My absolute favorite problem on the contest (I did not solve 20,22,24 but this is still nice enough), also a very nice choice for the 200th post The main idea is that the quadrilateral itself is quite irrelevant (in the sense of trigonometry) so we can draw a circle to intersect all those lines from $P$, which is amazing. We first show this lemma to eliminate the (very) annoying configuration issues: Lemma 1. (i)Suppose $ABCD$ is a cyclic convex quadrilateral with circumcenter $O$. Let $M$ and $N$ be the arc-midpoints of arc $AB$ and $CD$ respectively. Then $AC\cap BD$ lies the interior of $\angle MON$. (ii)Suppose $ABC$ is a triangle and $X_1, X_2$ are points in the interior of $\angle BAC$ such that $$\frac{\sin\angle BAX_1}{\sin\angle CAX_1}=\frac{\sin\angle BAX_2}{\sin\angle CAX_2}$$then $A,X_1,X_2$ are collinear. {Proof.} (i)[asy][asy] size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.735914763259068, xmax = 1.8777526696673268, ymin = -1.7688665514709507, ymax = 1.3008079345632986; /* image dimensions */ /* draw figures */draw(circle((0,0), 1), linewidth(0.8)); draw((-0.9994522386616171,-0.033094148036802204)--(0,0), linewidth(0.8)); draw((0,0)--(0.9809473781039253,-0.19427362505248777), linewidth(0.8)); draw((0.297117254826025,-0.9548410008397455)--(-0.7450550760389443,0.6670029487705454), linewidth(0.8)); draw((-0.6992994674694238,-0.7148288290192136)--(0.23330175500671707,0.972404386616384), linewidth(0.8)); draw((0.297117254826025,-0.9548410008397455)--(-0.3740601992436831,0.9274044249095299), linewidth(0.8)); /* dots and labels */dot((0,0),dotstyle); label("$O$", (0.012633994608542379,0.031491058481625825), NE * labelscalefactor); dot((0.9809473781039253,-0.19427362505248777),dotstyle); label("$M$", (1.006716343988832,-0.27936205402817155), NE * labelscalefactor); dot((-0.9994522386616171,-0.033094148036802204),dotstyle); label("$N$", (-1.1368749110266458,0.0023485791838323187), NE * labelscalefactor); dot((0.297117254826025,-0.9548410008397455),dotstyle); label("$A$", (0.3526295864161333,-1.1018275808770104), NE * labelscalefactor); dot((-0.6992994674694238,-0.7148288290192136),dotstyle); label("$D$", (-0.9450704671340888,-0.9039266813884547), NE * labelscalefactor); dot((0.6386218188575906,0.7695207420719878),dotstyle); label("$B$", (0.7055773912449658,0.8248141060326712), NE * labelscalefactor); dot((-0.7450550760389443,0.6670029487705454),dotstyle); label("$C$", (-0.8195456920062277,0.714720295352118), NE * labelscalefactor); dot((-0.20762891036568856,-0.16934757536590916),linewidth(4pt) + dotstyle); label("$G$", (-0.19460141373132253,-0.1433638173051352), NE * labelscalefactor); dot((-0.3740601992436831,0.9274044249095299),dotstyle); label("$D'$", (-0.4504076209008433,1.0093831415853634), NE * labelscalefactor); dot((0.23330175500671707,0.972404386616384),dotstyle); label("$A'$", (0.26844020177806316,1.0514778339043984), NE * labelscalefactor); dot((-0.30985614509608517,-0.01026004519200895),linewidth(4pt) + dotstyle); label("$F$", (-0.4212651416030498,0.018538845460384267), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $G=AC\cap BD$. Suppose the arc $MN$ containing $D,A$ is smaller than the arc $MN$ containing $B,C$. WLOG assume $\widehat{DN}<\widehat{AM}$, then $$\widehat{CA}=2\widehat{ND}+\widehat{DA}\leq \widehat{ND}+\widehat{DA}+\widehat{AM}\leq 180^{\circ}$$Hence let $A'$ be the reflection of $A$ about $ON$, then $AC$ and $A'D$ intersects on a point $F$ on line segment $ON$. \newline On the other hand, $$\widehat{AB}+\widehat{A'A}=2(\widehat{AN}+\widehat{AM})<360^{\circ}$$Therefore $DB$ lies between $DA'$ and $DA$ and so $G$ lies between $A$ and $F$, hence it lies in the segment $AF$, which obviously is in the interior of $\angle NOM$ as desired. (ii) [asy][asy] size(5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -25.202216841507315, xmax = 24.479090005620986, ymin = -26.407020814285335, ymax = 