Let ${A, B, C}$ and ${O}$ be four points in plane, such that $\angle ABC>{{90}^{{}^\circ }}$ and ${OA=OB=OC}$.Define the point ${D\in AB}$ and the line ${l}$ such that ${D\in l, AC\perp DC}$ and ${l\perp AO}$. Line ${l}$ cuts ${AC}$at ${E}$ and circumcircle of ${ABC}$ at ${F}$. Prove that the circumcircles of triangles ${BEF}$and ${CFD}$are tangent at ${F}$.
Problem
Source: 2009 JBMO Shortlist G5
Tags: geometry, JBMO
Davrbek
08.10.2017 18:53
So,It is easy to see $O$ circumcenter of $(ABC) \implies \angle OAC=B-90^\circ$. If $l \cap (ABC)=F$ at arc $AC$ containing $B$ and $l \cap AO=K,I\cap AC=E$. We need to show $\angle EBF+\angle FDC=\angle EFC \implies\angle EBF+\angle FDC+\angle FEC+\angle FCE=180^\circ$. Given $\angle AKD=\angle ACD=90^\circ$. So,quadrilateral $ADCK$ cyclic. So,$\angle OAC=\angle KDC=B-90^\circ$. In $\triangle ADK$ $\angle ADK=C$. So,quadrilateral $BDCE$ cyclic.Then,$\angle EBC=B-90^\circ$.In triangle $EDC$ $\angle CED=180^\circ-B$.So, $\angle EBF+\angle FDC+\angle FEC+\angle FCE=B-90^\circ+\angle CBF+B-90^\circ+180^\circ+C+\angle BCF=B+C+A=180^\circ$. Proved.
claserken
08.10.2017 19:52
Any solution with inversion?
MeineMeinung
09.10.2017 12:32
claserken wrote: Any solution with inversion?
Let $AA'$ be the diameter of the circumcircle of $ABC$. Since $AA'$ is the diameter of $(ABC)$, then $\angle ACA' = 90^{\circ}$. Since $AC \perp DC$, then $\angle DCA = 90^{\circ}$ and hence $\angle ACA' + \angle DCA = 180^{\circ}$ which implies $A',C',D$ are collinear.
Notice that $l$ and $AC$ are the altitude of the triangle $AA'D$ so $E$ is the orthocenter of triangle $AA'D$. Again, since $AA'$ is the diameter, then $\angle ABA' = 90^{\circ}$ and this again implies that $A',E,B$ are colinear.
Let the second intersection of $l$ with $(ABC)$ be $F'$. Since $E$ is the orthocenter of $AA'D$, then it is easy to see that we have $EA \cdot EC = ET \cdot ED = EA' \cdot EB = EF \cdot EF'$.
By performing negative inversion centered at $E$ with power $EA \cdot EC$, we obtain the following result :
$B$ is mapped to $A'$,
$F$ is mapped to $F'$
which implies that circle $BEF$ is mapped to line $A'F'$.
$C$ is mapped to $A$,
$D$ is mapped to $T$
which implies that the circle $CFD$ is mapped to circle $ATF'$.
Since both $\angle A'F'A$ and $A'TF'$ are right angle, we must have $\angle A'F'T = \angle F'AT$ which implies that $A'F'$ is tangent to $ATF'$ at $F'$. Therefore, circle $BEF$ is tangent to circle $CFD$ at $F$.
MathematicalPhysicist
15.10.2018 20:49
Nothing...
math_pi_rate
16.10.2018 10:04
This is just a restated problem. Anyway, here's my solution: Note that $O$ is the circumcenter of $\triangle ABC$. Let $A'$ be the antipode of $A$ in $\odot (ABC)$. Then $\angle A'CA=90^{\circ} \Rightarrow D,C,A'$ are collinear. Also, as $DE \perp AA'$ and $AE \perp A'D$, we get that $E$ is the orthocenter of $\triangle ADA'$. But $\angle A'BA=90^{\circ}$, giving $A',E,B$ are collinear. Thus $$\angle EFC=\angle CDF+\angle DCF=\angle CDF+\angle A'BF=\angle CDF+\angle EBF$$This easily gives that $\odot (CDF)$ and $\odot (BEF)$ are tangent to each other at $F$. REMARK: If one states the problem in terms of $\triangle ADA'$, then the problem becomes quite well known.
Assassino9931
20.10.2023 11:00
Fun fact, this is Balkan MO 2012/1