Let $ABCD$ be a cyclic quadrilateral. Let $DA$ and $BC$ intersect at $E$ and let $AB$ and $CD$ intersect at $F$. Assume that $A, E, F$ all lie on the same side of $BD$. Let $P$ be on segment $DA$ such that $\angle CPD = \angle CBP$, and let $Q$ be on segment $CD$ such that $\angle DQA = \angle QBA$. Let $AC$ and $PQ$ meet at $X$. Prove that, if $EX = EP$, then $EF$ is perpendicular to $AC$.
Problem
Source: RMM Shortlist 2023 G2
Tags: cyclic quadrilateral, RMM Shortlist, geometry, Hybrid
01.03.2024 00:35
Call $\omega=(ABCD)$ and $O$ its circumcenter. Construct $P'$ and $Q'$ on $CB$ and $AB$ respectively such that $EP=EP'=\text{Pow}_{\omega}(E)$ and $FQ=FQ'=\text{Pow}_{\omega}(Q)$. The claim is that $PQ\cap P'Q'\cap AC=\{X\}$ in general. This is equivalent to proving that $$\prod\frac{CP'}{BP'}=1.$$Notice that $\triangle EAP'\sim\triangle EP'D$ and $\triangle EAB\sim\triangle ECD$, so $\angle P'AB = \angle P'DC$, implying by the law of sines in triangles $\triangle P'BA$ and $\triangle P'DC$ that $$\frac{CP'}{BP'}=\frac{AP'}{DP'}\cdot\frac{\sin C}{\sin B}=\frac{EA}{EP'}\frac{\sin C}{\sin B}.$$Multiplying all of these relations finally gives $$\prod\frac{CP'}{BP'}=\frac{EA}{EP'}\cdot\frac{FQ'}{FD}\cdot\frac{EP}{EB}\cdot\frac{FA}{FQ}=1.$$ If $EP=EP'=EX$, $E$ is the circumcenter of $\triangle PP'X$, so $\frac{1}{2}\angle AFD=\frac{1}{2}\angle AEB=\angle PXP'=\angle QXQ'$, implying that $F$ is the circumcenter of $\triangle QQ'X$. Finally, $$EX^2-FX^2=\text{Pow}_{\omega}(F)-\text{Pow}_{\omega}(F)=EO^2-FO^2,$$so $OX\perp EF$. However, it's well-known that $OY\perp EF$ where $Y=AC\cap BD$, and since both $X$ and $Y$ are on $AC$, we obtain our conclusion. (Of course, a lot of unmentioned assumptions were made about the diagram.)
01.03.2024 13:43
Let $R,S$ be the reflections of $P,Q$ in $E,F$ respectively. We claim that $X$ lies on $RS$. Indeed, by alternate segment theorem, the angle condition $\angle CPD=\angle CBP$ is equivalent to line $AD$ being tangent to $\odot PBC$ hence: $$ ER^2=EP^2=EB \cdot EC=EA \cdot ED $$This means that $A,D$ are inverses with respect to the circle with diameter $PR$ which implies $(P,R;D,A)=-1$. Similarly $(Q,S;D,C)=-1$. Thus projecting through $X$ (which lies on $AC,PQ$) we get: $$ -1=\left(P,R;D,A\right) \stackrel{X}{=} \left(Q,RX \cap CD;D,C\right)=\left(Q,S;D,C\right) $$so $RX \cap CD \equiv S$ which shows that $X$ lies on $RS$. Let $\omega_1$, $\omega_2$ denote the circles with diameters $PR$, $QS$ respectively. The condition $EX=EP$ implies that $X$ lies on $\omega_1$ and so, as $X$ lies on $RS$: $$ \angle QXS=\angle PXS=180^{\circ}-\angle RXP=90^{\circ} $$which means that $X$ also lies on $\omega_2$. Now let $BD$ and $AC$ at $M$ and denote circle $\odot ABCD$ by $\Gamma$. From $EP^2=EA \cdot ED$ we get: $$ EP^2=\left(ED-AD\right) \cdot ED \Rightarrow DA \cdot DE=ED^2-EP^2=\left(ED+DP\right)\left(ED-EP\right)=DP \cdot DR $$hence $D$ lies on the radical axis of $\omega_1$ and $\odot EAC$. By a similar argument, we get $B$ also lies on this radical axis. Therefore, the radical axis is in fact line $BD$ and so the radical centre of $\omega_1,\Gamma,\odot EAC$ is $M$. Repeating the argument with $\omega_2$ we see $M$ is in fact the radical centre of $\omega_1,\omega_2,\Gamma$ and significantly lies on the radical axis of $\omega_1,\omega_2$. $X$ also lies on this line so the radical axis of $\omega_1,\omega_2$ is line $MX$ which is just line $AC$. Finally, line joining the centres of $\omega_1,\omega_2$ is $EF$ which is perpendicular to the radical axis $AC$ of these circles.
