 15 1 2 3 4 5 2 1 1 1

So we see that the sum is quadratic of the form 21( )2

S n n bn c⎛ ⎞= ⋅ + +⎜ ⎟⎝ ⎠

.

Substituting we find that 21(1) 1 12

S b c⎛ ⎞= ⋅ + + =⎜ ⎟⎝ ⎠

and 21(2) 2 2 32

S b c⎛ ⎞= ⋅ + + =⎜ ⎟⎝ ⎠

,

So 1 1,2 1 , 02 2

b c b c b c+ = + = ⇒ = = . So the desired sum is

( ) ( )1 21( ) 12

nS n n n+ ⎛ ⎞= − +⎜ ⎟⎝ ⎠

.

5. The Lucas sequence begins with 2, then 1, and then the remaining terms are found

as in the Fibonacci sequence. Write the recursive and closed forms for the terms of this sequence.

The recursive form for the Lucas sequence is simply

0 1 1 22, 1, , 2n n nt t t t t n− −= = = + ≥ , so the terms are 2, 1, 3, 4, 7, 11, 18, 29, 47, …. To find the closed formula, rewrite the recursive equation as 1 2 0n n nt t t− −− − =

and notice that this is exactly like the Fibonacci solution above, so

1 1 2 2 1 21 5 1 5

2 2

n n

n nnt c r c r c c

⎛ ⎞ ⎛ ⎞+ −= ⋅ + ⋅ = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠, but since the initial values are

different, we have 0 0

0 1 2 1 21 5 1 5 2

2 2t c c c c

⎛ ⎞ ⎛ ⎞+ −= + = + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ and

1 1

1 1 2 1 21 5 1 5 1

2 2t c c c c

⎛ ⎞ ⎛ ⎞+ −= + = + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

6. The first term of a sequence is 0 and the second is 1. For all

1 22, 3 2n n nn t t t− −≥ = ⋅ − ⋅ . Find the first 5 terms of the sequence, the recursive, and the closed form for the terms of this sequence.

The first five terms are 0, 1, 3, 7, and 15, so it looks like the function is simply

2 1.nnt = − Let’s use the method above to rewrite the recursive equation as

1 23 2 0n n nt t t− −− ⋅ + ⋅ = and proceed. We will have 1 23 2 0n n ncr cr cr− −− + = , so

( )2 2 3 2 0ncr r r− − + = . The solutions to the second factor are 1 21, 2r r= = . Now


4

we have ( ) ( )1 1 2 2 1 21 2n nn nnt c r c r c c= ⋅ + ⋅ = + . Now take the first two given values

to find the values for 1 2,c c . We get the system of equations:

( ) ( )( ) ( )

0 01 2 1 2 2 1

1 11 1 1 2 1 2

0 1 2

1 1 2 2 1 , 1

c c c c c c

c c c c c c

= + = + ⇒ = −

= − = − ⇒ − = ⇒ =.

This means then that ( ) ( )( )1 1 2 2 1n n n

nt = − − = − .

7. Sequence ( )1 2 3, , ,...a a a is defined recursively by 1 20, 100a a= = and

1 22 3n n na a a− −= − − . Find the greatest term in the sequence ( )1 2 3, , ,...a a a . {Duke Math Meet 1998 Team Problem #9)

Since 1 22 3n n na a a− −− + = − , this is a non-homogenous recurrence relation, so the above method does not work. However, finite difference will work. First, generate several values:

n 0 1 2 3 4 f(n) 0 100 197 291 382

1 100 97 94 91 2 -3 -3 -3

We see now that the function is quadratic, and in the form 23( )2

f n n bn c= − + + .

We also know that (0) 0 0f c= ⇒ = . Similarly we also know that

( )23(1) 100 1 (1) 101.5.2

f b b= = − + ⇒ = So we have 23( ) 101.52

f n n n= − + and

the maximum value of this function occurs when 101.5 33.832 3bxa

= − = = . Since

our function is a sequence (whose domain is the Whole Numbers), we need to check the values 33 and 34 to find the maximum. Since 34 is closer to the real-valued vertex, it yields 1717, which is the maximum value.

8. Find k given that (0)

and ( 50) 4000.( ) ( 1) 3 2

f kf

f n f n n=⎧

− =⎨ = + − −⎩

SMC 2003 Integer # 11

This problem can be done, among other ways, by using finite differences. First, it might help to rewrite the recursive equation as ( 1) ( ) 3 2f n f n n+ = + + . Now generate a table of the first several values.


5

x 0 1 2 3 4 y k k +

2 k + 7 k + 15 k + 26

1 2 5 8 11 2 3 3 3

We see now that the function is quadratic, and in the form 23( )2

f n n bn k= + + .

We also know that 23 1(1) 2 1 12 2

f k b k b= + = ⋅ + ⋅ + ⇒ = . Finally we are told that

( )23( 50) 4000 50 ( 50) 275.2

f k k− = = − + − + ⇒ =

Number Theory

Solutions:

1. The highest power of 6 that divides 40! The highest power of 6 that divides40! is also the lesser of the highest power of 2 and the highest power of 3that divides 40!. To find each, we count the number of factors of 40! that2 (3) divides and count multiplicity of 2’s (3’s) by counting the number offactors that powers of 2 (3) divide. b40

2 c+ b4022 c+ b40

23 c+ b4024 c+ b40

25 c =20+10+5+2+1 = 38 and b40

3 c+ b4032 c+ b40

33 c= 13+4+1 = 18, so thehighest power is 18.

