The Canadian Mathematical Society

in collaboration with

The CENTRE for EDUCATION

in MATHEMATICS and COMPUTING

presents the

Sun Life Financial

Canadian Open Mathematics Challenge
Wednesday, November 24, 2010

Solutions

©2010 Canadian Mathematical Society

2010 COMC Solutions Page 2

Part A

1. Solution 1
(9 + 5)2 − (9 − 5)2 142 − 42 196 − 16 180
Calculating, = = = = 4.
(9)(5) 45 45 45

Solution 2
For a general x and y with x and y not zero,
(x + y)2 − (x − y)2 (x2 + 2xy + y 2 ) − (x2 − 2xy + y 2 ) 4xy
= = =4
xy xy xy
(9 + 5)2 − (9 − 5)2
Since this expression equals 4 for any values of x and y, then = 4.
(9)(5)

Solution 3
For a general x and y with x and y not zero, we can factor as a difference of squares:
(x + y)2 − (x − y)2 [(x + y) + (x − y)][(x + y) − (x − y)] (2x)(2y)
= = =4
xy xy xy
(9 + 5)2 − (9 − 5)2
Since this expression equals 4 for any values of x and y, then = 4.
(9)(5)

Answer: 4

2. Simplifying both sides,

x − (8 − x) = 8 − (x − 8)
x−8+x = 8−x+8
3x = 24
x = 8

Therefore, x = 8.

Answer: x = 8

3. Solution 1
We call the ring between the middle and inner circles the “inner ring”.
We reflect the shaded portion of the inner ring across line segment CD. The area of the shaded
region does not change when we do this.
The shaded region is now the entire semi-circle to the right of CD.
Thus, the area of the shaded region is half of the area of the outer circle.
Since OC = 6, then the outer circle has radius 6 and so has area π62 = 36π.
Therefore, the area of the shaded region is 21 (36π) = 18π.
2010 COMC Solutions Page 3

Solution 2
We call the ring between the outer and middle circles the “outer ring”, and the ring between
the middle and inner circles the “inner ring”.
Since OC = 6, then the outer circle has radius 6 and so has area π62 = 36π.
Since OB = 4, then the middle circle has radius 4 and so has area π42 = 16π.
Since OA = 2, then the inner circle has radius 2 and so has area π22 = 4π.
Since the outer circle has area 36π and the middle circle has area 16π, then the area of the
outer ring is 36π − 16π = 20π.
Since the diameter CD divides each ring into two parts of equal area, then the shaded region
of the outer ring has area 21 (20π) = 10π.
Since the middle circle has area 16π and the inner circle has area 4π, then the area of the inner
ring is 16π − 4π = 12π.
Since the diameter CD divides each ring into two parts of equal area, then the shaded region
of the inner ring has area 12 (12π) = 6π.
Since the inner circle has area 4π and line segment CD passes through the centre of this circle,
then the shaded region of the inner circle has area 12 (4π) = 2π.
Therefore, the total shaded area is 10π + 6π + 2π = 18π.

Answer: 18π

4. First, we simplify the given expression:

(3.1 × 107 )(8 × 108 ) 3.1 × 8 107 × 108

3
= × 3
= 3.1 × 4 × 107+8−3 = 12.4 × 1012 = 124 × 1011
2 × 10 2 10
Therefore, this integer consists of the digits 124 followed by 11 zeroes, so has 14 digits.

Answer: 14

5. Solution 1
Let Q be the point on the line y = x that is closest to P (−3, 9).
Then P Q is perpendicular to the line y = x.
Since the line with equation y = x has slope 1 and P Q is perpendicular to this line, then P Q
has slope −1.
Note that a general point Q on the line with equation y = x has coordinates (t, t) for some real
number t.
t−9
For the slope of P Q to equal −1, we must have = −1 or t − 9 = −(t + 3) or 2t = 6
t − (−3)
or t = 3.
Therefore, the point on the line with equation y = x that is closest to P is the point (3, 3).
2010 COMC Solutions Page 4

Solution 2
Let Q be the point on the line y = x that is closest to P (−3, 9).
Then P Q is perpendicular to the line y = x.
Let R be the point on the line y = x so that P R is horizontal, as shown.

P R

Q
x

Since P R is horizontal and P has y-coordinate 9, then R has y-coordinate 9.

