In a triangle $ABC$, it is drawn a circumference with center in the incenter $I$ and that meet twice each of the sides of the triangle: the segment $BC$ on $D$ and $P$ (where $D$ is nearer two $B$); the segment $CA$ on $E$ and $Q$ (where $E$ is nearer to $C$); and the segment $AB$ on $F$ and $R$ ( where $F$ is nearer to $A$). Let $S$ be the point of intersection of the diagonals of the quadrilateral $EQFR$. Let $T$ be the point of intersection of the diagonals of the quadrilateral $FRDP$. Let $U$ be the point of intersection of the diagonals of the quadrilateral $DPEQ$. Show that the circumcircle to the triangle $\triangle{FRT}$, $\triangle{DPU}$ and $\triangle{EQS}$ have a unique point in common.
Problem
Source: Spanish Communities
Tags: geometry, incenter, circumcircle, inradius, trapezoid, geometric transformation, rotation
yetti
23.04.2006 01:34
Let r be the inradius and $r_0$ radius of the given circle centered at the incenter I. Let A', B', C' be the tangency points of the incircle (I, r) with the sides BC, CA, AB. The triangles $\triangle DEF \cong \triangle PQR$ are congruent and both are obtained from the contact triangle $\triangle A'B'C'$ by a spiral similarity with the center I, the same similarity coefficient $\frac{r_0}{r}$ and rotational angles $\angle DIA' = \angle EIB' = \angle FIC' = \phi$ resp. $\angle PIA' = \angle QIB' = \angle RIC' = -\phi$ of the same magnitude and opposite signs. Consequently, $2 \phi = \angle DIP = \angle EIQ = \angle FIR =$ $\angle (DE, PQ) = \angle (EF, QR) = \angle (FD, RP) =$ $= \angle DUP = \angle QSE = \angle RTF =$ $= \angle QUE = \angle RSF = \angle DTP$ Thus the circumcircles of the triangles $\triangle FRT, \triangle DPU, \triangle EQS$ all meet at the incenter I and moreover, these circumcircles are congruent to each other.
Luis González
31.12.2009 20:28
Since $ I$ is equidistant from $ BC,CA,AB,$ then it's clear that the intersected chords $ DP,EQ$ and $ FR$ are congruent, thus $EQFR , FRDP$ and $ DPEQ$ are isosceles trapezoids $ \Longrightarrow$ $ QR = EF,$ $ FD = RP ,$ $DE = PQ.$ This implies that $ \triangle DEF$ and $ \triangle PQR$ are congruent having $ I$ as their common circumcenter. Moreover, $ SFRT,$ $ TDPU$ and $ UEQS$ are cyclic due to $ \angle EFD = \angle QRP,$ $ \angle FDE = \angle RPQ$ and $ \angle DEF = \angle PQR.$ $ I$ is obviously center of the rotation taking $ \triangle DEF$ into $ \triangle PQR$ and since the midpoints $ M, N$ of the sides $ QR, EF$ are homologous under such rotation, the measure of the angle $ \angle MIN$ is precisely the rotational angle $\Longrightarrow$ $ \angle MIN = \angle RIF.$ But since $ IMSN$ is cyclic, we have $ \angle MIN = \angle FSR = \angle FTR = \angle FIR$ $ \Longrightarrow$ $ I \in \odot(FRT).$ By similar reasoning, we conclude that $ I \equiv \odot(FRT) \cap \odot(DPU) \cap \odot(EQS).$
Pashtet671
07.09.2024 10:04
Hi, @Luis González , can you please explain why can't we just prove that $SQEI$ is cyclic since $\angle QSE = \frac{1}{2}(\smile{QE} + \smile{FR}) = \smile{QE} = \angle QIE$? That would mean that every one of circumcircles would have a common point $I$.