## IGO 2016 Medium/5

For discussing Olympiad level Geometry Problems
Thamim Zahin
Posts: 98
Joined: Wed Aug 03, 2016 5:42 pm

### IGO 2016 Medium/5

5. Let the circles $w$ and $w'$ intersect in points $A$ and $B$. Tangent to circle $w$ at $A$ intersects $w'$ in $C$ and tangent to circle $w'$ at $A$ intersects $w$ in $D$. Suppose that the internal bisector of $\angle CAD$ intersects $w$ and $w'$ at $E$ and $F$, respectively, and the external bisector of $\angle CAD$ intersects $w$ and $w'$ in $X$ and $Y$, respectively. Prove that the perpendicular bisector of $XY$ is tangent to the circumcircle of triangle $BEF$.
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tafhim01
Posts: 3
Joined: Mon Aug 17, 2015 4:26 am

### Re: IGO 2016 Medium/5

[AST: Alternate Segment Theorem]

We assume that, WLOG radius of $\omega$ is less than radius of $\omega'$(equality case is trivial).

Let $P$ be the midpoint of $\widehat{EF}$ not containing $B$ of $\odot BEF$, thus $BP$ is the angle bisector of $\angle EBF$. Let the tangent at $P$ to $BEF$ intersect $XY$ at $M$.

By AST and similarity,
$\angle ABD = \angle ABC = \angle B.$
Now,
$\angle MPE = \angle PFE = \angle PEF \Rightarrow PM \parallel EF.$
Now,
$EF \perp XY \Rightarrow PM \perp XY.$
Extend $\overrightarrow{CA}$ to $C'$.
$\angle DAX = \angle C'AX \text{ [External angle bisector]}$
Again, by AST,
$\angle ABX = \angle C'AX.$
So,
\begin{align*} &\angle ABX = \angle DAX = \angle DBX \\ &\\ \Rightarrow &X\text{ is the midpoint of }\widehat{AD}\text{ not containing B.}\\ &\text{Similarly, }Y\text{ is the midpoint of arc }AC. \end{align*}
Now,
\begin{align*} &\angle ABE = \angle ADE = \angle CAF = \angle CBF\\ &\\ \Rightarrow &\angle EBF = \angle ABC - \angle ABE + \angle CBF = \angle ABC = \angle B. \end{align*}
So,
$\angle EFP = \frac{\angle B}{2}$
But,
$\angle AFC = \angle ABC = \angle B.$
So, $PF$ bisects $\angle AFC \Rightarrow Y, P, F$ are collinear. Similarly, $P, E, X$ are collinear.

By simple angle chasing,
\begin{align*} &\triangle ADE \sim \triangle CAF\\ &\\ \Rightarrow &\angle ACF = \angle ADE. \end{align*}
Now,
\begin{align*} &\angle PYX = \angle ACF = \angle ADE = \angle PXY.\\ &\\ \Rightarrow &\triangle PXY \text{ is isosceles.}\\ \Rightarrow &PM \text{ is the perpendicular bisector of } XY\text{ , and also tangent to } BEF\text{ at }P. \begin{align*} &\\ &\\ &\square \end{align*} \end{align*}