Since R lies on the line with equation y = x, then R has coordinates (9, 9).
Since P R is horizontal and QR has slope 1 (because it lies along the line with equation y = x),
then ∠P RQ = 45◦ .
Since ∠P QR = 90◦ , then 4P QR is isosceles and right-angled.
Let M be the midpoint of P R. Since P has coordinates (−3, 9) and R has coordinates (9, 9),
then M has coordinates (3, 9).
Since 4P QR is isosceles and M is the midpoint of P R, then QM is perpendicular to P R.
Thus, QM is vertical, so Q has x-coordinate 3.
Since Q lies on the line with equation y = x, then Q has coordinates (3, 3).

Solution 3
Note that a general point Q on the line with equation y = x has coordinates (t, t) for some real
number t.
p
Then P Q = (t − (−3))2 + (t − 9)2 or P Q2 = (t + 3)2 + (t − 9)2 .
Since we want to find the point on the line with equation y = x that is closest to P , then we
want to minimize the value of P Q, or equivalently to minimize the value of P Q2 .
In other words, we want to find the value of t that minimizes the value of

P Q2 = t2 + 6t + 9 + t2 − 18t + 81 = 2t2 − 12t + 90

Since this equation represents a parabola opening upwards, then its minimum occurs at its
−12
vertex, which occurs at t = − 2(2) = 3. Thus, t = 3 minimizes the length of P Q.
Therefore, the point on the line with equation y = x that is closest to P is the point (3, 3).

Answer: (3, 3)
2010 COMC Solutions Page 5

6. Let x be the number of people who studied for the exam and let y be the number of people
who did not study.
We assume without loss of generality that the exam was out of 100 marks.
Since the average of those who studied was 90%, then those who studied obtained a total of
90x marks.
Since the average of those who did not study was 40%, then those who did not study obtained
a total of 40y marks.
90x + 40y
Since the overall average was 85%, then = 85.
x+y
Therefore, 90x + 40y = 85x + 85y or 5x = 45y or x = 9y.
Therefore, x : y = 9 : 1 = 90 : 10. This means that 10% of the class did not study for the exam.

Answer: 10%

7. Solution 1
Since ABCD is a rectangle, then AD = BC = 10 and DC = AB = 20.
Since W A = 12, W B = 16, AB = 20, and 122 + 162 = 144 + 256 = 400 = 202 , then
W A2 + W B 2 = AB 2 . Thus, 4AW B is right-angled at W .
Note that 4CKD is congruent to 4AW B, so 4CKD is right-angled at K.
Extend W A and KD to meet at Y and W B and KC to meet at Z.

W
16
12
A B
20 Z
10
Y
D C

Suppose that ∠W AB = ∠KCD = θ. Then ∠W BA = ∠KDC = 90◦ − θ.

Now ∠Y AD = 180◦ − ∠W AB − ∠BAD = 180◦ − θ − 90◦ = 90◦ − θ.
Also, ∠Y DA = 180◦ − ∠KDC − ∠ADC = 180◦ − (90◦ − θ) − 90◦ = θ.
Therefore, 4Y DA is similar to 4W AB. This means that ∠DY A = 90◦ .
Also, since DA = 21 AB, then the sides of 4Y DA are half as long as the corresponding sides of
4W AB. Thus, Y D = 12 W A = 6 and Y A = 12 W B = 8.
Similarly, ∠BZC = 90◦ . Therefore, W Y KZ is a rectangle.
We have W Y = W A + AY = 12 + 8 = 20 and Y K = Y D + DK = 6 + 16 = 22.
2010 COMC Solutions Page 6

By the Pythagorean Theorem, since W K > 0, then

√ √ √ √ √
W K = W Y 2 + Y K 2 = 202 + 222 = 400 + 484 = 884 = 2 221

as required.

Solution 2
Since ABCD is a rectangle, then AD = BC = 10 and DC = AB = 20.
We coordinatize the diagram, putting D at the origin, A at (0, 10), C at (20, 0), and B at
(20, 10).
Since W A = 12, W B = 16, AB = 20, and 122 + 162 = 144 + 256 = 400 = 202 , then
W A2 + W B 2 = AB 2 . Thus, 4AW B is right-angled at W .
Note that 4CKD is congruent to 4AW B, so 4CKD is right-angled at K.
Since 4AW B is right-angled and we know its side lengths, then we can compute the trigono-
metric ratios of its angles.
WB 16 4 WA 12 3
In particular, sin(∠W AB) = = = and cos(∠W AB) = = = .
AB 20 5 AB 20 5
We drop perpendiculars from W to X on AB and from K to Y on DC.

