36th British Mathematical Olympiad 2000 Problems

36th British Mathematical Olympiad 2000 Problems

1.  Two circles meet at A and B and touch a common tangent at C and D. Show that triangles ABC and ABD have the same area.

2.  Find the smallest value of x² + 4xy + 4y² + 2z² for positive reals x, y, z with product 32.

3.  Find positive integers m, n such that (m^1/3 + n^1/3 - 1)² = 49 + 20 (6^1/3).

4.  Find a set of 10 distinct positive integers such that no 6 members of the set have a sum divisible by 6. Is it possible to find such a set with 11 members? 

Solutions

Problem 1

Two circles meet at A and B and touch a common tangent at C and D. Show that triangles ABC and ABD have the same area.

Solution

Let the common tangent meet the line AB at X. We have XC² = XA·XB = XD². So C and D are equidistant from the line AB. 

Problem 2

Find the smallest value of x² + 4xy + 4y² + 2z² for positive reals x, y, z with product 32.

Solution

Answer: 96.

We have (x + 2y)² ≥ 8xy with equality iff x = 2y. Hence (x + 2y)² + 2z² ≥ 256/z + 2z² with equality iff x = 2y. We have z² + 128/z - 48 = (z - 4)²(z + 8)/z ≥ 0 with equality iff z = 4 (for positive z). Hence 256/z + 2z² ≥ 96 with equality iff z = 4. So the expression given has minimum value 96, achieved only at x = 4, y = 2, z = 4. 

Problem 3

Find positive integers m, n such that (m^1/3 + n^1/3 - 1)² = 49 + 20 (6^1/3).

Solution

A little experimentation gives (2·6^1/3 + 2·6^2/3 - 1)² = 49 + 20·6^1/3. So we can take m = 48, n = 288. 

Problem 4

Find a set of 10 distinct positive integers such that no 6 members of the set have a sum divisible by 6. Is it possible to find such a set with 11 members?

Solution

Thanks to Arne Smeets

Take five integers = 0 mod 6 and five = 1 mod 6. Then any six must include at least one and at most five = 1 mod 6 and hence their sum cannot be 0 mod 6.

Amongst any 5 numbers there must be 3 with sum = 0 mod 3. (If three numbers have the same residue, then their sum = 0 mod 3. If not, then there must be at least one for each residue, and 0 + 1 + 2 = 0 mod 3.) So given 11 numbers, we can pick a₁, a₂, a₃ with a₁ + a₂ + a₃ = 0 mod 3. From the remaining 8, we can pick another three: b₁, b₂, b₃. From the remaining 5 we can pick another three: c₁, c₂, c₃. Two of the sums a₁ + a₂ + a₃, b₁ + b₂ + b₃, c₁ + c₂ + c₃ must have the same parity, so their sum is even. Thus we have 6 numbers with sum = 0 mod 6.