SOUTH AFRICAN TERTIARY MATHEMATICS OLYMPIAD

5 October 2013

Solutions

1. For which real value of x is log2(log3 x) = 1?

Solution: We have to have log3 x = 2 and thus x = 9.

2. Determine the value of the limit

limx→2013

sin(πx)

x− 2013.

Solution: L’Hospital’s rule yields

limx→2013

sin(πx)

x− 2013= lim

x→2013

π cos(πx)

1= π cos(2013π) = −π.

3. At the university where Peter studies, every student is friends on Facebook withprecisely 13 other students. How many students are there at most who are eitherPeter’s Facebook friends or friends with one of his friends?

Solution: Each of Peter’s 13 friends has 12 friends other than Peter. Thus thereare at most 13 · 12 students who are friends with one of his friends. Together withhis 13 friends, we get 13 · 13 = 169 people.

4. Iff(x) = sin(sin(sin(sin(sin(x))))),

what is f ′(0)?

Solution: By the chain rule,

f ′(x) = cos(sin(sin(sin(sin(x))))) cos(sin(sin(sin(x)))) cos(sin(sin(x))) cos(sin(x)) cosx

and thus f ′(0) = 1 · 1 · 1 · 1 · 1 = 1.

5. For which value of a does the following system of equations have no solution?

x+ 2y = 5,

−3x+ ay = 1.

Solution: By Cramer’s rule (or other methods), we get

x =5a− 2

a+ 6and y =

16

a+ 6.

Thus there is a solution in all cases except a = −6.

6. At a recent mathematics competition, 252 participants solved the first question, butonly 36 solved the last one. Surprisingly, only half of the participants who solvedthe last question could also solve the first one, but one third of the participants whodid not solve the first question could solve the last one. How many people took partin this competition?

Solution: Half the participants who solved the last question (18 people) did notsolve the first question. This is exactly one third of the participants who did notsolve the first question. Thus there were 252 people who solved the first question,and 54 who did not, which means there were 306 participants.

7. Thomas notices that in four hours, the clock’s hour hand will be in precisely thesame position where the minute hand is at the moment. How long will it take untilthe minute hand is in the exact same position where the hour hand is now?

Solution: Four hours correspond to an angle of 120◦ between the two hands. Thusthe minute hand has to perform a 240◦ rotation, which takes 40 minutes.

8. The complex number z satisfies Re z = 2. Determine the largest possible value ofRe(z2) + Im(z2).

Solution: Write z = 2 + iy. Then we have z2 = 4 + 4iy − y2 und thus

Re(z2) + Im(z2) = 4− y2 + 4y = 8− (y − 2)2,

which shows that the maximum is 8 (obtained for y = 2).

9. A positive integer a has precisely three positive divisors, while another positiveinteger b has precisely four positive divisors. What is the minimum possible numberof positive divisors of ab?

Solution: Let d be the divisor of a that is not 1 or a itself. We find that ab hasat least six divisors: all four divisors of b are divisors of ab, as are db and ab. Theexample a = 4, b = 8 shows that the number of divisors of ab can indeed be 6.

10. Let ‖x‖ be the distance of x from the nearest integer, e.g. ‖1.7‖ = 0.3 or ‖3.2‖ = 0.2.Determine ∫ 100

0

‖x‖ dx.

Solution: Note that∫ n+1

n

‖x‖ dx =

∫ n+1/2

n

(x− n) dx+

∫ n+1

n+1/2

(n+ 1− x) dx =

∫ 1/2

0

u du+

∫ 1

1/2

(1− u) du

=u2

2

∣∣∣1/20− (1− u)2

2

∣∣∣11/2

=1

8+

1

8=

1

4

for every integer n. Thus ∫ 100

0

‖x‖ dx = 100 · 1

4= 25.

11. Determine the sum1000∑n=1

ln

(1 +

1

n

).