W
12 16

A (0, 10) B (20, 10)

X 20
10
Y
D
x
C (20, 0)

Then AX = W A cos(∠W AB) = 12( 35 ) = 36 5

and W X = W A sin(∠W AB) = 12( 45 ) = 48 5
.
Since A has coordinates (0, 10), then W has coordinates ( 36 5
, 10 + 485
) = ( 36 , 98 ).
5 5
Since 4CKD is congruent to 4AW B, then in a similar way we can find that the coordinates
of K are (20 − 36
5
, − 48
5
) = ( 64
5
, − 48
5
).
Since we have the coordinates of W and K, then the distance between W and K is
q q
282 2
WK = ( 64
5
− 36 2
5
) + (− 48
5
− 98 2
5
) = 52
+ 146
52
√ √
= 25 142 + 732 = 52 196 + 5329
√ q
= 5 5525 = 2 5525
2
25
√
= 2 221
2010 COMC Solutions Page 7

√
Therefore, W K = 2 221.
√
Answer: W K = 2 221

8. Solution 1
First, we factor the first and third quadratic factors to obtain

(x + 1)(x + 2)(x2 − 2x − 1)(x − 3)(x − 4) + 24 = 0

Next, we rearrange the factors to obtain

(x + 1)(x − 3)(x2 − 2x − 1)(x + 2)(x − 4) + 24 = 0

and expand to obtain

(x2 − 2x − 3)(x2 − 2x − 1)(x2 − 2x − 8) + 24 = 0

Next, we make the substitution w = x2 − 2x to obtain

(w − 3)(w − 1)(w − 8) + 24 = 0

This is a cubic equation in w so we expand, simplify and factor:

(w2 − 4w + 3)(w − 8) + 24 = 0
w3 − 12w2 + 35w = 0
w(w2 − 12w + 35) = 0
w(w − 5)(w − 7) = 0

Therefore, the solutions in terms of w are w = 0 or w = 5 or w = 7.

If w = x2 − 2x = 0, then x(x − 2) = 0 which gives x = 0 or x = 2.
If w = x2 − 2x = 5, then x2 − 2x − 5 = 0. √
2 ± 24 √
The quadratic formula gives the roots x = = 1 ± 6.
2
If w = x2 − 2x = 7, then x2 − 2x − 7 = 0. √
2 ± 32 √
The quadratic formula gives the roots x = = 1 ± 8.
√ 2 √
Therefore, x = 0 or x = 2 or x = 1 ± 6 or x = 1 ± 8. (This last pair can be rewritten as
√
x = 1 ± 2 2.)

Solution 2
First, we factor the first and third quadratic factors and complete the square in the second
quadratic factor to obtain

(x + 1)(x + 2)((x − 1)2 − 2)(x − 3)(x − 4) + 24 = 0

2010 COMC Solutions Page 8

Next, we make the substitution y = x − 1 (which makes x + 1 = y + 2 and x + 2 = y + 3 and

x − 3 = y − 2 and x − 4 = y − 3) to obtain

(y + 2)(y + 3)(y 2 − 2)(y − 2)(y − 3) + 24 = 0

(We made this substitution because it made the algebra more “symmetric”; that is, after making
this substitution, factors of the form y − a are paired with factors of the form y + a.)
Next, we rearrange the factors to obtain

(y + 2)(y − 2)(y + 3)(y − 3)(y 2 − 2) + 24 = 0

Next, we multiply out pairs of factors to obtain

(y 2 − 4)(y 2 − 9)(y 2 − 2) + 24 = 0

Next, we make the substitution z = y 2 to obtain

(z − 4)(z − 9)(z − 2) + 24 = 0

This is a cubic equation in z so we expand and simplify:

(z 2 − 13z + 36)(z − 2) + 24 = 0
z 3 − 15z 2 + 62z − 48 = 0

By inspection, z = 1 is a solution so z − 1 is a factor of the cubic equation.

