Let $D$ be a point on the circumcircle of some triangle $ABC$. Let $E, F$ be points on $AC$, $AB$, respectively, such that $A,D,E,F$ are concyclic. Let $M$ be the midpoint of $BC$. Show that if $DM$, $BE$, $CF$ are concurrent, then either $BE \cap CF$ is on the circle $ADEF$, or $EF$ is parallel to $BC$. proposed by USJL
Problem
Source: 2022 IMOC G6 https://artofproblemsolving.com/community/c6h2918249p26069468
Tags: geometry, concurrency
supercarry
06.09.2022 09:38
Well, as a member of IMOC this year, I will post my originial thought and a very rapid way. Let $\Omega$ denote $(AEF), P=BE\cap CF, Pa=AP\cap BC$. We want to prove that $(A,D;E,F)_\Omega=(Pa,M;B,C)$. The way I use during the mock test was that I consider a spiral similarity that transform (AEF) to (ABC), then it reduced to IMOC problem last year. https://artofproblemsolving.com/community/c6h2645286p22890040 However, here is a much more efficient way to solve it. Since $D$ is Miquel point of $\{AB,AC,BC,EF\}$, we have $\triangle DFB \sim\triangle DEC$, which implies $\dfrac{DF}{DE}=\dfrac{FB}{EC}$. Hence $$(A,D;E,F)_\Omega=\dfrac{AE}{AF}\times\dfrac{DF}{DE}=\dfrac{AE}{AF}\times\dfrac{FB}{EC}=\dfrac{PaB}{PaC}=(Pa,M;B,C)$$The last equality is due to Ceva theorem.
i3435
06.09.2022 15:18
If $\overline{EF}||\overline{BC}$, then $D\equiv A$ and $\overline{DM},\overline{BE},\overline{CF}$ are concurrent by Ceva's theorem in $ABC$. If not, let $D'$ be the point such that $DBD'C$ is a parallelogram. Desargues on $\triangle D'BC,\triangle DEF$ means that $E'=\overline{D'B}\cap\overline{DE},T=\overline{BC}\cap\overline{EF},F'=\overline{D'C}\cap\overline{DF}$ are collinear. $\measuredangle BE'E=\measuredangle CDE=\measuredangle BTE$, so $B,E,E',T$ cyclic. Similarly $C,F',F,T$ are cyclic, so $\measuredangle DEB=\measuredangle E'EB=\measuredangle E'TB=\measuredangle F'TC=\measuredangle F'FC=\measuredangle DFC$. Thus $\overline{BE}\cap\overline{CF}$ is on $(ADEF)$, as desired.