Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Prove it Taiwan

Let the incenter of triangle ABCABC be II, and let the excenter inside angle AA be JJ. Let AA′‾\overline{AA'} be a diameter of the circumcircle of △ABC\triangle ABC, and let the points H1,H2H_1, H_2 be the orthocenters of △BIA′\triangle BIA', △CJA′\triangle CJA' respectively. Prove that: H1H2H_1H_2 is parallel to BCBC.

Solution

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

平移 △CJA′∪H2\triangle CJA' \cup H_2 至 △C1BA1′∪H2′\triangle C_1BA'_1 \cup H'_2 (即使得 JJ 平移至與 BB 重合)。由 IJ‾\overline{IJ} 中點位於 BC‾\overline{BC} 的中垂線上, 可得 C1I⊥BCC_1I \perp BC。令 I′I' 為 II 關於 ⊙(BIA′)\odot(BIA') 的對徑點、C1′C'_1 為 C1C_1 關於 ⊙(C1BA1′)\odot(C_1BA'_1) 的對徑點。則由 BI′‾=H1A′‾\overline{BI'} = \overline{H_1A'}, BC′‾=H2A′‾\overline{BC'} = \overline{H_2A'}, 知 BH1A′I′BH_1A'I', BH2A′C1′BH_2A'C'_1 為平行四邊形。因此原命題等價於證明 I′C1′I'C'_1 平行於 BCBC。

Figure 1

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