We factor out this linear factor to obtain (z − 1)(z 2 − 14z + 48) = 0.
The quadratic factor can be factored as (z − 6)(z − 8).
Therefore, we have (z − 1)(z − 6)(z − 8) = 0.
Therefore, the solutions in terms of z are z = 1 or z = 6 or z = 8.
√ √
Since z = y 2 , then the solutions in terms of y are y = ±1 or y = ± 6 or y = ± 8.
Since y = x − 1, then x = y + 1, and so the solutions in terms of x are x = 0 or x = 2 or
√ √ √
x = 1 ± 6 or x = 1 ± 8. (This last pair can be rewritten as x = 1 ± 2 2.)
√ √
Answer: x = 0, 2, 1 ± 6, 1 ± 2 2
2010 COMC Solutions Page 9

Part B

1. (a) Solution 1
From the first row, A + A = 50 or A = 25.
From the second column, A + C = 57. Since A = 25, then C = 57 − 25 = 32.

Solution 2
From the first row, A + A = 50 or A = 25.
From the first column, A + B = 37. Since A = 25, then B = 37 − 25 = 12.
From the second row, B + C = 44. Since B = 12, then C = 44 − 12 = 32.

(b) Solution 1
The sum of the nine entries in the table equals the sum of the column sums, or 50+n+40 =
90 + n. (This is because each entry in the table is part of exactly one column sum.)
Similarly, the sum of the nine entries in the table also equals the sum of the row sums, or
30 + 55 + 50 = 135.
Therefore, 90 + n = 135 or n = 45.

Solution 2
The sum of the nine entries in the table equals the sum of the row sums, or 30 + 55 + 50 =
135. (This is because each entry in the table is part of exactly one row sum.)
Since the entries in the table include three entries equal to each of D, E and F , then the
sum of the entries in the table is also 3D + 3E + 3F = 3(D + E + F ).
Therefore, 3(D + E + F ) = 135 or D + E + F = 45.
From the second column, D + E + F = n. Thus, n = 45.

Solution 3
From the first row, D + D + D = 30 or D = 10.
From the first column, D + 2F = 50. Since D = 10, then 2F = 50 − 10 and so F = 20.
From the third column, D + 2E = 40. Since D = 10, then 2E = 40 − 10 and so E = 15.
Therefore, n = D + E + F = 10 + 15 + 20 = 45.

(c) Solution 1
From the third row, 3R + T = 33.
From the fourth row, R + 3T = 19.
Adding these equations, we obtain 4R + 4T = 52 or R + T = 13.
From the first row, P + Q + R + T = 20.
Since R + T = 13, then P + Q = 20 − 13 = 7.
2010 COMC Solutions Page 10

Solution 2
The sum of the sixteen entries in the table equals the sum of the row sums, or 20 + 20 +
33 + 19 = 92. (This is because each entry in the table is part of exactly one row sum.)
The table includes two entries equal to each of P and Q and six entries equal to each of
R and T .
Therefore, 2P + 2Q + 6R + 6T = 92.
The last two rows of the table include four entries equal to each of R and T , so 4R + 4T =
33 + 19 = 52, or R + T = 13.
Therefore, 2P + 2Q = 92 − 6(R + T ) = 92 − 6(13) = 14, and so P + Q = 7.

Solution 3
From the third row, 3R + T = 33.
From the fourth row, R + 3T = 19.
Multiplying the first equation by 3 and subtracting the second equation gives
(9R + 3T ) − (R + 3T ) = 99 − 19 or 8R = 80 or R = 10.
Since 3R + T = 33, then T = 33 − 3(10) = 3.
From the first row, P + Q + R + T = 20.
Since R = 10 and T = 3, then P + Q = 20 − 10 − 3 = 7.

2. (a) To determine the coordinates of A and B, we equate values of y using the equations
y = x2 − 4x + 12 and y = −2x + 20 to obtain

x2 − 4x + 12 = −2x + 20
x2 − 2x − 8 = 0
(x − 4)(x + 2) = 0

Therefore, x = 4 or x = −2.
To determine the y-coordinates of points A and B, we can use the equation of the line.
If x = 4, then y = −2(4) + 20 = 12.
If x = −2, then y = −2(−2) + 20 = 24.
Therefore, the coordinates of A and B are (4, 12) and (−2, 24).

(b) Using the coordinates of A and B from (a), the coordinates of the midpoint M of AB are
( 21 (4 + (−2)), 12 (24 + 12)) or (1, 18).

(c) Solution 1
The line with equation y = −2x + 20 has slope −2.
Therefore, we have a line with slope −2 that intersects the parabola at points
2010 COMC Solutions Page 11

P (p, p2 − 4p + 12) and Q(q, q 2 − 4q + 12).

