ABC is an acute-angled triangle. There are points A1, A2 on the side BC, B1 and B2 on the side CA, and C1, C2 on the side AB such that the points are in the order: A, C1, C2, B; B, A1, A2, C; and C, B1, B2, A. Also angle AA1A2 = angle AA2A1 = angle BB1B2 = angle BB2B1 = angle CC1C2 = angle CC2C1. The three lines AA1, BB1 and CC1 meet in three points and the three lines AA2, BB2, CC2 meet in three points. Show that all six points lie on a circle.
Solution
Let AA1 and CC1 meet at U, AA1 and BB1 at V, BB1 and CC1 at W. Let AA2 and BB2 meet at X, BB2 and CC2 at Y, and CC2 and AA2 at Z.
The key is to show that AV = AZ and AU = AX (and similarly BW = BX, BV = BY etc). It follows that UX is parallel to BC and WX is parallel to CA. Hence ∠AUX = ∠AA1A2 (parallel) = ∠BB1B2 (given) = ∠BWX (parallel) = ∠VWX (same angle). So UXWV is cyclic. In other words, X lies on the circle UVW. But similarly Y and Z lie on the circle. Hence result.
AB2B and AC1C are similar, so ∠ABB2 = ∠ACC1. Similarly, ∠BAA1 = ∠BCC2. So ∠A1VB = ∠VAB + ∠VBA = ∠A1AB + (∠B1BB2 + ∠B2BA) = ∠BCC2 + ∠B1BB2 + ∠ACC1 = ∠BCC2 + ∠C2CC1 + ∠ACC1 = ∠C. Similarly, ∠A2ZC = ∠B.
Hence AV/sin ABV = AB/sin AVB (sine rule) = AB/sin A1VB = AB/sin C = AC/sin B (sine rule) = AC/sin A2ZC = AC/sin AZC = AZ/sin ACZ. But triangles ABB1 and ACC2 are similar, so ∠ABV = ∠ABB1 = ∠ACC2 = ∠ACZ. Hence AV = AZ.
We have AU/sin AC1U = AC1/sin AUC1 = AC1/sin B (that angle AUC1 = angle B follows in a similar way to A1VB = angle C) = (AC1/AC) AC/sin B = (AB2/AB) AC/sin B (trianges AC1C and AB2B similar) = (AB2/AB) (AB/sin C) = AB2/sin C = AB2/sin AXB2 (again follows similarly to ∠A1VB = ∠C) = AX/sin AB2X. But ∠AC1U = ∠AB2X, so AU = AX.
© John Scholes
jscholes@kalva.demon.co.uk
20 Sep 2002