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CMO 2018 题 3 - 语时lab
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<h1>CMO 2018 题 3</h1>
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<span class="date">2018/11/14</span>
<span>posted in </span>
<span class="posted-in"><a href='problemsolving.html'>解题</a></span>
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<p><mark>TODO: 微博图床已挂, 本篇缺图, 需要找回</mark></p>
<p>\(\triangle ABC\) 中, \(AB < AC\), \(O\) 为外心, \(D\) 是 \(\angle BAC\) 平分线上一点, \(E\) 在 \(BC\) 上, 满足 \(OE \parallel AD\), \(DE \perp BC\). 在射线 \(EB\) 上取点 \(K\) 满足 \(EK = EA\), \(\triangle ADK\) 外接圆与 \(BC\) 交于另一点 \(P \ne K\), \(\triangle ADK\) 外接圆与 \(\triangle ABC\) 外接圆交于另一点 \(Q \ne A\). 求证: \(PQ\) 与 \(\triangle ABC\) 外接圆相切.</p>
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<blockquote>
<p>题目来源: <a href="https://mp.weixin.qq.com/s/Lk68mFHaHnc0Rly4F2nEwQ">2018冬令营首日学生回忆版试题 (爱尖子发布)</a></p>
<p><img src="https://ws2.sinaimg.cn/large/006tNbRwly1fx7tzuhui2j30uk0i879u.jpg" alt=""/></p>
</blockquote>
<h2 id="toc_0">解答</h2>
<p>设 \(\triangle ABC\) 的外心为 \(O\), \(\triangle ADK\) 的外心为 \(O_1\). 连结 \(AO\), \(DO\), \(AO_1\), \(KO_1\) 和 \(EO_1\). 设 \(OD\) 与 \(BC\) 的交点为 \(F\).</p>
<p><img src="https://ws4.sinaimg.cn/large/006tNbRwly1fx7v0jv000j30hd0h7mye.jpg" alt=""/></p>
<p>不难证明 \(\angle OAD = \angle ADE = \dfrac{B-C}2\), 故 \(ADEO\) 是等腰梯形.</p>
<p>于是由 \(O_1A = O_1D\) 知 \(O_1O = O_1E\), 且 \(OD = AE = EK\). </p>
<p>因此, \(\triangle O_1KE \cong \triangle D_1DO\), 故 \(\angle KO_1D = \angle KFD = \angle EO_1O\).</p>
<p>因为 \(EO_1\) 垂直平分 \(AK\), 且 \(O_1O\) 垂直平分 \(AQ\), 所以 \(\angle EO_1O = \angle KAQ = 180^\circ - \angle KPQ\).</p>
<p>于是 \(\angle KFO = 180^\circ - \angle KFD = 180^\circ - \angle EO_1O = \angle KPQ\), 故有 \(OD \parallel PQ\). </p>
<p>另外, 由 \(\angle KFD = \angle KO_1D = 2 \angle KPD\) 知 \(\angle FPD = \angle FDP\), \(FP = FD\).</p>
<p>因此, 点 \(P\) 到 \(FD\) 的距离与点 \(D\) 到 \(FP\) 的距离相同, 都等于 \(DE = OA\). </p>
<p>即, 平行线 \(OD\) 与 \(PQ\) 之间的距离等于 \(\triangle ABC\) 外接圆 (圆 \(O\)) 的半径, 因此 \(PQ\) 与圆 \(O\) 相切.</p>
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