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Geometry Difficulty 8.5 Shortlist Prove it Hong Kong

In triangle $ABC$ with incentre $I$ , let $M_A, M_B$ and $M_C$ be the midpoints of $BC, CA$ and $AB$ respectively, and $H_A, H_B$ and $H_C$ be the feet of altitudes from $A, B$ and $C$ to the respective sides. Denote by $\ell_b$ the line being tangent to the circumcircle of triangle $ABC$ and passing through $B$ , and denote by $\ell'_b$ the reflection of $\ell_b$ in $BI$ . Let $P_B$ be the intersection of $M_A M_C$ and $\ell_b$ , and let $Q_B$ be the intersection of $H_A H_C$ and $\ell'_b$ . Define $\ell_c, \ell'_c, P_C$ and $Q_C$ analogously. If $R$ is the intersection of $P_B Q_B$ and $P_C Q_C$ , prove that $RB = RC$ .

Solution

Firstly, since $M_C M_A \parallel AC$ , we have $\angle M_C BP_B = \angle ACB = \angle M_C M_A B$ so that $\ell'_b$ is tangent to $(BM_A M_C)$ . This implies $P_B B^2 = P_B M_C \times P_B M_A$ , and hence $P_B$ lies on the radical axis of $B$ and the nine-point circle $\Gamma$ of $\triangle ABC$ .

Secondly, we have $\angle BH_C H_A = \angle ACB = \angle ABP_B = \angle Q_B BH_A$ so that $\ell'_b$ is tangent to $(BH_A H_C)$ . This implies $Q_B B^2 = Q_B H_A \times Q_B H_C$ , and hence $Q_B$ lies on the radical axis of $B$ and $\Gamma$ .

Therefore, $P_B Q_B$ is the radical axis of $B$ and $\Gamma$ . Similarly, $P_C Q_C$ is the radical axis of $C$ and $\Gamma$ . So $R$ is the radical centre of $B, C$ and $\Gamma$ . Thus, we have $RB^2 = RC^2$ , which implies $RB = RC$ as desired.

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