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Poncelet 定理: 圆和抛物线 `zhangboxin|20180409` - 语时lab
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<h1>Poncelet 定理: 圆和抛物线 `zhangboxin|20180409`</h1>
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<span class="date">2018/04/16</span>
<span>posted in </span>
<span class="posted-in"><a href='problemsolving.html'>解题</a></span>
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<p>已知圆 \(C_1 \colon x^2+y^2 = 1\) 和抛物线 \(C_2 \colon y = x^2 - 2\). \(P,Q,R\) 是抛物线 \(C_2\) 上的三个不同的点, 且直线 \(PQ\) 和 \(PR\) 都是圆 \(C_1\) 的切线. 求证: \(QR\) 也是圆 \(C_1\) 的切线.</p>
<span id="more"></span><!-- more -->
<blockquote>
<p>Qer: zhangboxin 20180409</p>
</blockquote>
<h2 id="toc_0">解答</h2>
<p>首先给出如下引理 (证明放在最后面): </p>
<blockquote>
<p>对于抛物线 \(C_2 \colon y = x^2 - 2\) 上的两个不同的点 \(M(s, s^2 - 2)\) 和 \(N(t, t^2-2)\) (其中 \(s \ne t\)), <br/>
直线 \(MN\) 与圆 \(C_1 \colon x^2+y^2 = 1\) 相切的充分必要条件是<br/>
\[<br/>
s^2 t^2 - s^2 - t^2 + 2st + 3 = 0<br/>
\]</p>
</blockquote>
<p><strong>原题证明如下:</strong></p>
<p>设 \(P,Q,R\) 三点的坐标分别为 \(P(p, p^2 - 2)\), \(Q(q, q^2 - 2)\) 和 \(R(r, r^2 - 2)\) (其中 \(p,q,r\) 两两互不相等).</p>
<p>则由 \(PQ,PR\) 与圆 \(C_1\) 相切得\[<br/>
\begin{cases}<br/>
p^2 q^2 - p^2 - q^2 + 2pq + 3 = 0 \\<br/>
p^2 r^2 - p^2 - r^2 + 2pr + 3 = 0<br/>
\end{cases}<br/>
\]<br/>
将这两个等式变形为<br/>
\[<br/>
\begin{cases}<br/>
(p^2 - 1) q^2+ 2pq - (p^2 - 3) = 0 \\<br/>
(p^2 - 1) r^2+ 2pr - (p^2 - 3) = 0<br/>
\end{cases}<br/>
\]<br/>
由于 \(q \ne r\), 故可将 \(q,r\) 看做二次方程<br/>
\[<br/>
(p^2 - 1) x^2+ 2px - (p^2 - 3) = 0 <br/>
\]<br/>
的两根, 因此由韦达定理<br/>
\[<br/>
\begin{cases}<br/>
q+r = -\dfrac{2p}{p^2-1} \\<br/>
qr = -\dfrac{p^2-3}{p^2-1}<br/>
\end{cases}<br/>
\]<br/>
于是<br/>
\[<br/>
\begin{aligned}<br/>
&\phantom{{}={}} q^2 r^2 - q^2 - r^2 + 2qr + 3 \\<br/>
&= (qr)^2 - (q+r)^2 + 4qr + 3 \\<br/>
&= \left( -\dfrac{p^2-3}{p^2-1} \right)^2 - \left( -\dfrac{2p}{p^2-1} \right)^2 + 4 \left( -\dfrac{p^2-3}{p^2-1} \right) + 3 \\<br/>
&= \cdots \\<br/>
&= 0<br/>
\end{aligned}<br/>
\]<br/>
因此根据引理知, 直线 \(QR\) 与圆 \(C_1\) 相切.</p>
<h3 id="toc_1">引理的证明</h3>
<p>对于抛物线 \(C_2 \colon y = x^2 - 2\) 上的两个不同的点 \(M(s, s^2 - 2)\) 和 \(N(t, t^2-2)\) (其中 \(s \ne t\)), 直线 \(MN\) 的方程为<sup id="fnref1"><a href="#fn1" rel="footnote">1</a></sup><br/>
\[<br/>
MN \colon (s+t)x - y - (st+2) = 0<br/>
\]</p>
<p>直线 \(MN\) 与圆 \(C_1 \colon x^2+y^2 = 1\) 相切的充分必要条件是<sup id="fnref2"><a href="#fn2" rel="footnote">2</a></sup><br/>
\[<br/>
\dfrac{|st+2|}{\sqrt{(s+t)^2+1}} = 1<br/>
\]</p>
<p>这个关系式可等价的变形为<br/>
\[<br/>
s^2 t^2 - s^2 - t^2 + 2st + 3 = 0<br/>
\]</p>
<p>即证.</p>
<div class="footnotes">
<hr/>
<ol>
<li id="fn1">
<p>写出两点式化简, 即可得到. <a href="#fnref1" rev="footnote">↩</a></p>
</li>
<li id="fn2">
<p>即圆心 \(O(0,0)\) 到直线 \(MN\) 的距离等于半径 \(1\). <a href="#fnref2" rev="footnote">↩</a></p>
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