02.03.2024 04:02
The angle condition implies that $(BCP)$ and $(ABQ)$ are tangent to $AD$ and $CD$, respectively. Thus, $EP^2 = EB \cdot EC = EA \cdot ED$ and $FQ^2 = FA \cdot FB = FC \cdot FD$, implying that $Pow_{X}(E) = Pow_{(ABCD)}(E)$. Also, $\tfrac{AP}{PD} = \sqrt{\tfrac{EA}{ED}}$ and $\tfrac{DQ}{QC} = \sqrt{\tfrac{FD}{FC}}$. If $Y = AC \cap BD$ then by Menelaus $\tfrac{EA}{ED} \cdot \tfrac{FD}{FC} \cdot \tfrac{YC}{YA} = 1$, so $\tfrac{CX}{XA} = \sqrt{\tfrac{CY}{AY}}$, implying that $XY = \sqrt{CY \cdot AY}$, and thus that $Pow_{X}(Y) = Pow_{(ABCD)}(Y)$. So $EY$ must be the radical axis of $X$ and $(ABCD)$, meaning that if $O$ is the center of $(ABCD)$ then $EY \perp OX$. By Brocard, $EF$ is perpendicular to the line through $O$ and $BD \cap AC$, so the only way for this to be true is if $O$ lies on $AC$, and $AC \perp EF$. $\square$
03.03.2024 00:11
The condition to points $P,Q$ is equivalent to $PE^2=EB \cdot EC$ and $FQ^2 = FA \cdot FB$. If $EX=EP$, then $E$ is center of Apollonius circle of $AXD$ (because $EX^2=EP^2=EB \cdot EC=EA \cdot ED$) and $\angle CXQ = \angle DXQ$. By similar reason we can get that $F$ is center of Apollonius circle of triangle $CXD$ and so $FQ=FX$. Now we have $EX^2=EB \cdot EC \Rightarrow \angle EXB = \angle BCA = \angle BDE$, so $BDXE$ is inscribed. By the same reason $BXFD$ is also inscribed. So, $FDBEX$ is inscribed. Let $EF \cap AC = T$. Then by Miquel ($CBTF,CDTE$ are inscribed), $FDAT, BATE$ are cyclic. So, $\angle DBA = \angle FEA = \angle TBA$ and $\angle CBD = \angle CAD = \angle TAE = \angle TBE$. From these conditions it's follows that $FB \perp CE$. By analogue, $DE \perp CF$, so $A$ is orthocenter of $FCE$ and $CA \perp EF$. $\blacksquare$
04.03.2024 01:19
Geo the mid The angle condition implies $(BCP)$ tangent to $\overline{DA}$ and $(ABQ)$ tangent to $\overline{DC}$, which by power of a point implies that the circle $\omega_1$ with center $E$ and radius $EP$, and the circle $\omega_2$ with center $F$ and radius $FQ$, are both orthogonal to $(ABCD)$. Let $\overline{DA}$ intersect $\omega_1$ again at $R$ and let $\overline{DC}$ intersect $\omega_2$ again at $S$, so $(A,D;P,R)=(C,D;Q,S)=-1$ and thus two applications of Ceva-Menelaus imply that $\overline{RS},\overline{PQ},\overline{AC}$ concur at $X$. Since $X \in \omega_1$, we have $\angle RXP=\angle SXQ=90^\circ$, so $X \in \omega_2$ as well. Then $\omega_1$ and $\omega_2$ both intersect $\overline{AC}$ again at the point $Y$ such that $(A,C;X,Y)=-1$, so $\overline{AC}$ is the radical axis of $\omega_1$ and $\omega_2$ and $\overline{EF} \perp \overline{AC}$. Remark: Somehow this solution is completely independent of the (strong) fact that $\overline{EF} \perp \overline{AC}$ would imply $\overline{AC}$ is a diameter by Brocard.
09.03.2024 19:18
I misread this question atleast 3 times lol. Introduce $P^{*}$ and $Q^{*}$ as points on opposite sides of $P,Q$ resp. such that $EP=EP^{*}$ and $EQ = EQ^{*}$, let $X = PQ \cap P^{*}Q^{*}$ Claim 1: $AC,PQ,P^{*}Q^{*}$ concur
So by Claim 1, and some angle chase we get $F$ is circumcenter of $QQ^{*}X$ so perpendicularity lemma + $\overline{OAC\cap BD} \perp EF$ gives $EF \perp AC$. @Assassino9931, are there only 2-3 questions in each section of RMMSL or are the rest not disclosed (due to permission from author etc.) ?