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National mathematical Olympiad Final round

Geometry Difficulty 6.3 National Olympiad Prove it Bulgaria

The incircle of $\triangle ABC$ has center $I$ and touches the sides $BC$ , $AC$ and $AB$ at points $A_1$ , $B_1$ and $C_1$ , respectively. An arbitrary line $\ell$ through $I$ is given and the points $A'$ , $B'$ and $C'$ are symmetric to $A_1$ , $B_1$ and $C_1$ , respectively, with respect to $\ell$ . Prove that the lines $AA'$ , $BB'$ and $CC'$ are concurrent.

Solution

Denote by $d_c(X)$ the distance from the point $X$ to the line $AB$ and analogously for the lines $BC$ and $CA$ . It is not difficult to see that the Sine version of the Ceva theorem implies that the equality
$\frac{d_b(A')}{d_c(A')} \cdot \frac{d_c(B')}{d_a(B')} \cdot \frac{d_a(C')}{d_b(C')} = 1$
is necessary and sufficient for the lines $AA'$ , $BB'$ and $CC'$ to be concurrent.

Note that $B_1A' = A_1B'$ . Moreover, since the lines $CB$ and $CA$ are tangent to the incircle, we have $\angle B'A_1B = \frac{1}{2} B'A_1 = \frac{1}{2} A'B_1 = \angle A'B_1C$ . Then $d_a(B') = A_1B' \sin \angle B'A_1B = B_1A' \sin \angle A'B_1C = d_b(A')$ . We analogously obtain $d_b(C') = d_c(B')$ and $d_c(A') = d_a(C')$ which completes the proof.

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