By ELMO 2011 Problem 1, quadrilateral $ABCD$ has incircle $\omega$. Now construct parallelograms $WPOS$, $XPOQ$. $YQOR$, $ZROS$. Now $AP-AS = WS-WP$, so we can draw a circle $\omega_A$ tangent to all four lines. Define $\omega_B$, $\omega_C$, $\omega_D$ similarly. Let $K$ be the exsimilicenter of $\omega_A$ and $\omega_C$. We claim this is the desired concurrency point. By Monge on $\omega_A$, $\omega_C$, $\omega$, we get $KAC$ collinear. By Monge on $\omega_A$, $\omega_B$, $\omega_C$ we get $KPQ$ collinear. By Monge on $\omega_A$, $\omega_D$, $\omega_C$ we get $KRS$ collinear. Thus done. Degenerate cases left as annoying exercise to reader. Also, I heard that you can use straight Cartesian coordinates on this?

The first step is to show that $ABCD$ also has an incircle.

We have that $PR=PM+MU+UR=PE+RL+OS$ and $PR=PN+NT+TR=PH+RI+OQ$, so $2PR=EH+LI+OQ+OS=PR+QS$. This means that $PR=QS$, so $EH+IL=FK+GJ\implies AB+CD=AD+BC$, as desired. Now, let $\omega$ be the incircle of $ABCD$, and let $Z$ be the exsimilicenter of $\omega_1$ and $\omega_3$. By Monge on $\omega_1$, $\omega_3$, and $\omega$, $Z$ lies on $AC$. Let $\ell_1$ and $\ell_2$ be the common external tangents of $\omega_1$ and $\omega_3$. Now, by the dual of Desargues' Involution Theorem on $Z$ and $APOS$, $ZA,ZO;ZP,ZS;\ell_1,\ell_2$ are pairs of an involution. Similarly, $ZC,ZO;ZQ,ZR;\ell_1,\ell_2$ are pairs of an involution. Since these two involutions share the pairs $ZAC,ZO$ and $\ell_1,\ell_2$, they must be the same involution $\Phi$. Let $\heartsuit$ (sorry out of letters) be the intersection of $PQ$ and $RS$. By the dual of Desargues' Involution Theorem on $Z$ and $PQSR$, $ZP,ZR;ZQ,ZS;ZO,Z\heartsuit$ are pairs of an involution. But they share the pairs $ZP,ZR$ and $ZQ,ZS$ with $\Phi$, so $ZO,Z\heartsuit$ is a pair of $\Phi$. But $ZO,ZAC$ is a pair of $\Phi$, so the result follows.

On an inspection,found out that my solution is almost the same solution as the first one, so not posting it again. Just wanted to remark that this problem is quite easy for a G7. But the configuration looks quite pretty and congrats to the problem creator(s) for finding such a beautiful property.

Yes, it is easy after one finds the construction of $W$, $X$, $Y$, and $Z$. However, how would one be motivated to construct these points and notice that it is the concurrency point of $AC, PQ$, and $RS$? Thank you very much!

Dear Mathlinkers, jus have a look at http://jl.ayme.pagesperso-orange.fr/Docs/Demir.pdf p. 26-27... Sincerely Jean-Louis

The motivation comes from just length chasing all the way.Doing ELMO problems helps very much.!! This was how I came to the solution after making a very nice diagram.

jayme wrote: Dear Mathlinkers, jus have a look at http://jl.ayme.pagesperso-orange.fr/Docs/Demir.pdf p. 26-27... Sincerely Jean-Louis Yeah,I have looked at .. but unfortunately , I understood nothing....Is there any copy of English version ?

Hallow ^^^^^^^^^^

v_Enhance wrote: Also, I heard that you can use straight Cartesian coordinates on this? Possibly. It is doable with complex numbers if you set it up correctly: Let $O$ be the origin, and let the internal and external angle bisectors of $\angle SOP$ be the real and imaginary axes. Then set the coordinates of the incenters as your variables, and let the equation $SO$ and $OP$ be $z/\overline{z} = t, z/\overline{z} = 1/t$ (so five variables in total). $DA$ is the reflection of $RP$ across the line joining the incenters of $ASOP$ and $DROS$ etc. so we can compute $A$ and the equation of $DA$. Intersect $DA$ with $SO$, which ends up being surprisingly easy to get $S$. Then compute $AS$. Similarly compute $AP, BP, BQ, CR, CQ, DS, DR$. Now verify that $\frac{AS}{AP}\cdot \frac{BP}{BQ} \cdot \frac{CR}{CQ} \cdot \frac{DS}{DR} = 1$, so we are done by two Menelaus'. It is probably doable with Cartesian with the same setup but would be messier I think.

