INMO $2020$ P1: Prove that $PQ$ is the perpendicular bisector of the line segment $O_1O_2$.

Question

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 $O_1O_2$.

My progress: This problem is really intimidating to me !

I observed that XBY is collinear, which can be proved by angle chase. Just note that $\angle BDA = \angle CBA$ and $\angle ACB = \angle ABD $ . Then $\Delta ABD \sim \Delta ACB$ . By cyclic quads, we get XBY collinear .

Then I was able to show $PO_1=PO_2$ by noticing that $\angle PO_1O_2 = 180- \angle DAB$ and $\angle O_1O_2P = 180-\angle BAC$ .

Then I am stuck. I also observed that $O_1,P,O_2,Q$ is cyclic but was not able to prove.

Here is a diagram:

I am also thinking of using spiral symmetry but I don't have any idea on how to use it ?

Please if possible send hints rather than solution. It helps me a lot . Thanks in advance.

Answer 1

Hints:

• Since angle bisector meets $\Gamma_1$ at $Y$ we see that $YB = YD$ and similary $XB = XC$. (Further more, since $\angle DYB = \angle DBC = \angle BXC$ (tangent chord) we have $\Delta BDY\sim \Delta CBX$. You don't need this.)
• You can prove $QO_1= Q_2O$ with spiral similarity around $A$ which takes $D$ to $O_1$ and $B$ to $O_2$. It takes $Y$ to $Q$ and since $YB = YD$ we have also $QO_1= Q_2O$.
• Let $\angle ABD = x$ and $\angle ABC = y$. Prove that $\angle O_2O_1P = \angle O_1O_2P = x+y$ and you are done.