Solution: The sum is a telescoping sum:

1000∑n=1

ln

(1 +

1

n

)=

1000∑n=1

(ln(n+ 1)− lnn)

= (ln 2− ln 1) + (ln 3− ln 2) + (ln 4− ln 3) + ·+ (ln 1001− ln 1000)

= ln 1001− ln 1 = ln 1001.

12. Each of the squares of an 8 × 8-board is coloured randomly black or white (eachwith probability 1

2). An “isolated” square is a square whose colour differs from all its

horizontally or vertically adjacent squares. What is the expected number of isolatedsquares?

Solution: For the four corners, the probability of being isolated is 122

= 14. For the

24 other squares on the edge of the board, the probability is 123

= 18, and for the 36

central squares it is 124

= 116

. Therefore, the expected number of isolated squares is

4

4+

24

8+

36

16=

25

4.

13. Given that 3 3 x−2 −3 −3−1 −2 0

2013

=

0 0 00 0 00 0 0

,determine x.

Solution: Taking the determinant on both sides, we find∣∣∣∣∣∣3 3 x−2 −3 −3−1 −2 0

∣∣∣∣∣∣2013

= 0,

thus ∣∣∣∣∣∣3 3 x−2 −3 −3−1 −2 0

∣∣∣∣∣∣ = x− 9 = 0,

which yields x = 9.

14. Determine all differentiable functions f : (0,∞)→ R such that

f ′(x

y

)=f(y)

f(x)

for all x, y > 0.

Solution: We set y = 1. This gives us

f ′(x) =f(1)

f(x),

which is a separable differential equation that can be solved explicitly:

d

dx

f(x)2

2= f(x)f ′(x) = f(1),

thusf(x)2

2= f(1)x+ C

for some constant C. This means that f(x) = ±√Ax+B, with A = 2f(1) and

B = 2C. We plug this back into the original equation:

± A

2√Ax/y +B

=

√Ay +B√Ax+B

for all x, y > 0. Squaring and multiplying out yields

A3x+A2B = A2(Ax+B) = 4(Ay+B)(Ax/y+B) = 4A2x+4ABy+4ABx/y+4B2.

Since this must hold for all y > 0, we have B = 0. This leaves us with A3 = 4A2

and consequently A = 4 (A = 0 yields f(x) = 0 for all x, in which case the righthand side of the original equation is undefined). f(x) = −

√4x is not a solution (the

signs do not match), but f(x) =√

4x = 2√x is.

15. Ifcos 3x

cosx=

1

3and 0 < x < π

2, determine

sin 3x

sinx.

Solution: We use the addition formula for the sine function: Since

sin 3x

sinx− cos 3x

cosx=

sin 3x cosx− cos 3x sinx

sinx cosx=

sin 2x

sinx cosx= 2,

we havesin 3x

sinx=

7

3.

16. What is the expected maximum of two independent uniformly distributed randomnumbers on [0, 1]?

Solution: If the first of the two numbers is x, then the second one will be smallerwith probability x (so that the maximum is x) and larger with probability 1 − x,yielding an average maximum of 1+x

2. Thus the expected value is

x2 + (1− x)1 + x

2=

1 + x2

2.

The average over all x is now∫ 1

0

1 + x2

2dx =

x

2+x3

6

∣∣∣10

=2

3.

17. Let R and S be points on the sides AB and AC, respectively, of triangle ABC, andlet P be the point of intersection of BS and CR. If the areas of triangles BPR,BPC and CPS are 5, 6, and 7, respectively, find the area of triangle ABC.

Solution: Let α = ∠BPC. Since ∠BPR = ∠CPS = 180◦ − α and ∠RPS = α,we have

Area(BPR) =BP ·RP

2sin(180◦ − α) =

BP ·RP2

sinα, Area(BPC) =BP · CP

2sinα,

Area(CPS) =CP · SP

2sin(180◦ − α) =

CP · SP2

sinα, Area(RPS) =RP · SP

2sinα.