In other words, line segment P Q has slope −2.
Therefore,

(p2 − 4p + 12) − (q 2 − 4q + 12)

= −2
p−q
p2 − q 2 − 4p + 4q
= −2
p−q
(p − q)(p + q) − 4(p − q)
= −2
p−q
(p + q) − 4 = −2 (since p 6= q)
p+q = 2

Therefore, p + q = 2, as required.

Solution 2
The line with equation y = −2x + 20 has slope −2.
Therefore, we have a line with slope −2 (say with equation y = −2x + b) that intersects
the parabola at points P and Q.
Since y = −2x + b and y = x2 − 4x + 12 intersect when x = p, then p2 − 4p + 12 = −2p + b,
which gives p2 − 2p + 12 − b = 0.
Since y = −2x + b and y = x2 − 4x + 12 intersect when x = q, then q 2 − 4q + 12 = −2q + b,
which gives q 2 − 2q + 12 − b = 0.
Since we have two expressions equal to 0, then

p2 − 2p + 12 − b = q 2 − 2q + 12 − b
p2 − 2p = q 2 − 2q
p2 − q 2 − 2p + 2q = 0
(p − q)(p + q) − 2(p − q) = 0
(p − q)(p + q − 2) = 0

Therefore, p − q = 0 or p + q − 2 = 0.
Since p 6= q, then p + q = 2.

(d) Since P has coordinates (p, p2 − 4p + 12) and Q has coordinates (q, q 2 − 4q + 12), then the
x-coordinate of the midpoint N of P Q is 21 (p + q).
Since p + q = 2 by (c), then the x-coordinate of N is 1.
Since the x-coordinate of M is 1 and the x-coordinate of N is 1, then line segment M N
is vertical.
2010 COMC Solutions Page 12

3. (a) Solution 1
Let U be the point in S vertically above O and let B be the point where AU intersects
the circle. (There will be one other point U in S with U O perpendicular to AC; this point
will be vertically below O. By symmetry, the length of U O is the same in either case.)
Join U O and BO.
U

A C
O

Let ∠BU O = θ.
Note that AO = BO = 1 since they are radii and BU = 1 by definition.
Therefore, 4U BO is isosceles and so ∠BOU = ∠BU O = θ.
Now ∠ABO is an exterior angle in this triangle, so ∠ABO = ∠BU O + ∠BOU = 2θ.
Since OB = OA, then 4ABO is isosceles and so ∠BAO = ∠ABO = 2θ.
But 4U AO is right-angled at O, and so ∠U AO + ∠AU O = 90◦ or 2θ + θ = 90◦ .
Therefore, 3θ = 90◦ or θ = 30◦ .
√ √
This tells us that 4U AO is a 30◦ -60◦ -90◦ triangle, and so U O = 3AO = 3.

Solution 2
Let U be the point in S vertically above O and let B be the point where AU intersects
the circle. (There will be one other point U in S with U O perpendicular to AC; this point
will be vertically below O. By symmetry, the length of U O is the same in either case.)
Join U O and BC.
U

A C
O
2010 COMC Solutions Page 13

Since AC is a diameter, then ∠ABC = 90◦ .

Therefore, 4ABC is similar to 4AOU since each is right-angled and each includes the
angle at A.
AB AO
Thus, = .
AC AU
Since U is in S, then BU = 1, so AU = AB + BU = AB + 1.
Also, AO = 1 and AC = 2 since the radius of the circle is 1.
AB 1
Therefore, = or AB 2 + AB − 2 = 0.
2 AB + 1
Factoring, we obtain (AB − 1)(AB + 2) = 0. Since AB > 0, then AB = 1 and so AU = 2.
√ √ √
By the Pythagorean Theorem, since U O > 0, then U O = AU 2 − AO2 = 22 − 12 = 3.
(b) As in (a), we can choose the point V in S that is vertically above C.
B is the point where AV intersects the circle. Note that BV = 1 by definition.
Join V C and BC.
Since AC is a diameter, then ∠ABC = 90◦ .
V

A C
O

Let V C = x.
Since 4V CA is right-angled at C and AV > 0, then by the Pythagorean Theorem,
√ √
AV = AC 2 + CV 2 = 4 + x2 .
Now 4V BC is similar to 4V CA since both are right-angled and the triangles share a
common angle at V .
Since these triangles are similar, then
VB VC
=
VC VA
1 x
= √
x x2 + 4
√
x2 + 4 = x2
x2 + 4 = x4
0 = x4 − x2 − 4
0 = (x2 )2 − x2 − 4
2010 COMC Solutions Page 14

