Let $\Gamma$ be a circle of center $O$, and $\delta$. be a line in the plane of $\Gamma$, not intersecting it. Denote by $A$ the foot of the perpendicular from $O$ onto $\delta$., and let $M$ be a (variable) point on $\Gamma$. Denote by $\gamma$ the circle of diameter $AM$ , by $X$ the (other than M ) intersection point of $\gamma$ and $\Gamma$, and by $Y$ the (other than $A$) intersection point of $\gamma$ and $\delta$. Prove that the line $XY$ passes through a fixed point.
Problem
Source: 2008 JBMO Shortlist G10
Tags: geometry, JBMO
RagvaloD
13.10.2017 02:58
Let $F$ - point of intersection of $AO$ and $XY$. We will prove, that $F$ - fixed point. Let $E$ - point of intersection of $AX$ and $\Gamma$ $\angle AXM=90$ so $EM$ is diameter of $\Gamma$ Let $L$ - point of intersection of $AO$ and $\gamma$. Easy to show, that $AL \perp AO$ and so $AY=AL$ Let $r$ -radius of $\Gamma$,$OA=a,OL=b$. $\angle MXA=90$ so $\triangle MXA \sim \triangle YAF$ $AF=\frac{AX*AY}{XM}=\frac{AX*AE*AY}{XM*AE}=\frac{(a^2-r^2)AY}{2S_{AME}}= \frac{(a^2-r^2)AY}{EM*AM *\sin \angle EMA}=\frac{(a^2-r^2)AY}{2r*AM *\sin \angle EMA}$ We will prove, that $\frac{AY}{AM *\sin \angle EMA}$ is const $AY^2=AL^2=r^2-b^2$ $MA^2=AL^2+LM^2=(a-b)^2+r^2-b^2=a^2-2ab+r^2$ $EA^2=ML^2+(AO+OL)^2=r^2-b^2+(a+b)^2=a^2+2ab+r^2$ $\cos \angle EMA = \frac{EM^2+MA^2-EA^2}{2EM*MA}=\frac{4r^2+a^2-2ab+r^2-a^2-2ab-r^2}{4r\sqrt{(a^2-2ab+r^2)}}=\frac{r^2-ab}{r\sqrt{(a^2-2ab+r^2)}}$ $\sin^2 \angle EMA = 1-(\frac{r^2-ab}{r\sqrt{(a^2-2ab+r^2)}})^2 = \frac{r^2(a^2-2ab+r^2)-(r^2-ab)^2}{r^2(a^2-2ab+r^2)}=\frac{r^2a^2-a^2b^2}{r^2(a^2-2ab+r^2)}$ So $\frac{AY}{AM *\sin \angle EMA} = \frac{\sqrt{r^2-b^2}}{\sqrt{a^2-2ab+r^2}\frac{a\sqrt{r^2-b^2}}{r\sqrt{a^2-2ab+r^2}}}=\frac{r}{a}$ And so $AF=\frac{(a^2-r^2)AY}{2r*AM *\sin \angle EMA}=\frac{(a^2-r^2)r}{2ra}=\frac{a^2-r^2}{2a}= const$