It follows that Area(BPR) · Area(CPS) = Area(BPC) · Area(RPS), which gives usArea(RPS) = 35

6. In the same way, let β = ∠BAC, and note that

Area(RAS) =AS · AR

2sin β, Area(RAC) =

AR · AC2

sin β,

Area(BAS) =AB · AS

2sin β, Area(BAC) =

AB · AC2

sin β,

from which we obtain Area(RAS) ·Area(BAC) = Area(RAC) ·Area(BAS). Now letx be the area of triangle ABC. We obtain(

x− 5− 6− 7− 35

6

)· x = (x− 5− 6) (x− 6− 7) ,

which reduces to x/6 = 143 or x = 858.

18. Determine the shortest distance between a point on the parabola y = 8x − x2 anda point on the parabola y = x2 + 15.

Solution: Let P1 = (x1, y1) and P2 = (x2, y2) be the points on the two parabolasthat are closest to each other (y1 = 8x1 − x21 and y2 = x22 + 15). The circle centredat P1 that passes through P2 has to touch the second parabola at that point with acommon tangent (if they were to intersect instead, one could replace P2 by a pointthat is closer to P1). Thus the line that connects P1 and P2 is perpendicular to theparabola tangent at P2. The same applies to P1, so the tangents are parallel.

The gradients of the two tangents are 8− 2x1 and 2x2 respectively, so we get x2 =4 − x1. Since they are also perpendicular to the line between P1 and P2, we alsohave

−x2 − x1y2 − y1

= 8− 2x1 = 2x2.

We plug in y1 and y2:

− (4− x1)− x1(4− x1)2 + 15− 8x1 + x21

= 8− 2x1,

which gives(2x1 − 4) = (8− 2x1)(2x

21 − 16x1 + 31)

or4x31 − 48x21 + 192x1 − 252 = 4(x1 − 3)(x21 − 9x1 + 21) = 0.

The second factor does not have any real zeros, hence x1 = 3, x2 = 1, y1 = 15,y2 = 16, and the shortest distance is

√5.

19. Let the sum of ten positive numbers x1, x2, . . . , x10 be equal to 1, and let z denotethe largest number in the sequence

x11 + x1

,x2

1 + x1 + x2,

x31 + x1 + x2 + x3

, . . . ,x10

1 + x1 + x2 + · · ·+ x10.

What is the smallest possible value of z?

Solution: Set yk = 1 + x1 + x2 + · · · + xk (and y0 = 1). Then xk = yk − yk−1 forall k, and the terms of the sequence can be rewritten as

yk − yk−1yk

= 1− yk−1yk

.

All elements of the sequence are ≤ z, thus

yk−1yk≥ 1− z,

which impliesy0y10

=y0y1· y1y2· y2y3· · · · · y9

y10≥ (1− z)10.

We are given that x1 + x2 + · · ·+ x10 = 1, thus y10 = 2 and consequently

1

2≥ (1− z)10,

from which we finally deduce z ≥ 1 − 110√2 . We get equality if yk = 2k/10 and thus

xk = 2k/10 − 2(k−1)/10. So 1− 110√2 is the smallest possible value of z.

20. Determine ∫ π4

0

ln(1 + tanx)dx.

Solution: Note that

1 + tan x =cosx+ sinx

cosx=

cosx+ cos(π/2− x)

cosx

=2 cos(π/4) cos(x− π/4)

cosx=

√2 cos(x− π/4)

cosx.

It follows that∫ π4

0

ln(1 + tan x)dx =

∫ π4

0

ln√

2 dx+

∫ π4

0

ln cos(x− π/4) dx−∫ π

4

0

ln cosx dx

=π ln 2

8+

∫ 0

−π4

ln cosx dx−∫ π

4

0

ln cosx dx

=π ln 2

8.

The two integrals cancel, since ln cos x is an even function.