APMO

We start with the following lemma. Lemma 1. Points $M,N,P,Q$ are concyclic. Point $M$ is the Miquel point of lines $AP=AB,PS=\ell ,AS=AD$, and $BR=BC$, and point $N$ is the Miquel point of lines $CQ=CD,RC=BC,QR=\ell$, and $DS=AD$. Both points $M$ and $N$ are on the circumcircle of the triangle determined by the common lines $AD,\ell$, and $BC$, which is $LRS$. Then, since quadrilaterals $QNRC,PMAS$, and $ABCD$ are all cyclic, using directed angles (modulo $180^{\circ }$ ) $$\begin{array}{rl}\measuredangle NMP& =\measuredangle NMS+\measuredangle SMP=\measuredangle NRS+\measuredangle SAP=\measuredangle NRQ+\measuredangle DAB=\measuredangle NRQ+\measuredangle DCB\\ & =\measuredangle NRQ+\measuredangle QCR=\measuredangle NRQ+\measuredangle QNR=\measuredangle NQR=\measuredangle NQP,\end{array}$$ which implies that $MNQP$ is a cyclic quadrilateral. Let $E$ be the Miquel point of $ABCD$ (that is, of lines $AB,BC,CD,DA$ ). It is well known that $E$ lies in the line $t$ connecting the intersections of the opposite lines of $ABCD$. Let lines $NQ$ and $t$ meet at $T$. If $T\ne E$, using directed angles, looking at the circumcircles of $LAB$ (which contains, by definition, $E$ and $M$ ), $APS$ (which also contains $M$ ), and $MNQP$, $$\measuredangle TEM=\measuredangle LEM=\measuredangle LAM=\measuredangle SAM=\measuredangle SPM=\measuredangle QPM=\measuredangle QNM=\measuredangle TNM,$$ that is, $T$ lies in the circumcircle $\omega$ of $EMN$. If $T=E$, the same computation shows that $\measuredangle LEM=\measuredangle ENM$, which means that $t$ is tangent to $\omega$. Now let lines $MP$ and $t$ meet at $V$. An analogous computation shows, by looking at the circumcircles of $LCD$ (which contains $E$ and $N$ ), $CQR$, and $MNQP$, that $V$ lies in $\omega$ as well, and that if $V=E$ then $t$ is tangent to $\omega$. Therefore, since $\omega$ meet $t$ at $T,V$, and $E$, either $T=V$ if both $T\ne E$ and $V\ne E$ or $T=V=E$. At any rate, the intersection of lines $MP$ and $NQ$ lies in $t$.

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Alternative solution.

Barycentric coordinates are a viable way to solve the problem, but even the solution we have found had some clever computations. Here is an outline of this solution. Lemma 2. Denote by ${\mathrm{pow}}_{\omega }X$ the power of point $X$ with respect to circle $\omega$. Let ${\Gamma }_1$ and ${\Gamma }_2$ be circles with different centers. Considering $ABC$ as the reference triangle in barycentric coordinates, the radical axis of ${\Gamma }_1$ and ${\Gamma }_2$ is given by $$\left({\mathrm{pow}}_{{\Gamma }_1}A-{\mathrm{pow}}_{{\Gamma }_2}A\right)x+\left({\mathrm{pow}}_{{\Gamma }_1}B-{\mathrm{pow}}_{{\Gamma }_2}B\right)y+\left({\mathrm{pow}}_{{\Gamma }_1}C-{\mathrm{pow}}_{{\Gamma }_2}C\right)z=0.$$ Proof: Let ${\Gamma }_i$ have the equation ${\Gamma }_i(x,y,z)=-a^{2}yz-b^{2}zx-c^{2}xy+(x+y+z)\left(r_{i}x+s_{i}y+t_{i}z\right)$. Then ${\mathrm{pow}}_{{\Gamma }_i}P={\Gamma }_i(P)$. In particular, ${\mathrm{pow}}_{{\Gamma }_i}A={\Gamma }_i(1,0,0)=r_i$ and, similarly, ${\mathrm{pow}}_{{\Gamma }_i}B=s_i$ and ${\mathrm{pow}}_{{\Gamma }_i}C=t_i$. Finally, the radical axis is $$\begin{array}{rl}& {\mathrm{pow}}_{{\Gamma }_1}P={\mathrm{pow}}_{{\Gamma }_2}P\\ ⟺& {\Gamma }_1(x,y,z)={\Gamma }_2(x,y,z)\\ ⟺& r_{1}x+s_{1}y+t_{1}z=r_{2}x+s_{2}y+t_{2}z\\ ⟺& \left({\mathrm{pow}}_{{\Gamma }_1}A-{\mathrm{pow}}_{{\Gamma }_2}A\right)x+\left({\mathrm{pow}}_{{\Gamma }_1}B-{\mathrm{pow}}_{{\Gamma }_2}B\right)y+\left({\mathrm{pow}}_{{\Gamma }_1}C-{\mathrm{pow}}_{{\Gamma }_2}C\right)z=0.\end{array}$$ We still use the Miquel point $E$ of $ABCD$. Notice that the problem is equivalent to proving that lines $MP,NQ$, and $EK$ are concurrent. The main idea is writing these three lines as radical axes. In fact, by definition of points $M,N$, and $E$ : - $MP$ is the radical axis of the circumcircles of $PAS$ and $PBR$; - $NQ$ is the radical axis of the circumcircles of $QCR$ and $QDS$; - $EK$ is the radical axis of the circumcircles of $KBC$ and $KAD$. Looking at these facts and the diagram, it makes sense to take triangle $KQP$ the reference triangle. Because of that, we do not really need to draw circles nor even points $M$ and $N$, as all powers can be computed directly from points in lines $KP,KQ$, and $PQ$. Associate $P$ with the $x$-coordinate, $Q$ with the $y$-coordinate, and $K$ with the $z$-coordinate. Applying the lemma, the equations of lines $PM,QN$, and $EK$ are - $MP:(KA\cdot KP-KB\cdot KP)x+(QS\cdot QP-QR\cdot QP)y=0$ - $NQ:(KC\cdot KQ-KD\cdot KQ)x+(PR\cdot PQ-PS\cdot PQ)z=0$ - $MP:(-QC\cdot QK+QD\cdot QK)y+(PB\cdot PK-PA\cdot PK)z=0$ These equations simplify to - $MP:(AB\cdot KP)x+(PQ\cdot RS)y=0$ - $NQ:(-CD\cdot KQ)x+(PQ\cdot RS)z=0$ - $MP:(CD\cdot KQ)y+(AB\cdot KP)z=0$ Now, if $u=AB\cdot KP,v=PQ\cdot RS$, and $w=CD\cdot KQ$, it suffices to show that $$\left|\begin{array}{ccc}u& v& 0\\ -w& 0& v\\ 0& w& u\end{array}\right|=0$$ which is a straightforward computation.