2. Let n! be a number ending in exactly 200 zeros. The number of endingzeros is also the highest power of 5 that divides n!. This number of 5’s isbn

5c+ bn52 c+ b n

53 c+ b n54 c. Setting this sum equal to 200 gives smallest and

largest solution for n are 805.

3. The smallest positive integer with 30 factors must either have some primeraised to the 29th power, one raised to the 14th and another to the first, oneraised to the 9th and another to the 2nd, one to the fifth and another to thefourth, or one each to the fourth, second, and first. The smallest of eachform are 229,214 ·3,29 ·32,25 ·34, and 24 ·32 ·5 respectively the smallest ofthese is the last with a value of 720.

4. 25!+26! = 27(25!) = 33(25!), so we need to add 3 to the highest power of3 that divides 25!, which is b25

3 c+ b259 c+3 = 8+2+3 = 13.

5. To count factors, we first find a prime factorization, in this case 425 =253575. Each prime factor can by of any integer power between 0 and 5, giv-ing 6 possibilities for each factor and multiplying these possibilities gives63 = 216 factors.

6. n is the largest power of 2 that divides 30!, which is b302 c+ b

3022 c+ b30

23 c+b30

24 c= 15+7+3+1 = 26.

1

7. How many distinct factors of 2004 are there? 2004 = 22 · 31 · 1671. Eachprime factor can be raised to any power between 0 and its power in 2004,giving 3 ·2 ·2 = 12 factors. Repeating this for other years gives:

n factorization factors n factorization factors n factorization factors2010 2 ·3 ·5 ·67 16 2040 23 ·3 ·5 ·17 32 2070 2 ·32 ·5 ·23 242011 2011 2 2041 13 ·257 4 2071 19 ·37 ·109 82012 22 ·503 6 2042 2 ·1021 4 2072 23 ·7 ·37 162013 5 ·11 ·61 8 2043 32 ·227 6 2073 3 ·691 42014 2 ·19 ·53 8 2044 22 ·7 ·73 12 2074 2 ·17 ·61 82015 5 ·13 ·31 8 2045 5 ·409 4 2075 52 ·83 62016 25 ·32 ·7 36 2046 2 ·3 ·11 ·31 16 2076 22 ·3 ·173 122017 2017 2 2047 23 ·89 4 2077 31 ·67 42018 2 ·1009 4 2048 211 12 2078 2 ·1039 42019 3 ·673 4 2049 3 ·683 4 2079 33 ·7 ·11 162020 22 ·5 ·101 12 2050 2 ·52 ·41 12 2080 25 ·5 ·13 242021 43 ·47 4 2051 7 ·293 4 2081 2081 22022 2 ·3 ·337 8 2052 22 ·33 ·19 24 2082 2 ·3 ·347 82023 7 ·172 6 2053 2053 2 2083 2083 22024 23 ·11 ·23 16 2054 2 ·13 ·79 8 2084 22 ·521 62025 34 ·52 15 2055 3 ·5 ·137 8 2085 3 ·5 ·139 82026 2 ·1013 4 2056 23 ·257 8 2086 2 ·7 ·149 82027 2027 2 2057 112 ·17 6 2087 2087 22028 22 ·3 ·13 12 2058 2 ·3 ·73 16 2088 23 ·32 ·29 242029 2029 2 2059 29 ·71 4 2089 2089 22030 2 ·5 ·7 ·29 16 2060 22 ·5 ·103 12 2090 2 ·5 ·11 ·19 162031 3 ·577 4 2061 32 ·229 6 2091 3 ·17 ·41 82032 24 ·127 10 2062 2 ·1031 4 2092 22 ·523 62033 19 ·107 4 2063 2063 2 2093 7 ·13 ·23 82034 2 ·32 ·113 12 2064 24 ·3 ·43 20 2094 2 ·3 ·349 82035 5 ·11 ·37 8 2065 5 ·7 ·59 8 2095 5 ·419 42036 22 ·509 6 2066 2 ·1033 4 2096 24 ·131 102037 3 ·7 ·97 8 2067 3 ·13 ·53 8 2097 32 ·233 62038 2 ·1019 4 2068 22 ·11 ·47 12 2098 2 ·1049 42039 2039 2 2069 2069 2 2099 2099 2

8. The probability that a random factor of 4200 is odd is the same as the prob-ability that the power of 2 in the factor is 0. 4200 = 23 ·3 ·52 ·7. 0 is one offour possible powers of 2, giving a probability of 1

4 .

9. 360 = 23 ·32 ·5 giving 4 ·3 ·2 = 24 factors.

10. The largest power of 12 that divides 100! is the smaller of the power of 3and half the power of 2. The power of 2 is 50 + 25 + 12 + 6 + 3 + 1 = 97and the power of 3 is 33+11+3+1 = 48, so the largest power of 12 is 48.

