Pan African Mathematical Olympiad 2000

Paper 1 Solutions

1. Solve the trigonometric equation

sin^{3} x(1 + cot x) + cos^{3} x(1 + tan x) = cos 2x.

Solution

sin^{3} x(1 + cot x) + cos^{3} x(1 + tan x) = cos 2x
⇒ sin^{2} x(sin x + cos x) + cos^{2} x(cos x + sin x) = cos^{2} x − sin^{2} x
⇔ (sin^{2} x + cos^{2} x)(sin x + cos x) = (cos x + sin x)(cos x − sin x)
⇔ 0 = (cos x + sin x)(cos x − sin x − 1)
⇔ 0 = (sin π/4 · cos x + cos π/4 · sin x)(sin π/4 · cos x − cos π/4 · sin x − \sqrt{2}/2)
⇔ 0 = [sin(π/4 + x)][sin(π/4 − x) − \sqrt{2}/2].

Thus sin(π/4 + x) = 0 or sin(π/4 − x) = \sqrt{2}/2 The sine function takes the value 0 at and only at multiples of π, so the solutions to the first case are of the form kπ − π/4, for k ∈ ℤ. These solutions give 1 + cot x = 0, 1 + tan x = 0 and cos 2x = 0, so they clearly satisfy the original problem.

We now examine the other candidate solutions. Now sin x = \sqrt{2}/2 at and only at π/4 and 5π/4 if x ∈ [0, 2π]. Thus the solutions of sin(π/4 − x) = \sqrt{2}/2 are of the form 2kπ and 2kπ − π/2 for k ∈ ℤ. None of these are solutions to the original problem, since they give cot x or tan x as undefined.

2. Let the polynomials P_{0}, P_{1}, P_{2}, … be defined by

P_{0}(x) = x^{3} + 213x^{2} − 67x − 2000

and

P_{n}(x) = P_{n−1}(x − n) for n = 1, 2, 3, …

What is the coefficient of x in P_{21}(x)? Solution

P_{21}(x) = P_{20}(x − 21)
= P_{19}(x − 21 − 20)
⋮
= P_{0}(x − 21 − 20 − ⋯ − 1)
= P_{0}(x − 231)
= (x − 231)^{3} + 213(x − 231)^{2} − 67(x − 231) − 2000.

The coefficient of x in each term can be found from the binomial theorem, so the coeffiient of x is

3 · 231^{2} − 213 · 2 · 213 − 67 = 61610.

3. Prove that if

\frac{p}{q} = 1 − \frac{1}{2} + \frac{1}{3} − \frac{1}{4} + … − \frac{1}{1334} + \frac{1}{1335}

where p and q are natural numbers, then 2003 divides p. Solution Firstly note that 2003 is a prime number (easily checked by trial division by all prime numbers less than \sqrt{2003}, i.e. all prime number up to 43.

1 − 1/2 + ⋯ + 1/1335 = 1 + 1/2 + ⋯ + 1/1335 − 2(1/2 + 1/4 + ⋯ + 1/1334)
= 1 + 1/2 + ⋯ + 1/1335 − (1 + 1/2 + ⋯ + 1/667)
= 1/668 + 1/669 + ⋯ + 1/1335
= (1/668 + 1/1335) + (1/669 + 1/1334) + ⋯ + (1/1001 + 1/1002)
= 2003(1/(668 × 1335) + 1/(669 × 1334) + ⋯ + 1/(1001 × 1002)).

Now the denomominator of

1/(668 × 1335) + 1/(669 × 1334) + ⋯ + 1/(1001 × 1002)

in simplest form is coprime to 2003. This is because 2003 is prime and the numbers 668 to 1335 are all smaller than 2003, so their product cannot be a multiple of 2003. Thus the sum is p/q in simplest form where 2003 | p.