Let $ABCD $ be cyclic quadrilateral with circumcircle $\omega $ and $M $ be any point on $\omega $. Let $E $ and $F $ be the intersection of $AB,CD $ and $AD,BC $ respectively. $ME $ intersects lines $AD,BC $ at $P,Q $ and similarly $MF$ intersects lines $AB,CD $ at $R,S $. Let the lines $PS $ and $RQ $ meet at $X $. Prove that as $M $ varies over $\omega $ $MX $ passes through fixed point. Proposed by Mehdi Etesami Fard
Problem
Source: Iran RMM TST 2019,day2 p6
Tags: geometry
30.07.2019 20:50
30.07.2019 22:28
Take a projective transformation that takes $AC\cap BD$ to the center of the circumcircle. Now $ABCD$ is a rectangle and $E$ and $F$ are points at infinity. We get that $MPDS$ and $MRBQ$ are rectangles as well, so $$\angle XPM=\angle CDM=\angle QBM=\angle XRM,$$so $PMXR$ is cyclic. By Thales, $A$ lies on this circle as well, so we get the circle $(ARXMP)$ and by symmetry $(CQMSX)$. By Thales, $\angle CXM=\angle MXA=90^\circ$, so $X$ lies on $AC$ and $MX$ is perpendicular to $AC$, so it passes through a fixed point at infinity.
30.07.2019 22:39
Any synthetic solution?
06.08.2020 23:18
L3435, i can finish your solution easier. Let $Y$ be the feet of perpendicular from $M$ to the $AC$. $Y$, $S$, $P$ on the Simson line of $M(\triangle{ACD})$; $Y$, $Q$, $R$ on the Simson line of $M(\triangle{ABC}) \implies X = Y \implies MX \perp AC$.
24.12.2020 11:33
Claim 1. $X$ lies on $AC$. Proof. Apply Pappus's theorem on $QCF$ and $ARE$. So, we get that $FR\cap EC, QE\cap AF, AC\cap QR$ are collinear. Thus, $PS$ passes through $AC\cap QR$. Thus, $X$ lies on $AC$. Let $AC\cap EF=J$. Now, observe that if we consider complete, quadrilateral $AEFC$, then by DDIT, there is a unique involution that swaps, $(MA,MF)$, $(ME,MC)$ and $(MJ, MB)$. Now, consider complete quadrilateral, $CAPS$. Then, again by DDIT, there is a unique involution that swaps $(MC,MP)$, $(MA,MS)$ and $(MX,MD)$ but as $MS=MF$ and $MP=ME$, we get that this involution also swaps $(MC,ME)$ and $(MA,MF)$ but since involutions are determined by two images, these involutions are the same. Thus, we get that a single involution exists which does the following $4$ swaps- $(MA,MF)$ $(MC, ME)$ $(MD, MX)$ $(MB, MJ)$ Now, we project this onto $EF$. Thus, $(A,C;B,D)=(F,E; J,MX\cap EF)$ but as $E,F,J$ are fixed, there's a unique point $MX\cap EF$ that satisfies this cross ratio and we are done. We can even characterize this point easily. We have $(AC;BD)=(AA\cap EF,J; E,F)=(F,E; J, AA\cap EF)$. Thus, this unique point is the intersection of the tangent to $(ABCD)$ from $A$ and $EF$. Also, this point can be similarly said to lie on the tangent from $C$. So, its the point at which the tangents from $A,C$ intersect. Thanks Wizard_32 for recommending! I spent many days on this trying to find a non MMP, non projective transform solution!
07.11.2023 09:20
Quite similar to the aforementioned solutions, but has some minor differences. Let $CX \cap AD={A'}, EX \cap AD={U}, CX \cap AB={V}$. $(E,P;M,Q)=(Q,M;P,E)=(RQ,RM;RP,RE)=(X,S;P,V)=(EX,ES;EP,EV)=(U,D;P,A)$ On the other hand, $(E,P;M,Q)=(E,D;S,C)=(XE,XD;XS,XC)=(U,D;P,A')$. Hence, $A'=A$, which means that $X$ is also on the line $AC$. Now, take the projective transformation which transforms $AC \cap BD$ into the circumcenter of $ABCD$. $ABCD$ is a square, thus $ME \perp AB, CD$, $MF \perp DA, BC$. Then, it turns out that $P,X,S$ is the Simson line with respect to $M$ and triangle $ADC$. Hence, $MX \perp AC$. This gives us that $MX$ always passes through $AA \cap CC$. (As a tiny remark, by Pascal on $AABCCD$, we know that this point is on $EF$.)
20.02.2024 22:51
Take a homography sending $ABCD$ to a rectangle while preserving the circumcircle. Then $E = \infty_E$ and $F = \infty_F$. $P$ and $Q$ are the projections of $M$ onto $AD$ and $BC$,and $R$ and $S$ are the projections of $M$ onto $AB$ and $CD$. Then let $X'$ be the projection of $M$ onto $AC$. By Simson line on $\triangle ABC$ and $\triangle ACD$ we get that $X'$ lies on $RQ$ and $PS$, so $X = X'$. Then all possible lines $MX$ concur at $\infty$ so we are done.