2

1 Written and Compiled Developed by John Goebel, NCSSM Problem Solving Course, 2006

Partial Fractions Solutions

1. Find the sum of the first 100 terms of the series 1 1 1 1 11 2 2 3 3 4 4 5 ( 1)n n

+ + + +⋅ ⋅ ⋅ ⋅ ⋅ +

.

The general term of the series 1( 1)n n +∑ can be broken into the two terms

( 1) ( )1 ( 1)

A B A n B nn n n n

+ ++ =

+ +, so we see that ( ) 0 1A B n A n+ + = + , so 0, 1A B A+ = = , so

1B = − and the series can be rewritten as 1 1 1 1 1 1 1 1 1 11 2 2 3 3 4 99 100 100 101⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + − + + − + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

. Notice now that at

each partial sum after the first, the terms collapse, so that

21 1 1 1 1 1 21 2 2 3 1 3 3

S ⎛ ⎞ ⎛ ⎞= − + − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

and 31 1 31 4 4

S = − = , etc. So the sum of the first 100

terms, so 1001 1 1001 101 101

S = − = .

2. Find the sum of the first 50 terms of the series

1 1 1 1 11 3 3 5 5 7 7 9 (2 1) (2 1)n n

+ + + +⋅ ⋅ ⋅ ⋅ − ⋅ +

.

( ) ( )2 1 2 11(2 1) (2 1) 2 1 2 1 (2 1) (2 1)

A n B nA Bn n n n n n

+ + −= + =

− ⋅ + − + − ⋅ +, so we know that 0A B+ =

and 1A B− = , so 1 1,2 2

A B= = − . This means our series can be rewritten as

1 1 1 1 1 1 1 1 1 1 12 1 3 3 5 5 7 2 3 2 1 2 1 2 1n n n n⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + − − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥− − − +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

, so all of the

terms except the first and the last drop out in any finite partial sum. So the sum of the

first 50 terms is 1 1 1 100 502 1 101 202 101⎛ ⎞− = =⎜ ⎟⎝ ⎠

.

3. Find the sum of the series 1 1 1 11 2 3 2 3 4 3 4 5 98 99 100

+ + + +⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

. FU_SR2001#23

Note that the general term of this series is 1 , 2 99( 1) ( 1)

nn n n

≤ ≤− +

. Break each term

into its partial fractions as follows:


1 ( 1) ( 1)( 1) ( 1)( 1) ( 1) 1 1 ( 1) ( 1)

A B C An n B n n Cn nn n n n n n n n n

+ + − + + −= + + =

− + − + − +.

From this we see that 2 2( ) ( ) 0 0 1A B C n A C n B n n+ + + + − = + + , so 11,2

B A C= − = = .

So we can now rewrite the series as follows: 1 1 1 1 1 1 1 1 1 11 1 1 1 12 2 2 2 2 2 2 2 2 21 2 3 2 3 4 3 4 5 97 98 99 98 99 100

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− + + − + + − + + − + + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠or

1 1 1 1 1 1 1 1 1 1 1 1 1 1 12 2 6 4 3 8 6 4 10 194 98 198 196 99 200

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + − + + − + + + − + + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Bu

t if we rearrange these terms we see that: 1 1 1 1 1 1 1 1 1 1 1 12 4 6 8 194 196 2 3 4 5 98 99

1 1 1 1 1 16 8 10 12 198 200

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 122 4 6 8 10 12 194 196 198 200 2 3 4 5 98 991 1 1 1 1 1 1 12 4 3 4 5 6 97 98

⎛ ⎞ ⎛ ⎞+ + + + + + − + + + + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞+ + + + + + + =⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞+ + + + + + + + + + − + + + + + + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞+ + + + + + + +⎜ ⎟⎝ ⎠

1 1 1 1 1 1 1 1198 200 2 3 4 5 98 99

1 1 1 1 49494 198 200 99 19800

⎛ ⎞+ + − + + + + + + =⎜ ⎟⎝ ⎠

+ + − =

4. How many terms of the series ( )( )

1 1 1 1 12 3 3 4 4 5 5 6 1 2n n

+ + + +⋅ ⋅ ⋅ ⋅ + +

must be added

to be greater than 0.4999?

( )( ) ( )( )1 ( 2) ( 1)

1 2 1 2 1 2A B A n B n

n n n n n n+ + +

= + =+ + + + + +

, so we know that

0, 2 1A B A B+ = + = . So 1, 1A B= = − , and the terms of our series, when written in

partial fractions, look like 1 1 1 1 1 1 1 12 3 3 4 4 5 1 2n n

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − + − + + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠. At

any partial sum, all of the terms except the first and last drop out, so 1 12 2nS

n= −

+.

We want this to be greater than 0.4999, so 1 1 10.4999 0.00012 2 2nS

n= − > = −

+, so we

need 1 1 99992 10,000

nn

< ⇔ ≥+

, so it would take 9999 terms for the sum to exceed

0.4999.


5. Find the sum of the first 50 terms of the series ( )2

1 1 1 1 , 2.3 8 15 1

nn

+ + + + + ≥−


( ) 22

1 ( 1) ( 1) .1 1 11

A B A n B nn n nn

− + += + =

+ − −−, so we know that 0A B+ = and 1B A− = .

