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
Answer: 120o.
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 = bB/(a+b), Y = cC/(a+c), X = (bB+cC)/(b+c). Hence Z-X = bB(c-a)/((a+b)(b+c)) - cC/(b+c), Y-X = -bB/(b+c) + cC(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 b2B2(a2-c2) + c2C2(a2-b2) + 2bcB.C (a2+bc) = 0. But B2 = c2, C2 = b2, so bc(2a2-b2-c2) + 2B.C (a2+bc) = 0.
But a2 = b2 + c2 - 2B.C (cosine rule), so 2a2-b2-c2 = a2-2B.C. Hence bc(a2-2B.C) + 2B.C(a2+bc) = 0, so B.C = -bc/2. Hence cos A = -1/2, so ∠A = 120o.
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 XY2 = CX2 + CY2 - 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 = (a2 + b2 - c2). So XY2 = (ab)2/(b+c)2 + (ab)2/(a+c)2 - ab(a2 + b2 - c2)/( (b+c)(a+c) ). So (a+b)2(b+c)2(c+a)2 XY2 = (ab)2(a+b)2(c+a)2 + (ab)2(a+b)2(b+c)2 - ab(a2+b2-c2)(a+b)2(b+c)(c+a) = -a6bc - a5bc2 - a5b2c + 2a4b3c + a4b2c2 + a4bc3 + 2a3b4c + 4a3b3c2 + 3a3b2c3 + a3bc4 - a2b5c + a2b4c2 + 3a2b3c3 + 2a2b2c4 - ab6c - ab5c2 + ab4c3 + ab3c4.
Similarly, we get (a+b)2(b+c)2(c+a)2ZX2 = (ac)2(b+c)2(c+a)2 + (ac)2(a+b)2(c+a)2 - ac(a2+c2-b2)(a+b)(b+c)(c+a)2 = -a6bc - a5b2c - a5bc2 + a4b3c + a4b2c2 + 2a4bc3 + a3b4c + 3a3b3c2 + 4a3b2c3 + 2a3bc4 + 2a2b4c2 + 3a2b3c3 + a2b2c4 - a2bc5 + ab4c3 + ab3c4 - ab2c5 - abc6.
Similarly, (a+b)2(b+c)2(c+a)2YZ2 = (bc)2(b+c)2(c+a)2 + (bc)2(a+b)2(b+c)2 - bc(b2+c2-a2)(b+c)2(a+b)(c+a) = a4b3c + 2a4b2c2 + a4bc3 + a3b4c + 3a3b3c2 + 3a3b2c3 + a3bc4 - a2b5c + a2b4c2 + 4a2b3c3 + a2b2c4 - a2bc5 - ab6c - ab5c2 + 2ab4c3 + 2ab3c4 - ab2c5 - abc6.
But YZ2 - XY2 - ZX2 = 0. So 2a6bc + 2a5b2c + 2a5bc2 - 2a4b3c - 2a4bc3 - 2a3b4c - 4a3b3c2 - 4a3b2c3 - 2a3bc4 - 2a2b4c2 - 2a2b3c3 - 2a2b2c4 = 0.
Factorising, we get 2a2bc(ab + bc + ca)(a2 - b2 - c2 - bc) = 0. Hence a2 = b2 + c2 + bc. But a2 = b2 + c2 - 2bc cos A. Hence cos A = -1/2, so A = 1200. Indeed we have established that this is a necessary and sufficient condition for ∠YXZ = 90o.
Does anyone have a geometric solution?
© John Scholes
jscholes@kalva.demon.co.uk
26 Aug 2002
Last corrected/updated 4 Feb 04