不定期传一些最近写的 MO 题.
-
如图,在锐角 \(\triangle ABC\) 中,\(O, H\) 分别是外心和垂心,\(K\) 是 \(AH\) 的中点,\(P\) 在 \(AC\) 上,且满足 \(\angle BKP = 90^\circ\).求证:\(OP \parallel BC\).
证明:
如图,作直线 \(BH\) 交 \(AC\) 于点 \(D\),连结 \(KD\);分别过 \(O, P\) 作 \(BC\) 的垂线交于 \(E, F\).
由 \(\angle BKP = \angle BDP = \angle BFP\),知 \(B, F, P, D, K\) 五点共圆.在 \(\text{Rt}\triangle HDA\) 中,\(KD = KH = \displaystyle\frac{AH}{2}\),\(\angle KDA = \angle KAD = 90^\circ - \angle C\),故 \(\angle KDP = 90^\circ + \angle C\).
又 \(\angle CPF = 90^\circ - \angle C\),知 \(\angle DPF = 90^\circ + \angle C = \angle KDP\),结合四点共圆,知四边形 \(PDKF\) 为等腰梯形.由此,\(PF = DK = \displaystyle\frac{AH}{2}\).又 \(OE = R \cdot \cos A, AH = AB \cdot \cos A \cdot \sec (90^\circ - C) = 2R \cos A \Rightarrow OE = \displaystyle\frac{AH}{2} = PF\),\(OE \perp BC, PF \perp BC \Rightarrow OE \parallel PF\),知四边形 \(OEFP\) 是平行四边形,于是 \(OP \parallel BC\),得证.