This gives us 1 1,2 2

B A= = − , so the nth partial sum of the series can be rewritten as

follows: 1 1 1 1 1 1 1 1 1 1 12 3 1 4 2 5 3 2 1 1n n n n⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + − + + − + + + − + + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥− + −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

. In

this series two terms near the front and two near the back remain, all others subtract out,

so the partial sum is 1 1 1 1 12 1 2 1n n⎡ ⎤+ − −⎢ ⎥+⎣ ⎦

. The sum of the first 50 terms would mean

51n = (since we are starting at 2) so this sum is 1 1 1 1 387512 2 51 52 5304⎡ ⎤+ − − =⎢ ⎥⎣ ⎦

. The limit

for the infinite series is three-fourths.

6. Find the sum of the first 50 terms of the series 2

1 1 1 1 , 2.6 11 20

nn n

+ + + + + ≥+


2 2

1 ( 1) , 2.1

A B An B n nn n n n n n

+ += + = ≥

+ + +, so 0, 1, 1A B B A+ = = ⇒ = − . The partial

sum whose last term is 2

1n n+

then is

1 1 1 1 1 1 1 1 1 1 1 13 2 4 3 5 4 1 1 2 1n n n n n

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + − + + − + + + − + + − + = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠, so the

sum of the first 50 terms (n = 51), would be 1 1 252 52 52− = . The limit of the infinite

series is one-half.

7. The series 22

1n n

∞

=∑ is sandwiched in between the two series in 5 and 6. Can you find the

sum of this infinite series by the methods of this section? Why or why not? The terms in this series do not telescope like the others, so this method does not help.

In fact, this is a special series which converges to 2

1 0.6456π

− ≈ , which is between the

two values in the previous two problems.

Perfect Squares

Solutions:

1. x2− y2 = 63⇔ (x+ y)(x− y) = 32 ·7. With x,y > 0, x+ y > x− y, so x+ yis one of 9, 21, or 63 with x− y = 7,3,1 respectively. There are then 3solutions.

2. 392 > 1500 > 382 and 232 > 500 > 222, so there are 16 perfect squaresbetween 500 and 1500.

3. If n2 + 2001 = k2, then k2− n2 = (k + n)(k− n) = 2001 = 3 · 23 · 29. Thisgives possible values for k−n of 1,3,23,29 which give n = 1000,332,32,20which sum to 1384.

4. If n2 +2004 = k2, then k2−n2 = (k +n)(k−n) = 2004 = 22 ·3 ·167. k−nand k +n are both even, so the possible values for k−n are 2,6 which given = 500,164 which sum to 664.

5. 1! · 2! · 3! · 4! · 5! · 6! · 7! · 8! · 9! = 2837465564738291 = 28+12+637+25573 =226395573 = (22)13(32)4(52)2(72) · 3 · 5 · 7. The power of (22) can be be-tween 0 and 13, of (32) between 0 and 4, of (52) between 0 and 2, of (72)between 0 and 1. This gives 14 ·5 ·3 ·2 = 420 perfect squares as divisors.

6. If n2− 19n + 99 = (n− k)2 = n2− 2nk + k2 ⇒ −19n + 99 = 2nk + k2 ⇒

k2−99 = n(2k−19)⇒ n =k2−992k−19

=2k +19

4− 35

4(2k−19). 2k−19 must

then divide 35, giving k =−8,6,7,9,10,12,13,27 and n = 1,9,10,18

7. If, n2− 2004n = m2, then n2−m2 = (n + m)(n−m) = 2004n is even, son−m is even. If n2−2004n = (n−2k)2 = n2−4nk+4k2⇒ n(4k−2004) =4k2

n =k2

k−501= k +501+

251001k−501

. 251001 = 32 ·1672, so for integer n,

k =−250500,−83166,−27388,−10020,334,492,498,500,502,504,510,668,1002,2004,28390,84168,251502. For the first nine k’s, this producesnon-positive n, for the other nine k’s, n = 252004,84672,28900,2672,2004,2672,28900,84672,252004, respectively, giving 5 distinct n.

1

Polynomials

Solutions:

1. (x− a)(x− b)(x− c) = x3− (a + b + c)x2 +(ab + ac + bc)x− abc, so 7 =a + b + c. If a and b are additive inverses, then c = 7. abc = −28, soa,b =−2,2, giving zeros of −2,2,7.

2. If p(x) = ax2 +bx+c leaves a remainder of 4 when divided by x, then c = 4.ax2+bx+c

x+1 = ax+(b−a)+ c−b+ax+1 , so c−b+a = 3. ax2+bx+c

x−1 = ax+(b+a)+c+b+a

x−1 , so c+b+a = 1. Combining the equations: c = 4, c−b+a = 3 andc+b+a = 1 gives b =−1, a =−2. Evaluating −2(2)2−1(2)+4 =−6.

3. If 0,1,√

2,1−√

3 are roots, then so are the conjugates −√

21 +√

3. Thisgives 6 roots, and so the polynomial is of degree at least 6.

4. x5−9x4 . . .+11 = 0. The roots’ product is (−1)511, and sum is 9. As theseare integers, one root is±11 and the others are±1. With a positive sum, the11 is positive and there are two more −1’s than 1’s so there are three −1’s,one 1 and one 11 as roots.