This is a quadratic equation in x2 . By the quadratic formula,

p √
2 1 ± (−1)2 − 4(1)(−4) 1 ± 17
x = =
2 2
√
2 2 1 + 17
Since x is positive, then x = .
2 r √
1 + 17
Since V C = x is positive, then V C = x = .
2
(c) We prove that such a circle does not exist by contradiction.
Suppose that there is a circle Z on which all of the points in S lie.
We coordinatize the original diagram, putting O at (0, 0), C at (1, 0), and A at (−1, 0).
For every point X in S that is above AC, there will be a corresponding point Y in S that
is below AC which is the reflection of X in AC.
Therefore, S is symmetric across the x-axis. Thus, Z is also symmetric across the x-axis
and so its centre lies on the x-axis.
Suppose Z has centre (p, 0) and radius r.
Then the equation of Z is (x − p)2 + y 2 = r2 .
√
From (a), the point (0, 3) lies on Z. Thus, p2 + 3 = r2 .
Also, the point W (2, 0) lies on Z. This point comes from choosing B to coincide with C
and extending AB horizontally by 1 unit. Thus, (2 − p)2 + 02 = r2 or p2 − 4p + 4 = r2 .
From the equations p2 + 3 = r2 and p2 − 4p + 4 = r2 , we equate values of r2 to obtain
p2 + 3 = p2 − 4p + 4 or 4p = 1 or p = 41 .
1
Thus, r2 = p2 + 3 = 16 + 3 = 49
16
and so r = 74 since r > 0.
Therefore, the equation
 q of Z must be (x − 14 )2 + y 2 = ( 47 )2 .
√
From (b), the point 1, 1+2 17 lies on the circle.
Therefore,
q √
2
1 2
(1 − 4
) + 1+ 17
2
= ( 74 )2
√
( 43 )2 + 1+2 17 = 49
16
√
9 8 17 49
16
+ 16
+ 2
= 16
√
17
2
= 2
√
17 = 4

This statement is false, so we have reached a contradiction.

Therefore, our assumption is false and there is no circle on which all of the points in S lie.
1 1 1
4. (a) First we note that if x > 0, then x + > 1, since if x ≥ 1, then x + ≥ 1 + > 1 and if
x x x
1 1
0 < x < 1, then > 1, so x + > 1.
x x
2010 COMC Solutions Page 15

We note that f (x) = x is equivalent to

   
1 1
x+ − x+ = x
x x
 
1 1
= x+
x x

In this last equation, the right side is a positive integer, so the left side is also a positive
integer.
1 1
Suppose that = n for some positive integer n. Then x = .
x   n  
1 1 1
Therefore, the equation = x + is equivalent to the equation n = +n .
x x n
 
1 1 1
Note that if n ≥ 2, then < 1, so n < n + < n + 1, which says that + n = n so
n n n
this result is true for all positive integers n ≥ 2.
 
1 1
Note also that if n = 1, then n + = 2, so + n 6= n.
n n
 
1
Therefore, if n is a positive integer, then n = + n if and only if n ≥ 2.
n
1
Therefore, the solution set of the equation f (x) = x is x = , where n is a positive integer
n
with n ≥ 2.
a
(b) Suppose that x = for some positive integer a > 1.
a+1

First, we calculate f (x).

Note that
1 a a+1 a2 + (a + 1)2 2a2 + 2a + 1 2(a2 + a) + 1 1
x+ = + = = 2
= 2
=2+ 2
x a+1 a a(a + 1) a +a a +a a +a
1 1 1 1
Since a > 1, then 2 < 2 = and so 2 < 2 + 2 < 3.
 a
 +a 1 +1 2 a +a
1 1
Therefore, x + = 2+ 2 = 2.
x a +a
     
a 1 1 1 1
Thus, if x = , then f (x) = x + − x+ = 2+ 2 −2= 2 .
a+1 x x a +a a +a

Second, we show that x 6= f (x).

a a2 1
Note that x = = and f (x) = , so x and f (x) would be equal if
a+1 a(a + 1) a(a + 1)
a2 1
and only if = which is true if and only if a2 = 1.
a(a + 1) a(a + 1)
Since a > 1, this is not true, so x 6= f (x).
1
(Alternatively, we could note that, from (a), x = f (x) if and only if x is of the form for
n
2010 COMC Solutions Page 16

a
some positive integer n > 1. Here, x = which is not of this form when a > 1, so
a+1
x 6= f (x).)

