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: enter image description here

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.

  • $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Aug 2 '20 at 16:46
  • $\begingroup$ Put new pictureDraw circle XAY $\endgroup$ – Aqua Aug 2 '20 at 17:00
  • $\begingroup$ Hmm..the three circles are concurring .. $\endgroup$ – Sunaina Pati Aug 2 '20 at 17:05
  • $\begingroup$ I think you mean $\angle PO_1O_2=180^\circ-\angle DAB$ etc. $\endgroup$ – user10354138 Aug 2 '20 at 17:09
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    $\begingroup$ 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 ! $\endgroup$ – 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.
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    $\begingroup$ This the fastest i ever accepted and voted answer ! Thanks for your help . I loved the spiral similarity part ! $\endgroup$ – Sunaina Pati Aug 2 '20 at 17:26
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    $\begingroup$ Perhaps you can wait with acception. Perhaps someone will ofer easier solution. $\endgroup$ – Aqua Aug 2 '20 at 17:27
  • $\begingroup$ If you don't mind , can you try this problem? math.stackexchange.com/questions/3774437/… ? $\endgroup$ – Sunaina Pati Aug 2 '20 at 17:27

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