5. 3√

x9−3x8 +18x7−28x6 +84x5042x4 +98x3 +72x2 +15x+1 = p(x) =x3 +ax2 +bx+ c, sox9−3x8 +18x7−28x6 +84x5 +42x4 +98x3 +72x2 +15x+1 =(x3 +ax2 +bx+ c)3 = x9 +3ax8 + . . .+(3bc2)x+ c3.

c3 = 1 ⇒ c = 1,3bc2 = 3b(1)2 = 15 ⇒ b = 5,3a =−3 ⇒ a =−1.

These give p(x) = x3−x2 +5x+1 and the sum of the squares of the coeffi-cients is 12 +(−1)2 +52 +12 = 28.

6. p(x) = a(x+1)(x−2)(x−3), p(0) = 1 = a(6), so a = 16 and p(x) = 1

6(x+1)(x− 2)(x− 3). We can either multiply this out and then divide by x− 1,or we can use the fact that adding or subtracting a multiple of (x−1) doesnot change the remainder to subtract x− 1 from each of the factors to getthat the remainder of p(x)

x−1 is 16(2)(−1)(−2) = 4

6 = 23

1

7. If 2 is a solution of x3 +hx+10 = 0, then x−2 divides x3 +hx+10.(x−2)(x2 +ax+b) = x3 +(−2+a)x2 +(−2a+b)x−2b = x3 +hx+10This equation implies a = 2, b =−5 and h =−9.

8. 4x3−8x2 +5x−1 A quick inspection yields x = 1 as a root and so x−1 asa factor.4x3− 8x2 + 5x− 1 = (x− 1)(4x2− 4x + 1) = (x− 1)(2x− 1)2. There arethus 2 real solutions: x = 1, 1

2 .

9. The roots of x4− 5x2 + 6 can be found by first substituting y = x2 into theequation: y2−5y+6 = (y−2)(y−3) = (x2−2)(x2−3) Each y factor givesa positive and a negative solution for x, so the sum of the squares of the rootsis 2+2+3+3 = 10.

10. N4 +6N < 6N3 +N2, is equivalent to N4−6N3−N2 +6N < 0. This factorsinto (N +1)N(N−1)(N−6) < 0. For this to be true, an odd number of thefactors must be negative, so N is in (−1,0)∪ (1,6). The integers in theseintervals are 2,3,4, and 5.

11. Applying Descartes’ Rule of Sign Changes, f (x) has 2 sign changes, so noor 2 positive real roots. f (−x) has 1 sign change, so f (x) has 1 negativereal root. If x > 1, then x3− x2 and 2x + 4 are both positive. If 0 < x ≤ 1,then x3 +4 and 2x− x2 are both positive, so there are no positive real roots.There is only one root, so only f (x) crosses the x axis exactly 1 time.

12. p(x) = x18− 3x17 + . . . with roots 1,2,3, . . . ,17. The −3 coefficient of x17

is the negative sum of the roots, so if r is the unknown root, −3 = −r−∑

17i=1 i =−r−153. Solving for r yields r =−150.

13.

0 = x4 + x3−3x2 + x = x(x3 + x2−3x+1) = x(x−1)(x2 +2x−1) =x(x−1)(x− −2+

√8

2 )(x− −2−√

82 ) = x(x−1)(x− (−1+

√2))(x− (−1−

√2))

The largest x so that |x|< 12 is√

2−1.

14. (x+2)3 +(3x−6)3 = (4x−4)3, expanding these givesx3 +6x2 +12x+8+27x3−162x2 +324x−216 = 64x3−192x2 +192x−640 = 36x3− 36x2− 144x + 90 = 36(x3− x2− 4x + 5

2) The negative sum ofthe roots is the coefficient of xn−1, in this case -1, so the sum of the roots is1.

2

15. The real x which satisfies x3 + (x− 1)3 + (x− 2)3 + (x− 3)3 + (x− 4)3 +(x− 5)3 = 33 is x = 3. We need one term to survive to produce the 33 andthe others to cancel out, so we need small enough x for some terms to bepositive and some negative.

16. The values of x for which (x− 3)4 +(x− 5)4 = −8 are 4 more than thosefor (x+1)4 +(x−1)4 =−8 or (x+1)4 +8+(x−1)4 = 0.Expanding this yields: 2x4+12x2+2+8 = 2(x4+6x2+5) = 2(x2+5)(x2+1) = 0The solutions of this are ±i

√5,±i, so the solutions to the original equation

are 4± i√

5,4± i.

17. The equations (x−2)4− (x−2) = 0 and x2− kx+ k = 0 share two roots.(x−2)4− (x−2) = (x−2)((x−2)3−1) =(x− 2)(x− 3)((x− 2)2 + (x− 2) + 1) = (x− 2)(x− 3)(x2− 3x + 3). Thefinal factor is of the proper form, so k = 3.

18. Find the remainder that results when (x + 1)5 +(x + 2)4 +(x + 3)3 +(x +4)2 +(x+5) is divided by x+2. Taking the remainder of each factor beforeexponentiating yields the same final remainder, so the remainder is (−1)5 +04 +13 +22 +3 =−1+0+1+4+3 = 7

19. a,3a,5a,b,b + 3,b + 5 are only 4 distinct numbers and a < b implies a isunique. 3a,5a are then equal to two of b,b+3,b+5. Computing the possi-ble differences gives 2a = 2,3,5 and the respective values of a are 1,1.5,2.5.

