对2022.07.21《我们爱几何》问题的证明
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2022.07.21《我们爱几何》问题是由严君啸命制的,问题如下:
问题:设$X,Y$是$\triangle ABC$外接圆$\odot O$与$A$-旁切圆$\odot I_A$的两个交点,$XX_1,YY_1$是$\odot I_A$的两条切线,且$X_1,Y_1$在$\odot O$上,$J$是弧$BAC$的中点. 求证:$\dfrac{AX}{AY}=\dfrac{JX_1}{JY_1}$.
Proof. 过$X$做$JX_1$平行线交$JY$于$Y_2$,过$Y$做$JY_1$平行线交$JX$于$X_2$,由对称性知$XYX_1Y_1$是等腰梯形,故$\measuredangle X_2YY_2=\measuredangle Y_1JY=\measuredangle XJX_1=\measuredangle X_2 XY_2$,故$X_2,Y_2,X,Y$共圆$c$. 由于$\measuredangle JY_2X_2=\measuredangle YXJ=90^\circ-\measuredangle OJY$,且$OJ\perp BC$,故$X_2Y_2//BC$. 由于$\measuredangle X_1XY_2=\measuredangle XX_1 J=90^\circ-\measuredangle OJX=\measuredangle JX_2Y_2$,故$XX_1$与$c$切于$X$,$YY_1$与$c$切于$Y$,于是$c$就是$\odot I_A$. 下面注意到由于$\measuredangle I_AAJ=90^\circ$,$A$在$\odot(AXY)$上且$X,Y,X_2,Y_2$共圆$\odot I_A$,故由Brocard定理知$A$是$XY,YY_2,X_2Y_2,XX_2$围成的完全四边形的Miquel点,故$\triangle AXX_2\stackrel{+}{\sim}\triangle AYY_2$. 设$XY_2$交$YX_2$于$Z$,故$\triangle ZXY$与$\triangle JX_1Y_1$位似,于是$\dfrac{AX}{AY}=\dfrac{XX_2}{YY_2}=\dfrac{XZ}{YZ}=\dfrac{JX_1}{JY_1}$. $\quad\Box$