第37届全国高中数学联赛加试第2题的复数法解答

预估时间：30min，加上写过程45min吧。几何法难度未知，不过联赛题应该不会难吧。联赛不推荐复数法。

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\centerline{\textbf{\st 第37届全国高中数学联赛加试第2题的复数法解答}}

\par$\bm{1}$．如图所示，在$\triangle ABC$中$M$是边$AC$的中点，$DE$是$\triangle ABC$的外接圆在点$A$处的切线上的两点，满足$MD\parallel AB$，且$A$是线段$DE$的中点，过$A$、$B$、$E$三点的圆与边$AC$相交于另一点$P$，过$A$、$D$、$P$三点的圆与$DM$的延长线相交于点$Q$\per 证明：$\angle BCQ=\angle BAC$\per（2021-联赛-13）

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\par\noindent\textbf{证明}\quad 本解答中，点与其所对应复数用同一字母表示，单位圆指圆$\abs Z=1$．

\kaishu

\par 引理1\quad$X$是复平面单位圆上一点\per 则单位圆$X$处的切线上的点$T$满足$$T+X^2\overline{T}=2X\mbox{\per}$$

\par 引理2\quad 复平面上$\triangle UVW$与$\triangle XYZ$顺向相似\per 则$$UY+VZ+WX=UZ+VX+WY\mbox{\per}$$

\par 引理3\quad 复平面上四点$W$、$X$、$Y$、$Z$共圆\per 则$$\frac{\left(W-Y\right)\left(X-Z\right)}{\left(W-Z\right)\left(X-Y\right)}\in\mathbb{R}\mbox{\per}$$

\songti

\par 引理的证明略\per

\par 回到原题\per

\par 以$\triangle ABC$的外接圆圆心为原点，半径为单位长建立复平面\per

\par $M=\frac{A+C}2$\per

\par $\because D$在$\triangle ABC$的外接圆$A$处的切线上\per

\par $\therefore$由{\kaishu 引理1}，$D+A^2\ov{D}=2A$\per (*)

\par $\because DM\parallel AB$\per

\par $\therefore\frac{D-M}{A-B}\in\mathbb R$\per

\par $\therefore\frac{\displaystyle D-\frac{A+C}2}{B-A}=\ov{\left(\frac{\displaystyle D-\frac{A+C}2}{B-A}\right)}=\frac{\displaystyle ABC\ov D-\frac{B\left(A+C\right)}2}{C\left(A-B\right)}$\per

\par $\therefore 2C\left(D+AB\ov D\right)=\left(A+C\right)\left(B+C\right)$\per ($**$)

\par 同理，$2C\left(D+AB\ov D\right)=\left(A+C\right)\left(B+C\right)$\per ($*$$*$$*$)

\par 联立($*$)、($**$)，解得$D=\frac{A\left(C^2+AC+AB-3BC\right)}{2C\left(A-B\right)}$\per

\par 又$\because A$为$DE$中点\per

\par $\therefore E=2A-D=\frac{A\left(3AC-BC-C^2-AB\right)}{2C\left(A-B\right)}$\per

\par $\because\angle BEA=\angle BPC$，$\angle EAB=\angle ACB=\angle PCB$\per

\par $\therefore\triangle BEA$与$\triangle BPC$顺向相似\per

{\fs\par 这个相似看不出来就自己算共圆去吧，我也没办法\per}

\par $\therefore$由{\kaishu 引理2}，$BP+EC+AB=BC+EB+AP$\per

\par 因此

\begin{align*}

P&=\frac{BC+EB-AB-EC}{B-A}=\frac{\displaystyle BC-AB+\frac{A\left(3AC-BC-C^2-AB\right)}{2C\left(A-B\right)}\left(B-C\right)}{B-A}\\

&=\frac{2B^2C^2-2ABC^2-AB^2C-2A^2BC+3A^2C^2-AC^3+A^2B^2}{2C\left(B-A\right)^2}\mbox{\per}

\end{align*}

\par 因此

\begin{align*}

P-D=&\frac{2B^2C^2-2ABC^2-AB^2C-2A^2BC+3A^2C^2-AC^3+A^2B^2}{2C\left(B-A\right)^2}\\

&-\frac{A\left(C^2+AC+AB-3BC\right)}{2C\left(A-B\right)}\\

=&\frac{\left(A-C\right)^2\left(2B^2-AB-AC\right)}{2C\left(B-A\right)^2}\mbox{\per}

\end{align*}

\par $\because Q$、$P$、$A$、$D$共圆\per

\par $\therefore$由{\kaishu 引理3}，$\frac{\left(Q-A\right)\left(P-D\right)}{\left(Q-D\right)\left(P-A\right)}\in\mathbb{R}$\per

\par 又$\because QD\parallel AB$，$A$、$P$、$C$共线\per

\par $\therefore\frac{\left(Q-A\right)\left(P-D\right)}{\left(B-A\right)\left(C-A\right)}=\frac{\left(Q-A\right)\left(P-D\right)}{\left(Q-D\right)\left(P-A\right)}\frac{Q-D}{B-A}\frac{P-A}{C-A}\in\mathbb{R}$\per

\par 代入$P-D$，有$\frac{\left(Q-A\right)\left(C-A\right)\left(2B^2-AB-AC\right)}{C\left(B-A\right)^3}=\frac{B\left(A\ov Q-1\right)\left(A-C\right)\left(2AC-BC-B^2\right)}{C\left(A-B\right)^3}$\per

\par 与($*$$*$$*$)联立，解得$Q=\frac{-AB^2+ABC-2AC^2+B^2C+BC^2}{2C\left(B-A\right)}$\per

\par $\therefore Q-C=\frac{-AB^2+ABC-2AC^2+B^2C+BC^2}{2C\left(B-A\right)}-C=\frac{B\left(B-C\right)\left(C-A\right)}{2C\left(B-A\right)}$\per

\par $\therefore\left.\frac{B-C}{Q-C}\middle/\frac{C-A}{B-A}\right.=\frac{2C\left(B-A\right)^2}{B\left(C-A\right)^2}\in\mathbb R$\per

\par $\therefore\angle BCQ=\angle BAC$\per 证毕\per

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