20. P(x) is a polynomial of degree 1996, To construct P(n), we can use termsthat are 0 for all positive integers j 6= i≤ 1997 and for integer i are precisely1i . To do this, we include x− j as a factor in each term, giving us terms of

the form1i(x− j)(i− j)

·1997

∏j=i+1

(x− j)(i− j)

. Then

P(n) =1997

∑i=1

1i(x− j)(i− j)

·1997

∏j=i+1

(x− j)(i− j)

and P(1998) =1997

∑i=1

1i(1998− j)

(i− j)·

1997

∏j=i+1

(1998− j)(i− j)

This simplifies to1997

∑i=1

1i

1997!(1998− i)

1(i−1)!(1997− i)!(−1)1997−i

Simplifying further gives1

1998

1997

∑i=1

(1998

i

)(−1)i−1.

Using the combinatorial identity thatn

∑i=0

(ni

)(−1)i = 0 gives that

P(1998)= 0− 11998

((−1)0−1

(1998

0

)+(−1)1998−1

(1998

0

))=

21998

=1

999

3

21. By the quadratic formula, r,s =√

5±12 ,

r8+s8 =(1+√

5)8 +(√

5−1)8

256=

2(54 +28 ·53 +70 ·52 +28 ·5+1)256

= 47

22. 2x3−9x2 +12x− k has a double root for two values ofk.(x−a)2(x−b) = x3− (2a+b)x2 +(a2 +2ab)x−a2b = x3−4.5x2 +6x− k

2This yields the equations:

2a+b = 4.5a2 +2ab = 6k = 2a2b

Solving the first equation for b gives b = 4.5− 2a and substituting intothe second gives a2 + 9a− 4a2 = 6 or 3a2− 9a + 6 = 3(a2− 3a + 2) =3(a−1)(a−2) = 0. a = 1,2 gives k = 5,12 respectively.

23. Determine the sum of the y-coordinates of the points of intersection ofy = x4− 5x2− x + 4 and y = x2− 3x. 0 = x4− 6x2 + 2x + 4. Let a,b,c,dbe the roots of this polynomial, that is, the x-coordinates of the points ofintersection. Then the coefficient of x3, which is 0, is −a− b− c− d andthe coefficient of x2 is the sum ab+ac+ad +bc+bd +cd. The sum of they-coordinates is a2−3a+b2−3b+ c2−3c+d2−3d = (a+b+ c+d)2−2(ab+ac+ad +bc+bd +cd)−3(a+b+c+d) = 02−2 ·−6−3 ·0 = 12.

24. Set the equations equal: x2− x− 5 = 1x . The x-coordinates then satisfy

x3− x2−5x−1 = 0. Let a,b,c be the roots of this polynomial, that is, thex-coordinates of the points of intersection. Then the coefficient of x, whichis -5, is bc + ac + ab, and the constant term, -1, is −abc. The sum of they-coordinates is 1

a + 1b + 1

c = bc+ac+ababc = −5

1 =−5.

4


Geometry Facts – Circles & Cyclic Quadrilaterals

Solutions:

1. We want sin(2 ).θ

( )

( )( ) sin

x rx R r R

r R r r R rxR r x R r

θ

= ⇒+ =

+ −= ⇒ = =

− +

2

2

4( )sin(2 ) 2sin( ) cos( ) 2 1( )

R r R r R r RrR r R r R r

θ θ θ − − −⎛ ⎞ ⎛ ⎞= = − =⎜ ⎟ ⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠.

Of course, using numbers is easier. For R = 4, r = 1, 24sin(2 ) 25θ = . Since the

angles can’t be equal for these two soluti0ons, they must be complementary. Placing a 4” and a 9” circle in line with and tangent to opposite sides of a 1” circle, then drawing the appropriate external tangents, would produce a quadrilateral that is cyclic. Potential there for interesting puzzles?

2. Two angles must be supplementary, but

they cannot be opposite. Thus the are consecutive, and the quadrilateral is a trapezoid, but not isosceles. Sliding triangles I and II together produces a 3-4-5 triangle , or area 6, with altitude to hypotenuse equal to 12/5 This makes the inner rectangle area equal to 2(1 / 5) = 36 / 5, for a total area of 6 + (36 / 5) = 66 / 5.

3. Remembering that Ptolemy’s Theorem says that ac bd pq+ = (where a,b,c, and d

are the lengths of the sides and p and q the diagonals), and using the inequalities ( )

( )( ) ( )

2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

22 2 2 2 2

0 ( ) 2( )( ) ( ) 0 ( ) ( ) 2( )( )

2 2

( ) .

ad bc ad ad bc bc ad bc ad bc

a d b c abcd a d b c b c b d b c abcd b d

a b c d ac bd pq

+ ≥ ⇒ + + ≥ ⇒ + ≥

⇒ + ≥ ⇒ + + + ≥ + +

⇒ + + ≥ + =

Since all of the quantities are positive, we have ( )( )2 2 2 2pq a b c d≤ + +

4. I. A. (7,24,25) and (15,20,25).


B. (35,120,125), (75,100,125) and (44,117,125). The first two come from the (7,24,25) and (3,4,5) triangles. A general approach would be to use the fact that

