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

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.

• This is screaming compute every power of a point you can! I'll check later if it actually works. Also a trigonometric/analytic approach looks doable. – Jack D'Aurizio Aug 2 '20 at 16:46
• Put new pictureDraw circle XAY – Aqua Aug 2 '20 at 17:00
• Hmm..the three circles are concurring .. – Sunaina Pati Aug 2 '20 at 17:05
• I think you mean $\angle PO_1O_2=180^\circ-\angle DAB$ etc. – user10354138 Aug 2 '20 at 17:09
• Oh, and since Y is the angle bisector of DAB implies Q is the angle bisector of O_1AO_2 and we are done since Q lies in the perpendicular bisector of O_1O_2 ! – Sunaina Pati Aug 2 '20 at 17:21

• 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.