15th Asian Pacific Mathematics Olympiad 2003 Problems

15th Asian Pacific Mathematics Olympiad 2003 Problems

1.  The polynomial a₈x⁸ +a₇x⁷ + ... + a₀ has a₈ = 1, a₇ = -4, a₆ = 7 and all its roots positive and real. Find the possible values for a₀.

2.  A unit square lies across two parallel lines a unit distance apart, so that two triangular areas of the square lie outside the lines. Show that the sum of the perimeters of these two triangles is independent of how the square is placed.

3.  k > 14 is an integer and p is the largest prime smaller than k. k is chosen so that p ≥ 3k/4. Prove that 2p does not divide (2p - k)!, but that n does divide (n - k)! for composite n > 2p.

4.  Show that (aⁿ + bⁿ)^1/n + (bⁿ + cⁿ)^1/n + (cⁿ + aⁿ)^1/n < 1 + (2^1/n)/2, where n > 1 is an integer and a, b, c are the sides of a triangle with unit perimeter.

5.  Find the smallest positive integer k such that among any k people, either there are 2m who can be divided into m pairs of people who know each other, or there are 2n who can be divided into n pairs of people who do not know each other.

15th APMO 2003 Problem 1

The polynomial a₈x⁸ +a₇x⁷ + ... + a₀ has a₈ = 1, a₇ = -4, a₆ = 7 and all its roots positive and real. Find the possible values for a₀.

Answer 1/2⁸

Solution

Let the roots be x_i. We have Sum x_i² = 4² - 2·7 = 2. By Cauchy we have (x₁·1 + ... + x₈·1) ≤ (x₁² + ... + x₈²)^1/2(1² + ... + 1²)^1/2 with equality iff all x_i are equal. Hence all x_i are equal. So they must all be 1/2.

15th APMO 2003 Problem 2

A unit square lies across two parallel lines a unit distance apart, so that two triangular areas of the square lie outside the lines. Show that the sum of the perimeters of these two triangles is independent of how the square is placed.

Solution

15th APMO 2003 Problem 3

k > 14 is an integer and p is the largest prime smaller than k. k is chosen so that p ≥ 3k/4. Prove that 2p does not divide (2p - k)!, but that n does divide (n - k)! for composite n > 2p.

Solution

Since k > p, we have p > 2p-k and hence p does not divide any of 1, 2, 3, ... , 2p-k. But p is prime, so it does not divide (2p - k)! So 2p does not either.

Now consider composite n > 2p. Note that k > 14, so p ≥ 13, so n > 26. Take q to be the largest prime divisor of n and put n = qr. We now have three cases.

(1) q > r ≥ 3. Then n > 2p ≥ 3k/2, so 2n/3 > k. Hence n-k > n-2n/3 = n/3 ≥ n/r = q > r. So q and r are distinct integers < n-k. Hence n = qr divides (n-k)!.

(2) r = 2. Then n = 2q > 2p. But p is the largest prime < k. Hence q ≥ k, so n-k ≥ n-q = q. Obviously q > 2 (since n > 26), so q and 2 are distinct integers < n-k. Hence n = 2q divides (n-k)! .

(3) The final case is n = q². We show that 2q ≤ n-k. Suppose not. Then 2q ≥ n-k+1 ≥ (2p+1)-k+1 ≥ 3k/2 + 1 - k + 1 = k/2 + 2. So q ≥ k/4 + 1. Also since 2q ≥ n-k+1, we have k ≥ n-2q+1 = (q-1)² ≥ k²/16. If k > 16, this is a contradiction. If k = 15 or 16, then p = 13 and k ≥ (q-1)² gives q ≤ 5, so n = q² ≤ 25. But n > 2p = 26. Contradiction. So we have established that 2q ≤ n-k. But that means that q and 2q are distinct integers ≤ n-k and so their product 2n divides (n-k)!.

Thanks to Gerhard Woeginger for this.

15th APMO 2003 Problem 4

Show that (aⁿ + bⁿ)^1/n + (bⁿ + cⁿ)^1/n + (cⁿ + aⁿ)^1/n < 1 + (2^1/n)/2, where n > 1 is an integer and a, b, c are the sides of a triangle with unit perimeter.

Solution

We may take a ≥ b ≥ c. Since a + b + c = 1 and a < b+c, we have b ≤ a < 1/2. Hence (a_n + b_n)^1/n < 2^1/n/2 (*).

We have (b + c/2)ⁿ = bⁿ + n/2 c b^n-1 + other positive terms > bⁿ + cⁿ. Hence (bⁿ + cⁿ)^1/n < b + c/2. Similarly, (cⁿ + aⁿ)^1/n < a + c/2. Adding we get (bⁿ + cⁿ)^1/n + (cⁿ + aⁿ)^1/n < a+b+c = 1. Adding to (*) gives the required result.

 

 

Let the lines be L, L'. Let the square be ABA'B', with A, A' the two vertices not between L and L'. Let L meet AB at X and AB' at Y. Let L' meet A'B' at X' and A'B at Y'. So AXY and A'X'Y' are similar. Suppose angle AXY = x. If we move L towards A by a distance d perpendicular to itself, then AX is shortened by d cosec x. If L' remains a distance 1 from L, then A'X' is lengthened by d cosec x. The new triangle AXY is similar to the old. Suppose that perimeter AXY = k·AX, then perimeter AXY is increased by kd cosec x. Since AXY and A'X'Y' are similar, perimeter A'X'Y' is shortened by kd cosec x, so the sum of their perimeters is unchanged. It remains to show that the sum of the perimeters does not depend on the angle x.

Let us move L towards A until L' passes through A', at which point the perimeter of A'X'Y' is zero. Now if h is the height of AXY (from the base XY), then 1 + h = AA' sin(45^o + x) = sin x + cos x. The perimeter of AXY is h/sin x + h/cos x + h/(sin x cos x) = h(sin x + cos x + 1)/(sin x cos x) = (sin x + cos x - 1)(sin x + cos x + 1)/(sin x cos x) = 2, which is independent of x.