2 2 2 2( ), (2 ), and ( )k m n k mn k m n− + , where k, m, n, are positive integers with m > n. produce all Pythagorean triplets. Setting 2 2 125m n+ = leads to m = 10, n = 5, [producing (75,100,125)] or m = 11, n = 2 [producing (44,117,125); setting

2 2 25m n+ = , leads to m = 4, n = 3 [producing 7,24,25) for which we use k = 5]. II. A. Ptolemy’s Theorem produces 7(5) (3)(3) (4)(4) 5FG FG+ = ⇒ = . The

answer is (17,7,15,25) [in any order]. B. Using sides a,x,a,25 (each diagonal is 2625 a− ) leads

2225 25ax = − .

Positive integral x’s come from a = 5, 19, or 15. The last leads to the solution in A. The other possibilities are (5,23,5,25) OR (10,17,10,25) [either answer in any order, is acceptable.]

C. Apply the theorem given in Part IA, for any appropriate d, to find n - 2 P3’s of

hypotenuse 2nd − . Build these triangles in a semicircle of diameter 2nd − . Connecting successive points on the semicircle produces a n-gon. Successive applications of Ptolemy’s Theorem to find each side of the n-gon shows each must be rational. Multiplying all sides by the least common denominator involved produces a Pn.

5. The sum of angles CDE, DCE, BAG, and ABG is

12 (360) 180o= , so angles E and G must be supplementary. Thus quadrilateral EFGH is cyclic. Let , , and .BAG a ABG b BCF c= = = Then

180a b G+ = − and 180b c F+ = − . Therefore (180 ) 180a c F G F E E F− = − = − − = + − , so

_ 2[ 180 ] 26o oA C E F= + − = In general, 2( ) 360A C E F− = + − and 2( )B D E F− = − .

6. If we arrange the sides of the hexagon, keeping the same outer segments of the

original circle attached, we must again produce a circle (of the same size)! Incidentally, the area of the hexagon will be preserved also. Arranging the sides in the order 2,7,11,2,7,11, shows that each triplet must fit in a semicircle. Applying Ptolemy’s Theorem to the quadrilateral, (in a semicircle) of sides 2,7,11,D (for the diameter), we have ( )( )2 2121 4 7 22D D D− − = + . This

implies 3 174 308 0,D D− − = or 3 (2 3 29) (4 7 11) 0D D− ⋅ ⋅ − ⋅ ⋅ = . We note that 11 20D< < ; also, if D is not irrational, it is integral. Hoping to find an integral D, we see that D would be divisible by 2 but not 3 or 4 (else 8 would divide the constant). The only possible integral D is then 14, which works.


7. Let the quadrilateral be ABCD. We will

prove that angle A, chosen randomly, is “nice.” Let AB = a, BC = b, CD = c and DA = d. Since ABCD is inscriptable, angles A and C are supplementary, so cos( ) cos( )C A= − . Now

( )

2 2 2

2 2 2 2

cos( )2

2 cos( )2

a d eAad

a d b c bc Aad

+ −=

+ − + −=

Clearing fractions and solving for cos(A), we

get ( ) ( )

( )

2 2 2 2

cos( )2

a d b cA

ad bc+ − +

=+

, which is

rational, since the sides are integers. Since the quadrilateral can be circumscribed about a circle, it must be true that a c b d+ = + . [This well-known fact can be proved by considering the lengths of the tangent segments to the circle from each vertex of ABCD.] This its semiperimeter s is equal to a+c (or b+d). Now Heros (actually Brahamgupta’s) formula for the area of the inscribed quadrilateral is

( )( )( )( )K s a s b s c s d= − − − − . Now we can replace s with either a+c or b+d, getting a wonderful formula for a quadrilateral that is both inscriptable and circumscriptable, K abcd= . Since we are given that abcd is a perfect square, our area is an integer. Finally noting that sin( ) sin( )C A= , we see that the area K

of ABCD is also given by ( )1 sin( )2

K ad bc A= + , so 2sin( ) KAad bc

=+

; since K is

an integer, sin(A) is rational. Thus angle A is “nice”, and our quadrilateral is “nice.” Incidentally, the quadrilateral with the smallest perimeter with these properties has sides 1,2,9,8 (in that order). For those interested, it can be proven that the only regular polygon that is “nice” is the square.


Triangles

Solutions: 1. Two sides of a triangle are 7 and 9 while the median to the third side has length 5.

Find the length of the third side.

The median theorem tells us that 2 2 2

2 4ca b cm +

= − , so

( )2 2 2 2

27 9 49 815 25 100 2 1302 4 2 4

c c c+ += − ⇒ = − ⇔ = − , so

2 160 4 10c c= ⇒ = 2. The sides of a triangle are consecutive integers and the area is an integer. Find

the triangle with the smallest perimeter that is not a right triangle. Are there others?

