IMO 1999 SL G4

Origin: GBR | Category: Geometry

Problem

For a triangle T = ABC we take the point X on the side (AB) such that AX/XB = 4/5, the point Y on the segment (CX) such that CY = 2Y X, and, if possible, the point Z on the ray (CA such that ∡CXZ = 180◦−∡ABC. We denote by \Sigma the set of all triangles T for which ∡XY Z = 45◦. Prove that all the triangles from \Sigma are similar and find the measure of their smallest angle.

Solution

We use the following lemma. Lemma. Let ABC be a triangle and X \inAB such that −−\to AX : −−\to XB = m : n. Then (m + n) cot \angleCXB = n cot A −m cot B and m cot \angleACX = (n + m) cot C + n cot A. Proof. Let CD be the altitude from C and h its length. Then using oriented segments we have AX = AD + DX = h cot A −h cot \angleCXB and BX = BD + DX = h cot B + h cot \angleCXB. The first formula in the lemma now follows from n \cdot AX = m \cdot BX. The second formula immediately follows from the first part applied to the triangle ACX and the point X′ \inAC such that XX′ \parallelBC. Let us set cot A = x, cot B = y, and cot C = z. Applying the second for- mula in the lemma to \triangleABC and the point X, we obtain 4 cot \angleACX = 9z+5x. Applying the first formula in the lemma to \triangleCXZ and the point Y and using \angleXY Z = 45◦and cot \angleCXZ = −y, we obtain 3 cot \angleXY Z = cot \angleACX −2 cot \angleCXZ = 9z+5x

• 2y ⇒5x + 8y + 9z = 12. We now use the well-known relation for cotangents of a triangle xy +yz + xz = 1 to get 9 = 9(x + y)z + 9xy = (x + y)(12 −5x −8z) + 9xy = 9 ⇒ (4y + x −3)2 + 9(x −1)2 = 0 ⇒x = 1, y = 1 2, z = 1

3. It follows that x, y, and z have fixed values, and hence all triangles T in \Sigma are similar, with their smallest angle A having cotangent 1 and thus being equal to \angleA = 45◦.