Birthday!

Today (March 18th) is my birthday! I was born on March 18th, 2009. 

Thanks to my parents for supporting me in my journey so much :)

I probably would've given up a long time back if it wasn't for them. 

Now it's time for 2009 IMO SL #18, which is 2009 IMO SL #G3. 

Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$.

Let the incircle touch $\overline{BC}$ at $X$ and let the $A$-excircle $\omega_A$ touch $\overline{BC}$ at $X'$ and $\overline{AC}$ at $Y'$. Denote $\omega_R$ and $\omega_S$ as the circles centered at $R$ and $S$ respectively, both with radius $0$. Notice that $BR=CY=CX=BX'$ and $YR=BC=AY'-AY=YY'$, so we have that $\overline{BY}$ is the radical axis of $\omega_A$ and $\omega_R$. Similarly, $\overline{CZ}$ is the radical axis of $\omega_A$ and $\omega_S$. Using both of these, we know that the intersection of $\overline{CZ}$ and $\overline{BY}$, namely $G$, is the radical center of $\omega_A$, $\omega_R$, $\omega_S$ and therefore equidistant from $R$ and $S$, as desired. $\square$