Let the sides be 1, , 1b b b− + , so the area is

3 3 3 3 3 2 2( 1) ( ) ( 1)2 2 2 2 2 2 2 2b b b b b b b bK b b b + −⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞= − − − − + =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠

. This

simplifies to ( )( )3 2 24b b b− + , so it looks like b has to be a multiple of 4 or one

of 2b − or 2b + must be a perfect square or 3 times a perfect square. If 4b = , then the area is an integer, (6), but the triangle is a 3-4-5 right triangle. The next value that works, is 14b = with the area

( )( ) ( )( ) ( )( )14 14 143 14 2 14 2 3 12 16 6 4 844 4 4

− + = = = . There are others. The

next two are 193,194,195 and 2710,2702,2703 with area 16296 and 316340. 3. Triangle ABC is reflected in (or about) its median AM (extended) as shown. If

6, 12, 10AE EC BD= = = and 3AB k= , compute k. 1987 ARML I8


The median AM divides triangle ABC into two triangles, ABM and CAM which both have the same area. Since B’AM is the reflection of BAM, and C’AM is the reflection of CAM, all of these triangles have the same area. Now look at triangles BAD and C’DM. They have the same area as well (take triangle ADM off of BAM and C’AM). the areas can be found using the formula

( )1 1sin( ) ' sin '2 2

BD DA ADB C D DM MDC⋅ ⋅ ∠ = ⋅ ⋅ ∠ , but since the angles are

equal, we have it has the same area 'BD DA C D DM⋅ ⋅ = ⋅ , so 6 10 12 5AD AD⋅ = ⋅ ⇒ = . So EM = 5 also. This makes CM = 15. Now the law

of cosines gives ( )2 2 212 15 5 344 43cos2(12)(15) 360 45

C + −∠ = = = . Now in triangle ABC,

2 2 2 4330 18 2(30)(18) 900 324 2(2)(6)43 19245

AB = + − = + − = , so

192 8 3AB = = . 4. Right triangle ABC (hypotenuse AB ) is inscribed in equilateral triangle PQR, as

shown. If 3PC = , and 2,BP CQ= = compute AQ. ARML 1991 I7

Using the law of cosines we can find BC, AC, and AB and then use the

Pythagorean Theorem to relate these. First let , 5AQ x AR x= = − and note that

3BR = . So, ( )2 2 2 12 3 2(2)(3) 4 9 6 72BC = + − = + − = ,

( )2 2 2 212 2(2)( ) 2 42AC x x x x= + − = − + , and

( )2 2 2 213 (5 ) 2(3)(5 ) 7 192AB x x x x= + − − − = − + . So now we have

( ) ( )2 27 19 2 4 7x x x x− + = − + + , so 8 35 8 15 5x x= ⇒ = = .

5. In triangle ABC, the perpendicular bisector of AC intersects AC at M and AB

at T. If the area of triangle AMT is 14

the area of triangle ABC, and o128A C+ = , compute the number of degrees in angle A. ARML 1988 I4


Since the area of AMT is one-fourth the area of ABC, the altitude MT must be half the altitude BP. Now, since MT is half of BP, AT = TB and AM = MP, but we know that AM = MC, so C must coincide with P, making triangle ABC a right triangle and since 128 , 38o om A m C m A∠ + ∠ = ∠ =

6. An isosceles triangle has a median equal to 15 and an altitude equal to 24. This

information determines exactly two triangles. Compute the area of each of these triangles. ARML 1994 T4

The two triangles are shown above, with BM the median with length 15 in the

first and AF the altitude with length 24. In the first triangle, we know that

( ) ( ) ( )2 2 22 22 2 2 115 4 8

2 4 2x y x

x y+

= − = + and ( )2 2 22 24x y= + , so

2 2900 4 8x y= + and 2 2576 4x y= − , which makes 2 2 2 2900 8 576 9 324 36 6y y y y y− = + ⇔ = ⇒ = ⇒ = making the area 144. In

the second triangle we have 2 2 215w z= + and 2 524 3015 24 4w z w z w z= ⇒ = ⇒ = ,


so 2

2 2 2 2 2 2 25 25 915 15 15 204 16 16

z z z z z z⎛ ⎞ = + ⇒ = + ⇒ = ⇒ =⎜ ⎟⎝ ⎠

and 25w = ,

which makes 7AF = . Thus the area of this triangle is 300. 7. Point P is inside ABC . Line segments

, , and APD BPE CPE are drawn with D on BC , E on CA , and F on AB . (See figure). Given that 6,AP = 9,BP =

6,PD = 3PE = , and 20CF = , find the area of ABC . AIME 1989 #15

8. The points (0,0), (a,11), and (b,37) are the vertices of an equilateral triangle. Find the value of ab. AIME 1994 # 8

9. A direct proof establishes that equal sides implies equal medians. If AB BC≅ , then AN CM≅ , and since MAC NCA∠ ≅ ∠ and AC AC≅ , we have

MAC NCA≅ , so AM CN≅ . To prove that equal medians implies equal sides, we can’t prove congruent triangles, so we will take an indirect approach. So we will assume the medians are equal, but the sides are not. So assume mAM mCN= , then

( ) ( )2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 22 22 4 2 4 2 4 2 4

b c a b a c b c a b a c b c a b a c+ + + +− = − ⇒ − = − ⇒ + − = + −

, so 2 2 2 2 2 2 2 22 2 2 2 3 3b c a b a c a c a c+ − = + − ⇒ = ⇒ = . This is really not an indirect proof, is it?

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