Let $ ABC$ be a triangle with $ \measuredangle{BAC} < \measuredangle{ACB}$. Let $ D$, $ E$ be points on the sides $ AC$ and $ AB$, such that the angles $ ACB$ and $ BED$ are congruent. If $ F$ lies in the interior of the quadrilateral $ BCDE$ such that the circumcircle of triangle $ BCF$ is tangent to the circumcircle of $ DEF$ and the circumcircle of $ BEF$ is tangent to the circumcircle of $ CDF$, prove that the points $ A$, $ C$, $ E$, $ F$ are concyclic. Author: Cosmin Pohoata
Problem
Source: Romanian TST 3 2008, Problem 1
Tags: geometry, circumcircle, geometry proposed
The QuattoMaster 6000
08.06.2008 00:55
pohoatza wrote: Let $ ABC$ be a triangle with $ \measuredangle{BAC} < \measuredangle{ACB}$. Let $ D$, $ E$ be points on the sides $ AC$ and $ AB$, such that the angles $ ACB$ and $ BED$ are congruent. If $ F$ lies in the interior of the quadrilateral $ BCDE$ such that the circumcircle of triangle $ BCF$ is tangent to the circumcircle of $ DEF$ and the circumcircle of $ BEF$ is tangent to the circumcircle of $ CDF$, prove that the points $ A$, $ C$, $ E$, $ F$ are concyclic.
Invert around $ F$, which maps point $ P$ to $ P'$. Now, since the circumcircles of $ \triangle FBE$ and $ \triangle DFC$ are tangent, we see that $ B'E'\parallel D'C'$. Similarly, $ E'D'\parallel B'C'$, so $ E'D'C'B'$ is a parallelogram. This means that $ \angle B'E'D' = \angle D'C'B'$. Also, the inversion maps $ \angle BED = \angle FEB + \angle FED = \angle FD'E' + \angle FB'E'$. Similarly,\[ \angle ACB = \angle DCB = \angle FCD + \angle FCB = \angle FB'C' + \angle FD'C'\]so\[ \angle FD'E' + \angle FB'E' = \angle BED = \angle ACB = \angle FB'C' + \angle FD'C'\]Also,\[ \angle B'FD' = 360 - (\angle B'E'D' + \angle FB'E' + \angle FD'E') = 360 - (\angle B'C'D' + \angle FB'C' + \angle FD'C') = 360 - \angle B'FD'\]Which implies that $ \angle B'FD' = 180$, so $ B'$, $ F$, and $ D'$ are collinear. Now, notice that $ A$ is the intersection between $ BE$ and $ CD$, so $ A'$ is the intersection between the circumcircles of $ \triangle FE'B'$ and $ \triangle FD'C'$. Now,\[ \angle FA'E' = 180 - \angle FB'E' = 180 - \angle FD'C' = 180 - \angle FA'C'\]so $ \angle FA'C' + \angle FA'E' = 180$, which means that $ E'$, $ A'$, and $ C'$ are collinear, so $ FCAE$ is cyclic.
plane geometry
08.06.2008 09:09
Could anyone solve it without using "inversion"?
Leonhard Euler
08.06.2008 17:26
From fact that circumcircle of triangle $ BCF$ is tangent to the circumcircle of triangle $ DEF$, we get $ \angle DFE=\angle FCD+\angle FBE$ and similarly $ \angle EFB=\angle FDE+\angle FCB$. By adding, $ \angle DFB=(\angle FCD+\angle FCB)+\angle FBE+\angle FDE=\angle DCB+\angle FBE+\angle FDE=\angle DEB+\angle FBE+\angle FDE$. Hence $ D,F,B$ are collinear. So $ \angle DFE=\angle FEB+\angle FBE$. Comparing with $ \angle DFE=\angle FCD+\angle FBE$, we get $ \angle FEB=\angle FCD$ which imply $ A,C,E,F$ are concyclic.
jatin
19.12.2011 17:16
Leonhard Euler wrote: $\cdots$ and similarly $ \angle EFB=\angle FDE+\angle FCB$$\cdots $ How ?
tenplusten
21.03.2017 18:24
Perform an inversion $\mathbf{I}(F,k)$ with arbitrary power $k>0$.(Define $X'$ as the inverse of $X$.) It's obvious that $E'C'D'B'$ is an parallelogram (because of tangencies) Let $\angle BEF=\alpha$, $\angle BEF=y$, $\angle ACF=x$. By angle-chasing we get $\angle E'B'C'=\alpha +y$ and $\angle E'D'C'=2x+y-\alpha$. Since $E'C'D'B'$ is a parallelogram. $\implies$ $x =\alpha$ $\implies$ $\angle BEF=\angle ACF$. So $AEFC$ is cyclic.
william122
04.04.2019 00:15
By the condition, we know that $\angle BEF+\angle FDC=\angle BFD$, and $\angle FED+\angle CBF=\angle CFD$, so $\angle DEB+\angle CBF+\angle FDC=\angle BFD=360-LHS$, and we must have $\angle BFD=180$. Now, $\angle FCD+\angle FDC=\angle CFB=\angle FDC+\angle FEB$, implying $\angle FCD=\angle FEB=180-\angle FEA$, as desired.
AwesomeYRY
23.08.2021 03:46
We will be using directed angles. Invert about $F$. Then in our inverted picture, it's clear that $B^*C^*D^*E^*$ is a parallelogram. Next note that \[\angle E^*B^*F +\angle FD^*E^*=\angle BEF + \angle FED =\angle BED = \angle ACB = \angle DCB = \angle DCF + \angle FCB= \angle C^*D^*F + \angle F B^*C^*\]From this it follows that \[\angle D^*FB^* = - \angle B^*E^*D^* - (\angle FB^*E^* + \angle E^*D^*F ) = -\angle D^*C^*B^* - (\angle FD^*C^* + \angle C^*B^*F) = \angle B^*FD^*\]Since $\angle D^*FB^* = \angle B^*FD^* = - \angle D^*FB^*$, we have that $D^*,F,B^*$ are collinear. Then, by Reim's, since $B^*E^*\parallel D^*C^*$, it follows that $A^*,C,E^*$ are concurrent. Inverting back, this means that $A,C,E,F$ are concyclic and we're done. $\blacksquare$.
rafaello
23.08.2021 04:51
We invert at $F$ with arbitrary radius. Then get the following problem. Inverted Romanian TST 3 2008, Problem 1 wrote: Let $EDCB$ be a parallelogram. Let $F$ be the point inside $EDCB$ such that $\measuredangle CBF+\measuredangle FDC=\measuredangle FBE+\measuredangle EDF$. Let $A$ be the intersection of $(FBE)$ and $(FCD)$. Prove that $A$ lies on $CE$. Note that as $\measuredangle CBF+\measuredangle FDC=\measuredangle FBE+\measuredangle EDF$ and $\measuredangle BED=\measuredangle DCB$, thus $B,F,D$ are collinear. Thus, $$\measuredangle FAC=\measuredangle FDC=\measuredangle FBE=\measuredangle FAE.$$We are done.