Third, we show that f (x) = f (f (x)).

1
We set y = f (x) = 2 .
a +a
1
Since a is a positive integer with a > 1, then y is of the form for some positive integer
n
n with n > 2.
Thus, y is of the form discovered in (a), so f (y) = y; in other words, f (f (x)) = f (x).

a
Therefore, if x = for some positive integer a > 1, then x 6= f (x), but f (x) = f (f (x)).
a+1
(c) Solution 1
We want to find an infinite family of rational numbers u with the properties that
• 0 < u < 1,
• u, f (u), and f (f (u)) are all distinct, and
• f (f (u)) = f (f (f (u))).
We will do this by finding an infinite family of rational numbers u with 0 < u < 1 with
a
the property that f (u) = for some positive integer a > 1.
a+1
1 1
In this case, (b) shows that f (f (u)) = 2 and that f (f (f (u))) = 2 .
a +a a +a
Thus, we will have f (f (u)) = f (f (f (u))) and f (u) 6= f (f (u)).
a 1
As long as we have u 6= and u 6= 2 , then we will have found a family of rational
a+1 a +a
numbers u with the required properties.
1
Note that in fact we cannot have u = 2 because in this case we would have f (u) = u
a +a
a
and so we would not have f (u) = .
a+1

We now show the existence of an infinite family of rational numbers u with 0 < u < 1
a
with f (u) = for some positive integer a > 1.
a+1
b
Let us consider candidate rational numbers u = with b and c positive integers and
b+c
c > 1.
Since b + c > b, then each is a rational number with 0 < u < 1.
1 b b+c b2 + (b + c)2 2b2 + 2bc + c2 c2
In this case, u + = + = = = 2 + .
u b+c b b(b + c) b2 + bc b2 + bc
c2 1 c2
If we suppose further that c2 < b2 + bc, then 2 < 1 and so u + = 2 + 2 < 3,
b + bc u b + bc
which gives
c2 c2
   
1 1
f (u) = u + − u+ =2+ 2 −2= 2
u u b + bc b + bc
2010 COMC Solutions Page 17

a c2
We want f (u) to be of the form . In other words, we want 2 to be of the form
a+1 b + bc
a
, which would be true if b2 + bc − c2 = 1.
a+1
Note that if b2 + bc − c2 = 1, then c2 = b2 + bc − 1 < b2 + bc, so the additional assumption
above is included in this equation. Also, if b2 + bc − c2 = 1, then b and c can have no
b
common divisor larger than 1 so u = is irreducible. Combining this with the fact
b+c
b a
that c 6= 1, we see that cannot be of the form .
b+c a+1
To summarize so far, if b2 + bc − c2 = 1 has an infinite family of positive integer solutions
b
(b, c), then the infinite family of rational numbers u = has the required properties.
b+c

Consider the equation b2 + bc − c2 = 1.

This is equivalent to the equations 4b2 + 4bc − 4c2 = 4 and 4b2 + 4bc + c2 − 5c2 = 4 and
(2b + c)2 − 5c2 = 4.
If we let d = 2b + c, we obtain the equation d2 − 5c2 = 4.
This is a version of Pell’s equation. It is known that if such an equation has one positive
integer solution, then it has infinitely many positive integer solutions.
Since d2 − 5c2 = 4 has one positive integer solution (d, c) = (7, 3), then it has infinitely
many positive integer solutions (d, c).
If d2 = 5c2 + 4 and c is odd, then c2 is odd, so d2 = 5c2 + 4 is odd, which means that d is
odd.
If d2 = 5c2 + 4 and c is even, then c2 is even, so d2 = 5c2 + 4 is even, which means that d
is even.
Therefore, if (d, c) satisfies d2 − 5c2 = 4, then d and c have the same parity so b = 21 (d − c)
is an integer.
In addition, since d2 = 5c2 + 4 > c2 then d > c which means that b = 12 (d − c) is a positive
integer.
Therefore, each positive integer solution (d, c) of the equation d2 − 5c2 = 4 gives a solution
(b, c) of the equation b2 + bc − c2 = 1 which is also a positive integer solution.

b
Therefore, there exists an infinite family of rational numbers u = with the required
b+c
properties.

