31st British Mathematical Olympiad 1995 Problems

31st British Mathematical Olympiad 1995 Problems

1.  Find all positive integers a ≥ b ≥ c such that (1 + 1/a)(1 + 1/b)(1 + 1/c) = 2.

2.  ABC is a triangle. D, E, F are the midpoints of BC, CA, AB. Show that ∠DAC = ∠ABE iff ∠AFC = ∠ADB.

3.  x, y, z are real numbers such that x < y < z, x + y + z = 6 and xy + yz + zx = 9. Show that 0 < x < 1 < y < 3 < z < 4.

4.  (1) How many ways can 2n people be grouped into n teams of 2?

(2) Show that (mn)! (mn)! is divisible by m!ⁿ⁺¹ n!^m+1 for all positive integers m, n.

Solutions

Problem

Find all positive integers a ≥ b ≥ c such that (1 + 1/a)(1 + 1/b)(1 + 1/c) = 2.

Solution

Answer: (a, b, c) = (5, 4, 3), (8, 3, 3), (7, 6, 2), (9, 5, 2), (15, 4, 2).

We have 1 + 1/a ≤ 1 + 1/b ≤ 1 + 1/c. So 2 ≤ (1 + 1/c)³. But (1 + 1/c)³ is a decreasing function of c and c = 4 gives 125/64 < 2. So the only possible values of c are 1, 2, 3. If c = 1, then (1 + 1/a)(1 + 1/b) = 1, which is obviously impossible. So c = 2 or 3.

We may write the expression as (c + 1)(a + b + 1) = (c - 1)ab. Suppose c = 2. Then (b - 3)a = 3b + 3. If b ≥ 7, then (b - 3)a ≥ 4a ≥ 4b ≥ 3b + 7 > 3b + 3. Contradiction. (b - 3)a = 3b + 3, so b - 3 must be positive. Thus we need only consider b = 4, 5 and 6. These give the three solutions with c = 2 shown above.

Suppose c = 3. Then 2ab = 4(a + b + 1), so ab = 2a + 2b + 2. So (b - 2)a = 2b + 2. If b ≥ 5, then (b - 2)a ≥ 3a ≥ 3b ≥ 2b + 5 > 2b + 2. Contradiction. But b - 2 must be positive, so we need only consider b = 3 and 4. That gives the two solutions shown above for c = 3. 

Problem 2

ABC is a triangle. D, E, F are the midpoints of BC, CA, AB. Show that ∠DAC = ∠ABE iff ∠AFC = ∠ADB.

Solution

The medians meet at the centroid G. DF is parallel to AC, so ∠FDG = ∠DAC (alternate angles) = ∠ABE (given) = ∠FBG (same angle). So FGDB is cyclic. So ∠ADB = ∠GDB (same angle) = 180^o - ∠GFB = ∠AFC. 

Problem 3

x, y, z are real numbers such that x < y < z, x + y + z = 6 and xy + yz + zx = 9. Show that 0 < x < 1 < y < 3 < z < 4.

Solution

(x + y) = 6 - z, so 9 = xy + yz + zx = xy + z(6 - z). Hence (z - 3)² = xy. Similarly, (x - 3)² = yz and (y - 3)² = zx. If any of x, y, z are zero, then two of (z - 3)², (y - 3)² and (x - 3)³ are zero, so two of x, y, z are equal, whereas we are told they are all unequal. Hence none of x, y, z are zero.

We also have that xy, yz and zx are all non-negative. So we cannot have x < 0 and y and z > 0 (for then xy < 0). Equally, we cannot have y < 0 and z > 0 (for then yz < 0). Finally, we cannot have x, y, z all negative, for then x + y + z would be negative. So x, y, z are all positive.

We have x + y = 6 - z. Also xy < (x + y)²/4 (strict inequality since x and y are unequal). So (6 - z)²/4 + (6 - z)z > 9 or 4z - z² > 0. Hence z < 4. If z ≤ 2, then since x < y < z, we have x + y + z < 6. Contradiction. hence z > 2. So |z - 3| < 1. But xy = (z - 3)², so xy < 1. So x < 1.

But (x - 3)² = yz, so yz > 4. But z < 4, so y > 1. If y ≥ 3, then x + y + z > y + z > 2y > 6. Contradiction. So y < 3.

Finally, (x - 3)(y - 3)(z - 3) = xyz - 3(xy + yz + zx) + 9(x + y + z) - 27 = xyz > 0. But (x - 3)(y - 3) > 0, so z > 3. 

Problem 4

(1) How many ways can 2n people be grouped into n teams of 2?
(2) Show that (mn)! (mn)! is divisible by m!ⁿ⁺¹ n!^m+1 for all positive integers m, n.

Solution

(1) Let the number of ways be t(n). Then t(1) = 1 and t(n) = (2n-1) t(n-1). So t(n) = 1.3.5 ... (2n-1). We can also write this as (2n)! /(2.4.6 ... 2n) = (2n)! / (2ⁿn!).

(2) The first part is evidently intended as a hint. Consider the number of ways of dividing mn people into m teams of n people. It is the number of ways of arranging mn people into a line divided by the number of ways of ordering the m teams times the number of ways or ordering each team = (mn)! / ( n! m!ⁿ ). So this must be integral. Similarly, (mn)! / ( m! n!^m ) is integral and hence their product.