## IGO 2016 Medium/5

- 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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### 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*}

\]

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*}

\]