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Taiwan IMO Selection Camp

Geometry Difficulty 5.2 AIME, harder Prove it Taiwan

Let the incenter of triangle $ABC$ be $I$ , and let the excenter inside angle $A$ be $J$ . Let $\overline{AA'}$ be a diameter of the circumcircle of $\triangle ABC$ , and let the points $H_1, H_2$ be the orthocenters of $\triangle BIA'$ , $\triangle CJA'$ respectively. Prove that: $H_1H_2$ is parallel to $BC$ .

Solution

(∠ denotes directed angles, $\pm$ denotes direct similarity.)

平移 $\triangle CJA' \cup H_2$ 至 $\triangle C_1BA'_1 \cup H'_2$ (即使得 $J$ 平移至與 $B$ 重合)。由 $\overline{IJ}$ 中點位於 $\overline{BC}$ 的中垂線上, 可得 $C_1I \perp BC$ 。令 $I'$ 為 $I$ 關於 $\odot(BIA')$ 的對徑點、 $C'_1$ 為 $C_1$ 關於 $\odot(C_1BA'_1)$ 的對徑點。則由 $\overline{BI'} = \overline{H_1A'}$ , $\overline{BC'} = \overline{H_2A'}$ , 知 $BH_1A'I'$ , $BH_2A'C'_1$ 為平行四邊形。因此原命題等價於證明 $I'C'_1$ 平行於 $BC$ 。

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.