Download Solutions 2013

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
sin(πx)
.
x→2013 x − 2013
lim
Solution: L’Hospital’s rule yields
π cos(πx)
sin(πx)
= lim
= π cos(2013π) = −π.
x→2013
x→2013 x − 2013
1
lim
3. At the university where Peter studies, every student is friends on Facebook with
precisely 13 other students. How many students are there at most who are either
Peter’s Facebook friends or friends with one of his friends?
Solution: Each of Peter’s 13 friends has 12 friends other than Peter. Thus there
are at most 13 · 12 students who are friends with one of his friends. Together with
his 13 friends, we get 13 · 13 = 169 people.
4. If
f (x) = sin(sin(sin(sin(sin(x))))),
what is f 0 (0)?
Solution: By the chain rule,
f 0 (x) = cos(sin(sin(sin(sin(x))))) cos(sin(sin(sin(x)))) cos(sin(sin(x))) cos(sin(x)) cos x
and thus f 0 (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+6
and
y=
Thus there is a solution in all cases except a = −6.
16
.
a+6
6. At a recent mathematics competition, 252 participants solved the first question, but
only 36 solved the last one. Surprisingly, only half of the participants who solved
the last question could also solve the first one, but one third of the participants who
did not solve the first question could solve the last one. How many people took part
in this competition?
Solution: Half the participants who solved the last question (18 people) did not
solve the first question. This is exactly one third of the participants who did not
solve 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 the
same position where the minute hand is at the moment. How long will it take until
the 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. Thus
the 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 of
Re(z 2 ) + Im(z 2 ).
Solution: Write z = 2 + iy. Then we have z 2 = 4 + 4iy − y 2 und thus
Re(z 2 ) + Im(z 2 ) = 4 − y 2 + 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 positive
integer b has precisely four positive divisors. What is the minimum possible number
of positive divisors of ab?
Solution: Let d be the divisor of a that is not 1 or a itself. We find that ab has
at least six divisors: all four divisors of b are divisors of ab, as are db and ab. The
example a = 4, b = 8 shows that the number of divisors of ab can indeed be 6.
10. Let kxk be the distance of x from the nearest integer, e.g. k1.7k = 0.3 or k3.2k = 0.2.
Determine
Z
100
kxk dx.
0
Solution: Note that
Z n+1
Z n+1/2
Z
kxk dx =
(x − n) dx +
n
n
2 1/2
n+1
Z
(n + 1 − x) dx =
n+1/2
for every integer n. Thus
Z
100
kxk dx = 100 ·
0
1
= 25.
4
Z
1
(1 − u) du
u du +
0
(1 − u)2 1
1 1
1
u = −
= + =
2 0
2
8 8
4
1/2
1/2
1/2
11. Determine the sum
1000
X
1
ln 1 +
n
n=1
.
Solution: The sum is a telescoping sum:
1000
X
X
1000
1
ln 1 +
=
(ln(n + 1) − ln n)
n
n=1
n=1
= (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 (each
with probability 12 ). An “isolated” square is a square whose colour differs from all its
horizontally or vertically adjacent squares. What is the expected number of isolated
squares?
Solution: For the four corners, the probability of being isolated is 212 = 14 . For the
24 other squares on the edge of the board, the probability is 213 = 18 , and for the 36
1
. Therefore, the expected number of isolated squares is
central squares it is 214 = 16
4 24 36
25
+
+
= .
4
8
16
4
13. Given that
2013 