Solution 2
As in Solution 1, we want to show the existence of an infinite family of rational numbers
a
u with 0 < u < 1 with f (u) = for some positive integer a > 1.
a+1
Consider the Fibonacci sequence which has F1 = 1, F2 = 1, and Fn = Fn−1 + Fn−2 for
n ≥ 3.
2010 COMC Solutions Page 18

F2n−1
Define un = for each positive integer n ≥ 2.
F2n+1
Note that 0 < F2n−1 < F2n+1 so 0 < u < 1.
F3 2
For example, u2 = = .
F5 5
In this case, f (u2 ) = ( 5 + 52 ) − b 25 + 52 c = 10
2 29
− b 29
10
9
c = 10 , which has the desired properties.
a 1
We must show that un is not of the form or of the form 2 :
a+1 a +a
F2n−1 a
If = , then aF2n−1 + F2n−1 = aF2n+1 or F2n−1 = a(F2n+1 − F2n−1 ) or
F2n+1 a+1
F2n−1 = aF2n . Since a is a positive integer and F2n > F2n−1 , this cannot be the
case.
F2n−1 1
If = 2 , then F2n+1 is divisible by F2n−1 . But Fj+1 and Fj−1 never
F2n+1 a +a
have a common divisor larger than 1, so this cannot be the case. (If Fj+1 and
Fj−1 have a common divisor larger than 1, then Fj = Fj+1 − Fj−1 also has this
divisor. We can continue this process using the equation Fj−2 = Fj − Fj−1 to
show that Fj−2 also has this divisor, and so on, until we obtain that F2 and F1
both have this divisor. Since F2 = F1 = 1, we have a contradiction.)
In general, note that
1 F2n−1 F2n+1
un + = +
un F2n+1 F2n−1
(F2n−1 )2 + (F2n+1 )2
=
F2n−1 F2n+1
(F2n−1 )2 + (F2n + F2n−1 )2
=
F2n−1 (F2n + F2n−1 )
2(F2n−1 )2 + 2F2n F2n−1 + (F2n )2
=
(F2n−1 )2 + F2n F2n−1
(F2n )2
= 2+
(F2n−1 )2 + F2n F2n−1
(F2n )2
= 2+
F2n−1 F2n+1

It is known that (F2n )2 − F2n−1 F2n+1 = −1 for all positive integers n. (See the end of the
solution for a proof of this.)
Set an = (F2n )2 , which is a positive integer.
1 an
Then, un + =2+ .
un an + 1
2010 COMC Solutions Page 19

Therefore,
   
1 1
f (un ) = un + − un +
un un
   
an an
= 2+ − 2+
an + 1 an + 1
 
an
= 2+ −2
an + 1
an
=
an + 1
Therefore, the infinite family of rational numbers un has the desired properties.

As a postscript, we prove that (Fm )2 − Fm−1 Fm+1 = (−1)m+1 for all positive integers
m ≥ 2.
We prove this result by induction on m.
When m = 2, we obtain (F2 )2 − F1 F3 = 12 − 1(2) = −1 = (−1)2+1 , as required.
Suppose that the result is true for m = k, for some positive integer k ≥ 2.
That is, suppose that (Fk )2 − Fk−1 Fk+1 = (−1)k+1 .
Consider m = k + 1. Then

(Fk+1 )2 − Fk Fk+2 = (Fk + Fk−1 )2 − Fk (Fk + Fk+1 )

= (Fk )2 + 2Fk Fk−1 + (Fk−1 )2 − (Fk )2 − Fk Fk+1
= 2Fk Fk−1 + (Fk−1 )2 − Fk Fk+1
= 2Fk Fk−1 + (Fk−1 )2 − Fk (Fk + Fk−1 )
= 2Fk Fk−1 + (Fk−1 )2 − (Fk )2 − Fk Fk−1
= Fk Fk−1 + (Fk−1 )2 − (Fk )2
= Fk−1 (Fk + Fk−1 ) − (Fk )2
= Fk−1 Fk+1 − (Fk )2
= (−1)((Fk )2 − Fk−1 Fk+1 )
= (−1)(−1)k+1 (by our inductive assumption)
= (−1)(k+1)+1

as required.
Therefore, (Fm )2 − Fm−1 Fm+1 = (−1)m+1 for all positive integers m ≥ 2 by induction,
which shows that (F2k )2 − F2k−1 F2k+1 = (−1)2k+1 = −1.