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

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2020 INMO Problems/Problem 1

PROBLEM

SOLUTION