16th USAMO 1987 Problem 2 The feet of the angle bisectors of the triangle ABC form a right-angled triangle. If the right-angle is at X, where AX is the bisector of angle A, find all possible values for angle A. Solution
: 120 o . Thanks to Richard Rusczyk for this vector solution Use vectors origin A. Write the vector AB as B etc. Using the familiar BX/CX = AB/AC etc, we have Z = b B /(a+b), Y = c C /(a+c), X = (b B +c C )/(b+c). Hence Z - X = b B (c-a)/((a+b)(b+c)) - c C /(b+c), Y - X = - b B /(b+c) + c C (b-a)/((a+c)(b+c)). We have ( Z - X ) . ( Y - X ) = 0, so after multiplying through by (a+b)(a+c)(b+c) 2 , we get b 2 B 2 (a 2 -c 2 ) + c 2 C 2 (a 2 -b 2 ) + 2bc B.C (a 2 +bc) = 0. But B 2 = c 2 , C 2 = b 2 , so bc(2a 2 -b 2 -c 2 ) + 2 B.C (a 2 +bc) = 0. But a 2 = b 2 + c 2 - 2 B.C (cosine rule), so 2a 2 -b 2 -c 2 = a 2 -2 B.C . Hence bc(a 2 -2 B.C ) + 2 B.C (a 2 +bc) = 0, so B.C = -bc/2. Hence cos A = -1/2, so ∠A = 120 o . One can get there by straight algebra, but my first attempt was a horrendous slog: Put a = BC, b = CA, c = AB as usual. We have XY 2 = CX 2 + CY 2 - 2 CX.CY cos C. Using the fact that CY/AY = BC/AB, we have CY = ab/(a + c). Similarly, CX = ab/(b + c). Also 2ab cos C = (a 2 + b 2 - c 2 ). So XY 2 = (ab) 2 /(b+c) 2 + (ab) 2 /(a+c) 2 - ab(a 2 + b 2 - c 2 )/( (b+c)(a+c) ). So (a+b) 2 (b+c) 2 (c+a) 2 XY 2 = (ab) 2 (a+b) 2 (c+a) 2 + (ab) 2 (a+b) 2 (b+c) 2 - ab(a 2 +b 2 -c 2 )(a+b) 2 (b+c)(c+a) = -a 6 bc - a 5 bc 2 - a 5 b 2 c + 2a 4 b 3 c + a 4 b 2 c 2 + a 4 bc 3 + 2a 3 b 4 c + 4a 3 b 3 c 2 + 3a 3 b 2 c 3 + a 3 bc 4 - a 2 b 5 c + a 2 b 4 c 2 + 3a 2 b 3 c 3 + 2a 2 b 2 c 4 - ab 6 c - ab 5 c 2 + ab 4 c 3 + ab 3 c 4 . Similarly, we get (a+b) 2 (b+c) 2 (c+a) 2 ZX 2 = (ac) 2 (b+c) 2 (c+a) 2 + (ac) 2 (a+b) 2 (c+a) 2 - ac(a 2 +c 2 -b 2 )(a+b) (b+c)(c+a) 2 = -a 6 bc - a 5 b 2 c - a 5 bc 2 + a 4 b 3 c + a 4 b 2 c 2 + 2a 4 bc 3 + a 3 b 4 c + 3a 3 b 3 c 2 + 4a 3 b 2 c 3 + 2a 3 bc 4 + 2a 2 b 4 c 2 + 3a 2 b 3 c 3 + a 2 b 2 c 4 - a 2 bc 5 + ab 4 c 3 + ab 3 c 4 - ab 2 c 5 - abc 6 . Similarly, (a+b) 2 (b+c) 2 (c+a) 2 YZ 2 = (bc) 2 (b+c) 2 (c+a) 2 + (bc) 2 (a+b) 2 (b+c) 2 - bc(b 2 +c 2 -a 2 )(b+c) 2 (a+b) (c+a) = a 4 b 3 c + 2a 4 b 2 c 2 + a 4 bc 3 + a 3 b 4 c + 3a 3 b 3 c 2 + 3a 3 b 2 c 3 + a 3 bc 4 - a 2 b 5 c + a 2 b 4 c 2 + 4a 2 b 3 c 3 + a 2 b 2 c 4 - a 2 bc 5 - ab 6 c - ab 5 c 2 + 2ab 4 c 3 + 2ab 3 c 4 - ab 2 c 5 - abc 6 . But YZ 2 - XY 2 - ZX 2 = 0. So 2a 6 bc + 2a 5 b 2 c + 2a 5 bc 2 - 2a 4 b 3 c - 2a 4 bc 3 - 2a 3 b 4 c - 4a 3 b 3 c 2 - 4a 3 b 2 c 3 - 2a 3 bc 4 - 2a 2 b 4 c 2 - 2a 2 b 3 c 3 - 2a 2 b 2 c 4 = 0. Factorising, we get 2a 2 bc(ab + bc + ca)(a 2 - b 2 - c 2 - bc) = 0. Hence a 2 = b 2 + c 2 + bc. But a 2 = b 2 + c 2 - 2bc cos A. Hence cos A = -1/2, so A = 120 0 . Indeed we have established that this is a necessary and sufficient condition for ∠YXZ = 90 o . Does anyone have a geometric solution? 16th USAMO 1987 © John Scholes jscholes@kalva.demon.co.uk 26 Aug 2002 Last corrected/updated 4 Feb 04