Let $B$ and $D$ be points on segments $[AE]$ and $[AF]$ respectively. Excircles of triangles $ABF$ and $ADE$ touching sides $BF$ and $DE$ is the same, and its center is $I$. $BF$ and $DE$ intersects at $C$. Let $P_1, P_2, P_3, P_4, Q_1, Q_2, Q_3, Q_4$ be the circumcenters of triangles $IAB, IBC, ICD, IDA, IAE, IEC, ICF, IFA$ respectively. a) Show that points $P_1, P_2, P_3, P_4$ concylic and points $Q_1, Q_2, Q_3, Q_4$ concylic. b) Denote centers of theese circles as $O_1$ and $O_2$. Prove that $O_1, O_2$ and $I$ are collinear.
Problem
Source: Turkey NMO 2012
Tags: geometry, circumcircle, trigonometry, geometry proposed
tobash_co
27.11.2012 09:56
Firstly we prove that $P_1,P_3$ lie on $IF$, $P_2,P_4$ lie on $IE$. It suffices to prove one of them; the others follow by symmetry. We have $\angle BP_1I=2\angle BAI=\angle BAF$. But since $\angle AP_1B=2\angle BIA=\angle EBF-\angle BAF=\angle AFB$, $A,F,P_1,B$ are concyclic. It follows that $P_1\in IF$, as desired. So $P_4,P_2,I$ and $P_1,P_3,I$ are collinear. Letting $r_i$ be the radius of circle $(P_i)$. By considering $AI$ and $CI$ as the radical axes of $(P_1),(P_4)$ and $(P_2),(P_3)$, we now have \[\frac{r_1}{r_4}=\frac{\sin \angle ABI}{\sin \angle ADI}=\frac{\sin \angle CBI}{\sin \angle CDI}=\frac{r_2}{r_3}\] which proves that $P_1,P_2,P_3,P_4$ are concyclic, as desired. Analogously, so are $Q_1,Q_2,Q_3,Q_4$. Now for the second part. Note that as previously established, $A,F,P_1,B$ are concyclic. Since $\angle AQ_4F=2\angle AIF=\angle ABF$, $Q_4$ also lies on this circle. It follows that \[\angle AP_1Q_4=\angle EBQ_4=\angle EBI=\frac{\angle AP_1I}{2}=\angle AP_1P_4\] which proves that $Q_4$ lies on $P_1P_4$. Analogously, we see that $P_1,P_4,Q_1,Q_4$ and $P_2,P_3,Q_2,Q_3$ are collinear. Let these two lines intersect at $G$, and $P_1P_2\cap P_3P_4=S$, $Q_1Q_2\cap Q_3Q_4=R$. By Brokard's theorem, we know $GS$ is the polar of $I$ w.r.t. $(O_1)$, and $GR$ is the polar of $I$ w.r.t. $(O_2)$. But we also know that $GP_1,GS,GP_3,GI$ and $GQ_1,GR,GQ_3,GI$ are both harmonic pencils. Thus $G,S,R$ are collinear. It follows that $IO_1\perp GS\equiv GR\perp IO_2$, which shows that $I\in O_1O_2$, as desired.
Pathological
26.09.2019 03:34
Here's a different way to solve the second part. We'll show that $I$ and $O_1$ are collinear with the center of mass of $ABCD.$ This would imply the problem by symmetry. Call $\omega$ the common $A-$excircle of $\triangle ADE, \triangle ABF.$ Invert about $\omega$. Let $P_1, P_2, P_3, P_4$ be the points where $AE, DE, BF, AF$ touch $\omega.$ Observe that $A, B, C, D$ invert to the midpoints of $P_1P_4, P_1P_3, P_2P_3, P_2P_4$ respectively, call these points $A', B', C', D'$ respectively. Therefore, $P_1, P_2, P_3, P_4$ map to the reflections of $I$ over $A'B', B'C', C'D', D'A'.$ The circumcenter $O'$ of these four points is exactly the reflection of $I$ over the center of mass of $A'B'C'D'.$ However, the center of mass of $A'B'C'D'$ is just the center of mass of $ABCD$. Since $I, O', O_1$ are collinear, we have shown the desired result. With a similar result, we are done. $\square$