ABCDE wrote: But they share the pairs $ZP,ZR$ and $ZQ,ZS$ with $\Phi$ But why are $ZP,ZR$ and $ZQ,ZS$ (reciprocal)pairs of $\Phi$? $ZP,ZS$ and $ZQ,ZR$ indeed are but i don't see how this implies the statement. Im very sorry if this is a silly question...

The trig bash is fairly straightforward. [asy][asy] unitsize(75); pair A, B, C, D, P, Q, R, S, O, Ia, Ib, Ic, Id, X = (-0.8, 1), Y = reflect((0, 0), (1, 0)) * X; real ra, rb, rc, rd; O = origin; Ia = 0.7 * (0, 1); Ib = 0.6 * (-1, 0); Ic = 0.9 * (0, -1); Id = 1.0 * (1, 0); ra = abs(foot(Ia, O, X)-Ia); rb = abs(foot(Ib, O, X)-Ib); rc = abs(foot(Ic, O, X)-Ic); rd = abs(foot(Id, O, X)-Id); P = extension(O, X, reflect(Ia, Ib) * O, reflect(Ia, Ib) * Y); Q = extension(O, Y, reflect(Ib, Ic) * O, reflect(Ib, Ic) * X); R = extension(O, X, reflect(Ic, Id) * O, reflect(Ic, Id) * Y); S = extension(O, Y, reflect(Id, Ia) * O, reflect(Id, Ia) * X); A = extension(P, reflect(P, Ia) * O, S, reflect(S, Ia) * O); B = extension(Q, reflect(Q, Ib) * O, P, reflect(P, Ib) * O); C = extension(R, reflect(R, Ic) * O, Q, reflect(Q, Ic) * O); D = extension(S, reflect(S, Id) * O, R, reflect(R, Id) * O); draw(circle(Ia, ra)^^circle(Ib, rb)^^circle(Ic, rc)^^circle(Id, rd), gray(0.6)); draw(A--B--C--D--cycle, gray(0.4)); draw(P--R^^Q--S); draw(Ia--P--Ib--Q--Ic--R--Id--S--cycle); draw(Ia--Ib--Ic--Id--cycle); draw(Ia--Ic^^Ib--Id); dot(A^^B^^C^^D^^P^^Q^^R^^S^^O^^Ia^^Ib^^Ic^^Id); label("$A$", A, dir(A - Ia)); label("$B$", B, dir(B - Ib)); label("$C$", C, dir(C - Ic)); label("$D$", D, dir(D - Id)); label("$P$", P, rotate(90, O) * dir(A - B)); label("$Q$", Q, rotate(90, O) * dir(B - C)); label("$R$", R, rotate(90, O) * dir(C - D)); label("$S$", S, rotate(90, O) * dir(D - A)); label("$O$", O, dir(290)); label("$I_A$", Ia, dir(90)); label("$I_B$", Ib, dir(180)); label("$I_C$", Ic, dir(270)); label("$I_D$", Id, dir(0)); label("$\alpha$", Ia, 4*dir(dir(Ib-Ia)+dir(O-Ia))); label( "$\beta$", Ib, 3*dir(dir(Ic-Ib)+dir(O-Ib))); label("$\gamma$", Ic, 3*dir(dir(Id-Ic)+dir(O-Ic))); label("$\delta$", Id, 5*dir(dir(Ia-Id)+dir(O-Id))); label("$\kappa$", O, 3*dir(dir(Ia-O)+dir(S-O))); label("$\overline{\alpha}$", Ib, 3*dir(dir(Ia-Ib)+dir(O-Ib))); label("$\overline{\alpha}$", P, 3*dir(dir(Ia-P)+dir(O-P))); label("$\delta$", S, 4*dir(dir(Ia-S)+dir(O-S))); label("$\kappa$", O, 3*dir(dir(Ia-O)+dir(P-O))); label("$\overline{\kappa}$", O, 3*dir(dir(Id-O)+dir(S-O))); [/asy][/asy] By two applications of Menelaus, the conclusion is equivalent to \[\frac{AP}{PB} \cdot \frac{BQ}{QC} \cdot \frac{CR}{RD} \cdot \frac{DS}{SA} = 1 \iff \frac{AS}{AP} \cdot \frac{BP}{BQ} \cdot \frac{CQ}{CR} \cdot \frac{DR}{DS} = 1.