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Let ''ABC'' be a triangle, ''I'' its in-centre; $ A_1, B_1, C_1 $ be the reď¬‚ections of I in BC, CA, AB respectively. Suppose the circumcircle of triangle $ A_1 B_1 C_1 $ passes through A. Prove that $ B_1, C_1, I, I_1 $ are concyclic, where $ I_1 $ is the incentre of triangle $ A_1 B_1 C_1 $.

**Basic Sketch:**

**Hint 1:** I is indeed the circumcenter of $ A A_1 B_1 C_1 $ with circum radius = 2r. $ B_1 C_1 $ is the radical axis of the two circles concerned hence the other center has to lie of IA (since IA is perpendicular to radical axis and I is one of the centers hence IA is the line joining the centers).

**Hint 2:** We prove that A is the center of the circle $ B_1, C_1, I, I_1 $ . Using cosine we show that $ \Delta AIH $ is a 30-60-90 triangle as IH = r and IA = 2r. This implies $ \angle B_1 I C_1 $ is $ 120^o $ . Hence it is sufficient to show $ \angle B_1 I_1 C_1 $ is $ 120^o $ . A simple angle chasing in triangle $ A_1 I_1 C_1 $ and $ B_1 I_1 C_1 $ completes the proof (we observe that $ \angle A_1 C_1 B_1 = \frac {A+B} {2} $ and similarly with the other angles).

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