## For Raiyan

- ahmedittihad
**Posts:**161**Joined:**Mon Mar 28, 2016 6:21 pm

### For Raiyan

Let $\omega $ be the incircle of $\triangle ABC$. $X$, $Y$ lies on $AB$ and $AC$ such that $XY \parallel BC$ and $XY$ is tangent to $\omega$. Let $E$, $F$ be the touchpoint of $\omega$ with $AB$ and $AC$. $EF \cap BC = P$. $M$ is the midpoint of $XY$. Prove that $MP$ is tangent to $\omega$.

Frankly, my dear, I don't give a damn.

- Raiyan Jamil
**Posts:**138**Joined:**Fri Mar 29, 2013 3:49 pm

### Re: For Raiyan

We use complex numbers. Set the incircle as the unit circle. Let $D$ be the touchpoint of $\omega$ with $BC$ and $Q$ be a point on $\omega$ such that $QEDF$ is harmonic.Now, since $(P,D;B,C)=-1$, we get that $PQ$ is tangent to $\omega$.Now, since $XY||BC$ and $XY$ is tangent to $\omega$, let at $Z$, then we get that $D$ and $Z$ are diametrically opposite. Now, with the intersection formula, we get,

$Q=\frac{2ab-ac-bc}{a+b-2c}$

$X=\frac{2ac}{c-a}$

$Y=\frac{2bc}{c-b}$

$M=\frac{ac}{c-a}+\frac{bc}{c-b}$

Now,let the intersection of tangents at $Q$ and $Z$ meet at $M'$. Again using the intersection formula we get that $M'=\frac{ac}{c-a}+\frac{bc}{c-b}$. So, $M'$ coincides with $M$, which concludes the proof.

$Q=\frac{2ab-ac-bc}{a+b-2c}$

$X=\frac{2ac}{c-a}$

$Y=\frac{2bc}{c-b}$

$M=\frac{ac}{c-a}+\frac{bc}{c-b}$

Now,let the intersection of tangents at $Q$ and $Z$ meet at $M'$. Again using the intersection formula we get that $M'=\frac{ac}{c-a}+\frac{bc}{c-b}$. So, $M'$ coincides with $M$, which concludes the proof.

A smile is the best way to get through a tough situation, even if it's a fake smile.

- ahmedittihad
**Posts:**161**Joined:**Mon Mar 28, 2016 6:21 pm

### Re: For Raiyan

For those who love synthetic,

Let $\omega$ touch $BC$ at $D$ and $PQ$ touch $\omega$ at $D'$. Also let $AD \cap PQ = X$, and $AD \cap \omega$ for the second time at $Y$.

By homothety, we can say that $X$ is the touchpoint of the incircle of $\triangle APQ$ with $PQ$. As $\omega$ is the A-excirce of $\triangle APQ$, it's well known that $MD'=MX$. Now, $\angle D'YD = 90^{\circ} \Longrightarrow \angle D'YX = 90^{\circ}$. So, $M$ is the circumcenter of $\triangle D'XY$. So, $MD'=MY \Longrightarrow MY$ is tangent to $\omega$ at $Y$. Now, $P$ lies on the polar of $A$ $wrt$ $\omega$. From La Hire, $A$ lies on the polar of $P$ $wrt$ $\omega$. So, $AD$ is the polar of $P$. Implying that $PY$ is tangent to $\omega$ at $Y$.

We get our conclusion.

$Q.E.D$

Let $\omega$ touch $BC$ at $D$ and $PQ$ touch $\omega$ at $D'$. Also let $AD \cap PQ = X$, and $AD \cap \omega$ for the second time at $Y$.

By homothety, we can say that $X$ is the touchpoint of the incircle of $\triangle APQ$ with $PQ$. As $\omega$ is the A-excirce of $\triangle APQ$, it's well known that $MD'=MX$. Now, $\angle D'YD = 90^{\circ} \Longrightarrow \angle D'YX = 90^{\circ}$. So, $M$ is the circumcenter of $\triangle D'XY$. So, $MD'=MY \Longrightarrow MY$ is tangent to $\omega$ at $Y$. Now, $P$ lies on the polar of $A$ $wrt$ $\omega$. From La Hire, $A$ lies on the polar of $P$ $wrt$ $\omega$. So, $AD$ is the polar of $P$. Implying that $PY$ is tangent to $\omega$ at $Y$.

We get our conclusion.

$Q.E.D$

Frankly, my dear, I don't give a damn.