\]Let $I_A$, $I_B$, $I_C$, $I_D$ denote the incenters of $APOS$, $BQOP$, $CROQ$, $DSOR$ respectively. Introduce $\alpha = \angle OI_AI_B$, $\beta = \angle OI_BI_C$, $\gamma = \angle OI_CI_D$, $\delta = \angle OI_DI_A$, and $\kappa = \angle I_AOS$. For convenience, let $\overline{\phi}$ denote the complement of any angle $\phi$. Since $\angle I_AOI_B = \angle I_API_B = 90^{\circ}$, quadrilateral $I_AOI_BP$ is cyclic, so $\angle I_APO = \angle I_AI_BO = \overline{\alpha}$; similarly $\angle I_ASO = \delta$. It follows that \[\angle I_AAP = \angle I_AAS = 180^{\circ} - \overline{\alpha} - \delta - \kappa.\]Letting $r_A$ denote the inradius of $APOS$, observe $AP = r_A(\cot \overline{\alpha} + \cot (180^{\circ} - \overline{\alpha} - \delta - \kappa))$ and $AS = r_A(\cot \delta + \cot (180^{\circ} - \overline{\alpha} - \delta - \kappa))$, so \begin{align*} \frac{AS}{AP} & = \frac{\cot \overline{\alpha} + \cot (180^{\circ} - \overline{\alpha} - \delta - \kappa)}{\cot \delta + \cot (180^{\circ} - \overline{\alpha} - \delta - \kappa)}\\ & = \frac{\cot \overline{\alpha} - \cot (\overline{\alpha} + \delta + \kappa)}{\cot \delta - \cot (\overline{\alpha} + \delta + \kappa)}\\ & = \frac{\cos \overline{\alpha} \sin (\overline{\alpha} + \delta + \kappa) - \sin \overline{\alpha} \cos (\overline{\alpha} + \delta + \kappa)}{\sin \overline{\alpha} \sin (\overline{\alpha} + \delta + \kappa)} \div \frac{\cos \delta \sin (\overline{\alpha} + \delta + \kappa) - \sin \delta \cos (\overline{\alpha} + \delta + \kappa)}{\sin \delta \sin (\overline{\alpha} + \delta + \kappa)}\\ & = \frac{\sin(\delta + \kappa)}{\sin \overline{\alpha} \sin (\overline{\alpha} + \delta + \kappa)} \div \frac{\sin(\overline{\alpha} + \kappa)}{\sin \delta \sin (\overline{\alpha} + \delta + \kappa)}\\ & = \frac{\sin(\delta + \kappa)}{\sin(\overline{\alpha} + \kappa)} \cdot \frac{\sin \delta}{\cos \alpha}. \end{align*}Similarly, \[\frac{BP}{BQ} = \frac{\sin(\alpha + \overline{\kappa})}{\sin(\overline{\beta} + \overline{\kappa})} \cdot \frac{\sin \alpha}{\cos \beta}, \frac{CQ}{CR} = \frac{\sin(\beta + \kappa)}{\sin(\overline{\gamma} + \kappa)} \cdot \frac{\sin \beta}{\cos \gamma}, \frac{DR}{DS} = \frac{\sin(\gamma + \overline{\kappa})}{\sin(\overline{\delta} + \overline{\kappa})} \cdot \frac{\sin \gamma}{\cos \delta}.\]Since $\overline{\alpha} + \kappa$ and $\alpha + \overline{\kappa}$ are supplementary, $\sin(\overline{\alpha} + \kappa) = \sin(\alpha + \overline{\kappa})$. Recognizing similar equalities, \begin{align*} \frac{AS}{AP} \cdot \frac{BP}{BQ} \cdot \frac{CQ}{CR} \cdot \frac{DR}{DS} & = \tan \alpha \tan\beta \tan \gamma \tan \delta\\ & = \frac{OI_B}{OI_A} \cdot \frac{OI_C}{OI_B} \cdot \frac{OI_D}{OI_C} \cdot \frac{OI_A}{OI_D}\\ & = 1 \end{align*}as desired.