11.392239932967978; /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */draw((-6.621567571233008,2.2614058495069975)--(-11.645519949032499,-8.783314854068072), linewidth(0.8)); draw((-11.645519949032499,-8.783314854068072)--(4.821879511532499,-9.461149698691813), linewidth(0.8)); draw((4.821879511532499,-9.461149698691813)--(-6.621567571233008,2.2614058495069975), linewidth(0.8)); draw((-4.21369830164493,-12.520236055464252)--(-6.621567571233008,2.2614058495069975), linewidth(0.8) + linetype("4 4") + qqwuqq); /* dots and labels */dot((-6.621567571233008,2.2614058495069975),dotstyle); label("$A$", (-8.136727812156662,2.022170021992736), NE * labelscalefactor); dot((-11.645519949032499,-8.783314854068072),dotstyle); label("$B$", (-13.439788655389458,-8.18522528528242), NE * labelscalefactor); dot((4.821879511532499,-9.461149698691813),dotstyle); label("$C$", (4.981370063208673,-9.062423319501379), NE * labelscalefactor); dot((-5.186152606147439,-6.550447130601631),dotstyle); label("$X_1$", (-4.30895457192848,-6.032102837654067), NE * labelscalefactor); dot((-4.21369830164493,-12.520236055464252),dotstyle); label("$Y_1$", (-3.072902796438129,-13.049687111405737), NE * labelscalefactor); dot((-4.776368794736306,-9.066064417319975),linewidth(4pt) + dotstyle); label("$X_2$", (-4.6279356752808285,-8.743442216149031), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Suppose $AX_1$ and $AY_1$ intersect $BC$ at $X_2$ and $Y_2$ resecpectively, then $$\frac{BX_2}{X_2C}=\frac{BX_2}{X_2A}\cdot\frac{X_2A}{X_2C}=\frac{\sin\angle BX_2A}{\sin\angle X_2BA}\cdot\frac{\sin\angle X_2CA}{\sin\angle X_2AC}=\frac{\sin\angle BAY_2}{\sin\angle Y_2BA}\cdot\frac{\sin\angle Y_2CA}{\sin\angle CAY_2}=\frac{BY_2}{CY_2} $$since both $X_2$ and $Y_2$ lies inside segment $BC$ we have $X_2=Y_2$ as desired. $\blacksquare$ Now here comes the interesting part. Let $I_1,I_2,I_3,I_4$ be the incentres of $\triangle PAB, \triangle PBC,\triangle PCD$ and $\triangle PDA$ respectively. Now construct a circle $\Gamma$ with centre at $P$. For every existing point $X$ let $X'$ be the intersection of the ray $PX$ and $\Gamma$. Let $M,N$ be the midpoint of $C'D'$ and $A'B'$. Suppose $C'A'\cap B'D'=L$ and $E$ is the miquel point of $A'B'C'D'$ then by a well-known fact, $E$ and $L$ are image of inversion w.r.t. $\Gamma$ and hence $P,L,E$ are collinear. Let $PR$ intersect $I_1'I_3'$ at $J$. We now show that $P,J,E$ are collinear. From the lemma, $E$ lies in the interior of $\angle I_3'PI_1'$. Moreover $J$ lies in the interior of $\angle I_3'PI_1'$ since $R$ lies between $I_1$ and $I_3$. Therefore, in view of lemma 1 it suffices to show $$\frac{\sin\angle I_1'PJ}{\sin\angle I_3'PJ}=\frac{\sin\angle I_1'PE}{\sin\angle I_3'PE}\hspace{50pt}(1)$$On one hand, letting $r_1$ and $r_3$ be the inradius of $\triangle PAB$ and $\triangle PCD$ then \begin{align*} \frac{\sin\angle I_1'PJ}{\sin\angle I_3'PJ}=\frac{\sin\angle I_1PR}{\sin\angle I_3PR}&=\frac{\sin\angle I_1PR}{\sin\angle I_1RP}\cdot\frac{\sin\angle I_3RP}{\sin\angle I_3PR}\\ &=\frac{I_1R}{I_1P}\cdot\frac{I_3P}{I_3R}\\ &=\frac{r_1}{r_3}\cdot\frac{I_3P}{I_1P}\\ &=\frac{I_1P\sin\angle I_1PB}{I_3P\sin\angle CPI_3}\cdot\frac{I_3P}{I_1P}\\ &=\frac{\sin\angle I_1'PB}{\sin\angle I_3'PC}\\ &=\frac{\sin\angle A'C'B'}{\sin\angle C'B'D'}\\ &=\frac{A'B'}{C'D'} \end{align*}[asy][asy] size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.9630821987450637, xmax = 1.4943773625694903, ymin = -1.6755582349551141, ymax = 1.3217358451772825; 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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Meanwhile, notice that $E$ is the center of spiral similarity sending $\overline{MN}$ to $\overline {D'A'}$. Therefore, $$\angle EMP=90^{\circ}+\angle EMD'=90^{\circ}=\angle ENA'=\angle ENI_1'$$hence $\sin\angle EMP=\sin\angle ENP$. Therefore, by Ceva's theorem we have $$\frac{\sin\angle I_1'PE}{\sin\angle I_3'PE}=\frac{\sin\angle NPE}{\sin\angle MPE}=\frac{\sin\angle EMN}{\sin\angle ENM}\cdot\frac{\sin\angle ENP}{\sin\angle PME}=\frac{EN}{EM}=\frac{A'N}{MD'}=\frac{A'B'}{C'D'}$$as desired. Therefore, $P,E,R$ are collinear, by symmetry $P,E,Q$ are also collinear. So $P,Q,R$ are collinear as desired.

