# Two constructed triangles with common vertex have tangent circumcircles at this vertex. Here's a complicated-sounding geometry problem, Any help would be appreciated :)

Let $\triangle ABC$ be an obtuse-angled triangle with circumcentre $O$, circumcircle $\Gamma$ and $\angle ABC > 90^\circ$. Let $AB$ intersect the line through $C$ perpendicular to $AC$ in $D$. Let $l$ be the line through $D$ perpendicular to $AO$, and let $E$ be the intersection of $l$ and $AC$, and let $F$ be the point between $D$ and $E$ where $l$ intersects $\Gamma$. Can you prove that the circumcircles of triangles $\triangle BFE$ and $\triangle CFD$ are tangent at $F$?

Thanks :D

• Well, I wasn't able to solve it and I have no more time today. So I'll list some simple observations which may or may not help anyone solving the problem. First, it seems that the centers of the two circles and $F$ are collinear on the line $AF$, proving this may be one possible approach to tackling the problem. Secondly, the triangles $\triangle ABC$, $\triangle AED$ and $\triangle BEC$ are all similar. Finally, $A$, $C$, $D$ and the intersection of $l$ with $AO$ are all concyclic on the circle with diameter $AD$. – EuYu Feb 24 '13 at 21:17
• Woah, that's much further than I got. Thanks for the hints, now maybe I can get something going! :D Thanks a lot!!! – BittersweetNostalgia Feb 24 '13 at 21:22
• Oh thanks for the edit, it looks cooler and simpler now. I haven't had a chance to look at it yet, but has anyone made any progress? – BittersweetNostalgia Feb 25 '13 at 13:32

Let $l$ intersect $AO$ at $H$, and $AO$ intersect the circle again at $P$. Since $AP$ is a diameter, $PC \perp AC$, so $P, C, D$ are collinear. Now the 2 altitudes $AC, DH$ of $\triangle APD$ intersect at $E$, so $E$ is the orthocenter of $\triangle APD$. Thus $PE \perp AB$. Since $AP$ is a diameter, $PB \perp AB$, so $P, E, B$ are collinear.

$$\angle{PFE}=\angle{PFH}=90^{\circ}-\angle{HPF}=90^{\circ}-\angle{APF}=\angle{PBD}-\angle{DBF}=\angle{PBF}=\angle{EBF}$$

$$\angle{PFC}=\angle{PAC}=90^{\circ}-\angle{APC}=90^{\circ}-\angle{HPD}=\angle{HDP}=\angle{FDC}$$

Thus $PF$ is tangent to both the circumcircle of $\triangle BFE$ and the circumcircle of $\triangle CFD$, so the 2 circumcircles are tangent at $F$. Hint: The diagram looks somewhat like this. You can solve the problem using simple geometry, I suppose.

• Thanks for the diagram Inceptio. Could you please prove the question for me? To me the geometry isn't so simple :/ Thanks! – BittersweetNostalgia Feb 24 '13 at 10:54
• Well, I'm on to it. Will post the answer once done. – Inceptio Feb 24 '13 at 10:55
• Hey Inceptio, not rushing you or pushing you or anything, but if it's fine with you, I kinda need the answer quite soon... Could you maybe post a solution quite soon? Thanks... :P – BittersweetNostalgia Feb 24 '13 at 14:07
• I'm trying it since then bro. I'm ain't really getting it. I got what to prove . But I'm not getting how to start it. -_- – Inceptio Feb 24 '13 at 16:46
• @Inceptio "You can solve the problem using simple geometry, I suppose." But most IMO geometry questions can be solved using "simple geometry". That doesn't necessarily mean that the question is easy to solve. – Ivan Loh Mar 29 '13 at 8:46