rmtf1111 wrote: ABCDE wrote: But they share the pairs $ZP,ZR$ and $ZQ,ZS$ with $\Phi$ But why are $ZP,ZR$ and $ZQ,ZS$ (reciprocal)pairs of $\Phi$? $ZP,ZS$ and $ZQ,ZR$ indeed are but i don't see how this implies the statement. Im very sorry if this is a silly question... I think it's just typos in that solution. When I apply DDIT I get the following: Through $Z$ to $APOS$ one gets $ZAC \leftrightarrow ZO$, $ZS \leftrightarrow ZP$, and the two tangents. Through $Z$ to $CROQ$ one gets $ZAC \leftrightarrow ZO$, $ZQ \leftrightarrow ZR$, and the two tangents. Through $Z$ to $PQRSOK$ one gets $Z\heartsuit \leftrightarrow ZO$, $ZQ \leftrightarrow ZR$, $ZS \leftrightarrow ZP$.

Here's a straightforward solution using Monge and Pappus. Let $\omega_A$, $\omega_B$, $\omega_C$, $\omega_D$ be the incircles of $APOS$, $BQOP$, $CROQ$, $DSOR$. Claim: $ABCD$ has an incircle $\omega$. Proof. Verify that $AB+CD = AD+BC$ by many applications of Pitot's theorem, as in ELMO 2011. $\blacksquare$ [asy][asy] size(350); defaultpen(fontsize(10pt)); pair A, B, C, D, P, Q, R, S, O, Ia, Ib, Ic, Id, I, E, F, G, H, X, Y, Z; pair M = (-0.8, 1), N = reflect((0, 0), (1, 0)) * M; real ra, rb, rc, rd; O = origin; Ia = 0.6 * (0, 1); Ib = 0.6 * (-1, 0); Ic = 0.9 * (0, -1); Id = 1.0 * (1, 0); ra = abs(foot(Ia, O, M)-Ia); rb = abs(foot(Ib, O, M)-Ib); rc = abs(foot(Ic, O, M)-Ic); rd = abs(foot(Id, O, M)-Id); P = extension(O, M, reflect(Ia, Ib) * O, reflect(Ia, Ib) * N); Q = extension(O, N, reflect(Ib, Ic) * O, reflect(Ib, Ic) * M); R = extension(O, M, reflect(Ic, Id) * O, reflect(Ic, Id) * N); S = extension(O, N, reflect(Id, Ia) * O, reflect(Id, Ia) * M); A = extension(P, reflect(P, Ia) * O, S, reflect(S, Ia) * O); B = extension(Q, reflect(Q, Ib) * O, P, reflect(P, Ib) * O); C = extension(R, reflect(R, Ic) * O, Q, reflect(Q, Ic) * O); D = extension(S, reflect(S, Id) * O, R, reflect(R, Id) * O); E = extension(A, B, Q, S); F = extension(B, C, P, R); G = extension(C, D, Q, S); H = extension(D, A, P, R); X = extension(E, F, G, H); Y = extension(E, H, F, G); Z = extension(F, S, P, G); I = extension(B, incenter(A, B, C), D, incenter(A, D, C)); draw(circle(Ia, ra)^^circle(Ib, rb)^^circle(Ic, rc)^^circle(Id, rd), mediumblue+linewidth(0.4)); draw(A--B--C--D--cycle, heavygreen); draw(P--R^^Q--S, heavygreen); draw(A--E--S^^Q--G--C, heavygreen); draw(A--H--P^^B--F--P, heavygreen); draw(G--X^^E--F, dashed+lightblue); draw(E--Y^^F--G, dashed+lightblue); draw(CP(I, foot(I, A, B)), lightblue); draw(C--X^^D--Y, heavycyan+dashed); draw(F--S^^G--Z, purple+dotted); dot("$A$", A, dir(120)); dot("$B$", B, dir(260)); dot("$C$", C, dir(300)); dot("$D$", D, dir(D)); dot("$O$", O, dir(270)); dot("$P$", P, dir(90)); dot("$Q$", Q, dir(180)); dot("$R$", R, dir(R)); dot("$S$", S, dir(100)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$G$", G, dir(G)); dot("$H$", H, dir(90)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot(Z); [/asy][/asy] Let $E = \overline{AB}\cap \overline{QS}$, $F = \overline{BC}\cap \overline{PR}$, $G = \overline{CD}\cap\overline{QS}$, $H = \overline{DA}\cap\overline{PR}$; furthermore, let $X$ be the exsimilicenter of $\omega_A$ and $\omega_C$ and $Y$ be the exsimilicenter of $\omega_B$ and $\omega_D$. Now by Monge, we have: $\overline{XAC}$ collinear from $\omega$, $\omega_A$, and $\omega_C$; similarly, $\overline{YBD}$ collinear. $\overline{EXF}$ collinear from $\omega_A$, $\omega_B$, and $\omega_D$. Similarly, $\overline{GHX}$, $\overline{EHY}$, $\overline{FYG}$ all collinear. Let $Z = \overline{PQ}\cap \overline{RS}$; we will show that $Z$ lies on $\overline{AC}$. Pappus is actually enough to finish: From $\overline{PRF}$ and $\overline{SQG}$, we get that $\overline{FS}\cap \overline{GP}$ lies on $\overline{ZC}$. From $\overline{FHP}$ and $\overline{GES}$, we get that $\overline{FS}\cap \overline{GP}$ lies on $\overline{AX}$. Since $\overline{XAC}$ collinear, it follows that $\overline{ZAC}$ collinear, as desired.

