# 2020 INMO Problems/Problem 1

## PROBLEM

Let $\Gamma_1$ and $\Gamma_2$ be two circles of unequal radii, with centres $O_1$ and $O_2$ respectively, intersecting in two distinct points $A$ and $B$. Assume that the centre of each circle is outside the other circle. The tangent to $\Gamma_1$ at $B$ intersects $\Gamma_2$ again in $C$, different from $B$; the tangent to $\Gamma_2$ at $B$ intersects $\Gamma_1$ again at $D$, different from $B$. The bisectors of $\angle DAB$ and $\angle CAB$ meet $\Gamma_1$ and $\Gamma_2$ again in $X$ and $Y$, respectively. Let $P$ and $Q$ be the circumcentres of triangles $ACD$ and $XAY$, respectively. Prove that $PQ$ is the perpendicular bisector of the line segment $O_1O_2$.

$\emph{Proposed by Prithwijit De}$

## SOLUTION

Let $O, S, T$ be the circumcenters of triangles $BCD$, $XAY$ and $ACD$.

$\textbf{Claim:}$ $A, C, O, D$ are concyclic.

$\emph{Proof.}$ Note that $$\angle (DA, AC)=360^{\circ}-(\angle DAB+\angle BAC) =360^{\circ}-2\cdot(180^{\circ}-\angle DBC)=\angle DOC$$ where $\angle DAB=\angle BAC=180^{\circ}-\angle DBC$ follows from the tangency, proving the claim. $\blacksquare$

Now remark that $\triangle ACB \sim \triangle ABD \sim \triangle AO_2O_1$, where $O_1, O_2$ are the centers of $\omega_1, \omega_2$. By looking at the spiral similarity pivoted at $A$, taking $CB$ to $BD$, we conclude that $Y$ goes to $X$, so, the quadrilaterals $ACYB, AO_2SO_1,$ and $ABXD$ are all similar, yielding $SO_1=SO_2$.

Finally, note that $\triangle DO_1O \sim \triangle DBC \sim \triangle OO_2C$. Let $O'$ be the reflection of $O_2$ in $TO$, clearly, $O_1, O'$ are reflections in the perpendicular bisector of $DO$, thus, $TO_2=TO_1$ follows. $\blacksquare$ ~anantmudgal09