I guess the quadrilateral is not cyclic @ above

gabrupro wrote: I guess the quadrilateral is not cyclic @ above Are you referring to the quadrilateral in the lemma? The labelling in the lemma is irrelevant to the labelling of points in the problem. So by the statement of the lemma I am referring to any cyclic quadrilateral and I apply it to the cyclic quadrilateral $A'B'C'D'$ in the main proof.

Nice problem! We use cross product of vectors for linearity. It is sufficient to show that $\overrightarrow{PQ} \times \overrightarrow{PR} = \overrightarrow{0}$. Let vertices $ A $, $ B $, $ C $ and $ D $ are located in the clockwise direction in the plane. Let the incenters of triangles $\bigtriangleup PAB$, $\bigtriangleup PBC$, $\bigtriangleup PCD$ and $\bigtriangleup PDA$ be $I_1$, $I_2$, $I_3$ and $I_4$ respectively, and their radii are $r_1$, $r_2$, $r_3$ and $r_4$ respectively. Consider for $i=1, 2, 3, 4$, $\overrightarrow{v_i}$ is the unit vector starting from $ P $ toward $ I_i $. We show equal angles $\angle API_1$ and $\angle BPI_1$ by $\angle P_1$. Angles $\angle P_2$, $\angle P_3$ and $\angle P_4$ are defined in a similar way. Clearly $\frac{QI_1}{QI_3} = \frac{r_1}{r_3}$. So $$\frac{r_1 + r_3}{r_1 r_3} \overrightarrow{PQ} = \frac{\overrightarrow{PI_1} }{r_1} + \frac{\overrightarrow{PI_3} }{r_3} = \frac{\overrightarrow{v_1} }{\sin \angle P_1} + \frac{\overrightarrow{v_3} }{\sin \angle P_3}.$$In the same manner, we have $$\frac{r_2 + r_4}{r_2 r_4} \overrightarrow{PR} = \frac{\overrightarrow{PI_2} }{r_2} + \frac{\overrightarrow{PI_4} }{r_4} = \frac{\overrightarrow{v_2} }{\sin \angle P_2} + \frac{\overrightarrow{v_4} }{\sin \angle P_4}.$$If we show the unit vector perpendicular to the plane outwards by $\overrightarrow{u}$, then obviously $$ \overrightarrow{v_1} \times \overrightarrow{v_2} = - \sin \left( \angle P_1 + \angle P_2 \right) \overrightarrow{u}, \quad \overrightarrow{v_1} \times \overrightarrow{v_4} = \sin \left( \angle P_1 + \angle P_4 \right) \overrightarrow{u}, $$$$ \overrightarrow{v_3} \times \overrightarrow{v_2} = \sin \left( \angle P_3 + \angle P_2 \right) \overrightarrow{u}, \quad \overrightarrow{v_3} \times \overrightarrow{v_4} = - \sin \left( \angle P_3 + \angle P_4 \right) \overrightarrow{u}. $$By above descriptions, we simply conclude that $$ \frac{r_1 + r_3}{r_1 r_3} \overrightarrow{PQ} \times \frac{r_2 + r_4}{r_2 r_4} \overrightarrow{PR} = \overrightarrow{0}.$$And finally $\overrightarrow{PQ} \times \overrightarrow{PR} = \overrightarrow{0}$. So $ P $, $ Q $ and $ R $ are colinear and we are done. $\blacksquare$