Fun problem. Posting for storage. $\textbf{Claim 01.}$ $ABCD$ has an incircle. $\textit{Proof.}$ To prove this let the incircle of $APOS$ tangent to $AD$ at $A_D$ and $AB$ at $A_B$. Define others similarly. Then, we have \begin{align*} (AB + CD) - (AD + BC) &= (AP - AS) + (BP - BQ) + (CR - CQ) + (RD - RS) \\ &= 2(OP - OS) + 2(OR - OQ) \\ &= 2(PR - QS) \end{align*}It suffices to prove $PR = QS$. However, \[ PR = PA_B + A_D D_A + RD_C = PB_A + B_C C_B + RC_D \]Therefore, we have \[ PR = \frac{1}{2} (A_B B_A + B_C C_B + C_D D_C + D_A A_D) \]Similarly, we will then get $PR = QS$. Now, let the incircle of $APOS$, $BPOQ, CQOR, RODS$ be $\omega_1, \omega_2, \omega_3, \omega_4$ respectively. Let the incircle of $ABCD$ be $\omega$. Monge Theorem on $\omega_1, \omega_3, \omega$ gives us $A, C$ and the exsimilicenter of $\omega_1$ and $\omega_3$ collinear. Let the exsimilicenter of $\omega_1$ and $\omega_3$ be $N$. Furthermore, let the two tangents of $\omega_1$ and $\omega_3$ be $\ell_a, \ell_b$. By Dual Desargues Involution Theorem on $N,APOS$, we have an involution $\varphi$ swapping $(NA, NO); (NP,NS); (\ell_a, \ell_b)$. Similarly, by DDIT on $N,CROQ$, we have an involution $\varphi'$ swapping $(NO, NC); (NQ, NR); (\ell_a, \ell_b)$. However, $NC \equiv NA$. Therefore, $\varphi = \varphi'$ as both of them swaps $(\ell_a, \ell_b)$ and $(NO, NA \equiv NC)$. Now, let $PQ \cap RS = M$. We have DDIT on $N,PQSR$ would gives us an involution $\varphi''$ which swaps $(NP,NS), (NR, NQ), (NO,NM)$. Now, notice that involution $\varphi$ swaps $(NQ,NR)$ and $(NP,NS)$. Therefore, $\varphi = \varphi''$. This means that $\varphi$ swaps $(NO,NM)$ as well. However, we know that $\varphi$ swaps $(NO, AC)$. Therefore, $N,M,A,C$ collinear, which proves that $M$ lies on $AC$.

