INMO 2008 problem 1

Let ”ABC” be a triangle, ”I” its in-centre; $$A_1, B_1, C_1$$ be the reﬂections 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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