32nd British Mathematical Olympiad 1996 Problems

32nd British Mathematical Olympiad 1996 Problems

1.  Find all non-negative integer solutions to 2^m + 3ⁿ = k².

2.  The triangle ABC has sides a, b, c, and the triangle UVW has sides u, v, w such that a² = u(v + w - u), b² = v(w + u - v), c² = w(u + v - w). Show that ABC must be acute angled and express the angles U, V, W in terms of the angles A, B, C.

3.  The circles C and C' lie inside the circle S. C and C' touch each other externally at K and touch S at A and A' respectively. The common tangent to C and C' at K meets S at P. The line PA meets C again at B, and the line PA' meets C' again at B'. Show that BB' is a common tangent to C and C'.

4.  Find all positive real solutions to w + x + y + z = 12, wxyz = wx + wy + wz + xy + xz + yz + 27. 

Solutions

Problem 1

Find all non-negative integer solutions to 2^m + 3ⁿ = k².

Solution

Answer: 2⁰ + 3¹ = 2², 2³ + 3⁰ = 3² , 2⁴ + 3² = 5².

If m = 0, then 3ⁿ = k² - 1 = (k + 1)(k - 1). But if 3 divides both k + 1 and k - 1, then it divides their difference 2, which is impossible, so k - 1 = 1 and we have the solution m = 0, n = 1, k = 2.

If m = 1, then k² - 2 = 3ⁿ. But squares are 0 or 1 mod 3, so k² - 2 is 1 or 2 mod 3, so n must be 0. But that does not give a solution. So we may assume m > 1. Hence k² = 3ⁿ = (4 - 1)ⁿ = (-1)ⁿ mod 4. But squares are 0 or 1 mod 4, so n must be even. Put n = 2s. Then 2^m = k² - 3^2s = (k + 3^s)(k - 3^s). So both k + 3^s and k - 3^s are powers of 2. Suppose k - 3^s = 2^a and k + 3^s = 2^b, so that m = a + b. Then 2.3^s = 2^b - 2^a. So a = 1 and 3^s = 2^b-1 - 1. But 2^b-1 - 1 = (-1)^b-1 - 1 mod 3, so b - 1 must be even or s = 0.

If s = 0, then b - 1 = 1, so m = 3 and we have the solution 2³ + 3⁰ = 3².

So suppose b -1 is even. Put b - 1 = 2r, then 3^s = (2^r - 1)(2^r + 1). But 3 cannot divide both 2^r - 1 and 2^r + 1 or it divides their difference, which is 2. So we must have 2^r - 1 = 1 and hence r = s = 1, b = 2r+1 = 3, n = 2s = 2, m = a + b = 4 and we have the solution 2⁴ + 3² = 5².   

Problem 2

The triangle ABC has sides a, b, c, and the triangle UVW has sides u, v, w such that a² = u(v + w - u), b² = v(w + u - v), c² = w(u + v - w). Show that ABC must be acute angled and express the angles U, V, W in terms of the angles A, B, C.

Solution

Thanks to Arne Smeets

a² + b² - c² = u(v + w - u) + v(w + u - v) - w(u + v - w) = w² - (u - v)² = 2uv(1 - cos W) > 0. Similarly, b² + c² - a² > 0, and c² + a² - b² > 0. So ABC is acute-angled.

We have ab = √(uv(w² - u² - v² + 2uv) ) = uv √( 2(1 - cos W) ), so cos C = (a² + b² - c²)/2ab = √ ((1 - cos W)/2 ) = sin W/2 = cos(90^o - W/2). Thus W = 180^o - 2C. Similarly for the other angles. 

Problem 3

Thanks to Arne Smeets

The circles C and C' lie inside the circle S. C and C' touch each other externally at K and touch S at A and A' respectively. The common tangent to C and C' at K meets S at P. The line PA meets C again at B, and the line PA' meets C' again at B'. Show that BB' is a common tangent to C and C'.

Solution

 

Let the centers of C, C' be O, O'. We have PA·PB = PK2 = PA'·PB', so triangles PAA' and PB'B are similar. Hence ∠PA'A = ∠PBB'. Let W be the center of S. A homothecy center A takes O to W, C to S and B to P, so ∠BOA = ∠PWA. Hence ∠PBB' = ∠PA'A = ½ ∠PWA = ½ ∠BOA = ∠BOM, where M is the midpoint of AB.

∠OMB = 90^o, so ∠OBB' = 180^o - ∠PBB' - ∠MBO = 90^o. Hence BB' is tangent to C. Similarly, it is tangent to C'. 

Problem 4

Find all positive real solutions to w + x + y + z = 12, wxyz = wx + wy + wz + xy + xz + yz + 27.

Answer

w = x = y = z = 3 is the only solution.

Solution

Put p = √(wxyz). By AM/GM applied to wx, wy, wz, xy, xz, yz we have p ≤ (wx + wy + wz + xy + xz + yz)/6 = p²/6 - 27/6, so (p - 3)² ≥ 36. Hence |p - 3| ≥ 6. But p is positive, so p ≥ 9. Hence wxyz ≥ 81. But applying AM/GM to w, x, y, z we have √p ≤ 3, with equality iff w = x = y = z. So wxyz ≤ 81 with equality iff w = x = y = z.

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