3
3
x
0 0 0
−2 −3 −3
= 0 0 0 ,
−1 −2 0
0 0 0

determine x.
Solution: Taking the determinant
3
−2
−1
thus
on both sides, we find
2013
3
x −3 −3
= 0,
−2 0
3
3
x
−2 −3 −3 = x − 9 = 0,
−1 −2 0 which yields x = 9.
14. Determine all differentiable functions f : (0, ∞) → R such that
x
f (y)
0
f
=
y
f (x)
for all x, y > 0.
Solution: We set y = 1. This gives us
f 0 (x) =
f (1)
,
f (x)
which is a separable differential equation that can be solved explicitly:
d f (x)2
= f (x)f 0 (x) = f (1),
dx 2
thus
f (x)2
= f (1)x + C
2
√
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:
√
Ay + B
A
± p
=√
Ax + B
2 Ax/y + B
for all x, y > 0. Squaring and multiplying out yields
A3 x+A2 B = A2 (Ax+B) = 4(Ay +B)(Ax/y +B) = 4A2 x+4ABy +4ABx/y +4B 2 .
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 right
hand 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.
1
cos 3x
sin 3x
= and 0 < x < π2 , determine
.
cos x
3
sin x
Solution: We use the addition formula for the sine function: Since
15. If
sin 3x cos 3x
sin 3x cos x − cos 3x sin x
sin 2x
−
=
=
= 2,
sin x
cos x
sin x cos x
sin x cos x
we have
sin 3x
7
= .
sin x
3
16. What is the expected maximum of two independent uniformly distributed random
numbers on [0, 1]?
Solution: If the first of the two numbers is x, then the second one will be smaller
with probability x (so that the maximum is x) and larger with probability 1 − x,
yielding an average maximum of 1+x
. Thus the expected value is
2
x2 + (1 − x)
The average over all x is now
Z 1
0
1+x
1 + x2
=
.
2
2
1 + x2
x x3 1 2
dx = + = .
2
2
6 0 3
17. Let R and S be points on the sides AB and AC, respectively, of triangle ABC, and
let P be the point of intersection of BS and CR. If the areas of triangles BP R,
BP C and CP S are 5, 6, and 7, respectively, find the area of triangle ABC.
Solution: Let α = ∠BP C. Since ∠BP R = ∠CP S = 180◦ − α and ∠RP S = α,
we have
BP · RP
BP · CP
BP · RP
sin(180◦ − α) =
sin α, Area(BP C) =
sin α,
Area(BP R) =
2
2
2
CP · SP
CP · SP
RP · SP
Area(CP S) =
sin(180◦ − α) =
sin α, Area(RP S) =
sin α.
2
2
2
It follows that Area(BP R) · Area(CP S) = Area(BP C) · Area(RP S), which gives us
. In the same way, let β = ∠BAC, and note that
Area(RP S) = 35
6
AS · AR
AR · AC
sin β, Area(RAC) =
sin β,
2
2
AB · AS
AB · AC
Area(BAS) =
sin β, Area(BAC) =
sin β,
2
2
from which we obtain Area(RAS) · Area(BAC) = Area(RAC) · Area(BAS). Now let
x be the area of triangle ABC. We obtain
35
x−5−6−7−
· x = (x − 5 − 6) (x − 6 − 7) ,
6
Area(RAS) =
which reduces to x/6 = 143 or x = 858.
18. Determine the shortest distance between a point on the parabola y = 8x − x2 and
a point on the parabola y = x2 + 15.
Solution: Let P1 = (x1 , y1 ) and P2 = (x2 , y2 ) be the points on the two parabolas
that are closest to each other (y1 = 8x1 − x21 and y2 = x22 + 15). The circle centred
at P1 that passes through P2 has to touch the second parabola at that point with a
common tangent (if they were to intersect instead, one could replace P2 by a point
that is closer to P1 ). Thus the line that connects P1 and P2 is perpendicular to the
parabola 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 also
have
x2 − x1
= 8 − 2x1 = 2x2 .
−
y2 − y1
We plug in y1 and y2 :
−
(4 − x1 ) − x1
= 8 − 2x1 ,
(4 − x1 )2 + 15 − 8x1 + x21
which gives
(2x1 − 4) = (8 − 2x1 )(2x21 − 16x1 + 31)
or
4x31 − 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 denote
the largest number in the sequence
x1
x2
x3
x10
,
,
,...,
.
1 + x1 1 + x1 + x 2 1 + x1 + x2 + x3
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 for
all k, and the terms of the sequence can be rewritten as
yk − yk−1
yk−1
=1−
.
yk
yk
All elements of the sequence are ≤ z, thus
yk−1
≥ 1 − z,
yk
which implies
y0
y0 y1 y2
y9
=
·
·
· ··· ·
≥ (1 − z)10 .
y10
y1 y2 y3
y10
We are given that x1 + x2 + · · · + x10 = 1, thus y10 = 2 and consequently
1
≥ (1 − z)10 ,
2
1
from which we finally deduce z ≥ 1 − 10√
. We get equality if yk = 2k/10 and thus
2
1
is the smallest possible value of z.
xk = 2k/10 − 2(k−1)/10 . So 1 − 10√
2
20. Determine
Z
π
4
ln(1 + tan x)dx.
0
Solution: Note that
cos x + cos(π/2 − x)
cos x + sin x
=
cos x
cos √
x
2 cos(π/4) cos(x − π/4)
2 cos(x − π/4)
=
.
=
cos x
cos x
1 + tan x =
It follows that
Z π
Z
4
ln(1 + tan x)dx =
0
π
4
√
Z
π
4
ln 2 dx +
ln cos(x − π/4) dx −
0
Z 0
Z π
4
π ln 2
=
+
ln cos x dx −
ln cos x dx
8
− π4
0
0
=
π ln 2
.
8
The two integrals cancel, since ln cos x is an even function.
Z
π
4
ln cos x dx
0