When you realize that Menelaus is sufficient to solve a G7... [asy][asy] size(8cm); 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 = -28.457435948614894, xmax = 24.665018487480715, ymin = -27.96024239952826, ymax = 12.618241020753967; /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0); pen zzttff = rgb(0.6,0.2,1); /* draw figures */draw(circle((-5.400054273068851,-7.700989482364915), 4.345816949812666), linewidth(0.8)); draw((-16.244173370203114,-12.01688134684045)--(4.726075344215098,-0.48369824585354604), linewidth(0.8) + blue); draw((-4.552623445137903,-14.206195513368455)--(-11.70228968055285,3.7580711078066984), linewidth(0.8) + qqwuqq); draw((-1.7999789027426196,-10.13524137755767)--(-9.412652225566063,-6.032102837654061), linewidth(0.8) + zzttff); draw((-8.2563457259138,-10.976309939615467)--(-6.526533602176052,-3.503708304939073), linewidth(0.8) + zzttff); draw((-16.244173370203114,-12.01688134684045)--(-6.526533602176052,-3.503708304939073), linewidth(0.8) + blue); draw((-8.59157022251616,-4.057929232327401)--(-9.412652225566063,-6.032102837654061), linewidth(0.8)); draw((-8.59157022251616,-4.057929232327401)--(-6.526533602176052,-3.503708304939073), linewidth(0.8)); draw((-9.412652225566063,-6.032102837654061)--(-10.6139120965913,-8.920359457220973), linewidth(0.8)); draw((-10.6139120965913,-8.920359457220973)--(-4.552623445137903,-14.206195513368455), linewidth(0.8)); draw((-4.552623445137903,-14.206195513368455)--(4.726075344215098,-0.48369824585354604), linewidth(0.8)); draw((-6.526533602176052,-3.503708304939073)--(4.726075344215098,-0.48369824585354604), linewidth(0.8)); draw((-16.244173370203114,-12.01688134684045)--(-1.7999789027426196,-10.13524137755767), linewidth(0.8) + blue); draw((-8.2563457259138,-10.976309939615467)--(-11.70228968055285,3.7580711078066984), linewidth(0.8) + qqwuqq); draw((-11.70228968055285,3.7580711078066984)--(-1.7999789027426196,-10.13524137755767), linewidth(0.8) + qqwuqq); /* dots and labels */dot((-6.526533602176052,-3.503708304939073),dotstyle); label("$A$", (-6.358158152299367,-2.6618310555556275), NE * labelscalefactor); dot((-9.412652225566063,-6.032102837654061),dotstyle); label("$B$", (-10.651732124154956,-5.5663075659285255), NE * labelscalefactor); dot((-8.2563457259138,-10.976309939615467),dotstyle); label("$C$", (-9.01007148785723,-12.511794873341977), NE * labelscalefactor); dot((-1.7999789027426196,-10.13524137755767),dotstyle); label("$D$", (-1.180613068591158,-10.533383337290873), NE * labelscalefactor); dot((-8.59157022251616,-4.057929232327401),linewidth(4pt) + dotstyle); label("$E$", (-8.420757413288817,-3.7141776172849386), NE * labelscalefactor); dot((-10.6139120965913,-8.920359457220973),linewidth(4pt) + dotstyle); label("$F$", (-10.441262811809093,-8.59706566370894), NE * labelscalefactor); dot((-4.552623445137903,-14.206195513368455),linewidth(4pt) + dotstyle); label("$G$", (-4.463934341186608,-15.458365246184048), NE * labelscalefactor); dot((4.726075344215098,-0.48369824585354604),linewidth(4pt) + dotstyle); label("$H$", (4.880903126969672,-0.1361993074052816), NE * labelscalefactor); dot((-11.70228968055285,3.7580711078066984),linewidth(4pt) + dotstyle); label("$I$", (-11.535703236007576,4.115280801981134), NE * labelscalefactor); dot((-16.244173370203114,-12.01688134684045),linewidth(4pt) + dotstyle); label("$J$", (-16.839529907123303,-11.459448311612666), NE * labelscalefactor); dot((-7.367051894868528,-7.134656592188692),linewidth(4pt) + dotstyle); label("$K$", (-7.200035401682816,-6.787029577534526), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Lemma. Suppose $EFGH$ is a circumscribed quadrilateral and the touching points on side $EF,FG,GH,HE$ are $B,C,D,A$ respectively, then $$\frac{EK}{KG}=\frac{BE}{CG}$$Proof. By applying Pascal's theorem to $BBDCCA$ and $AACDDB$ we obtain that $J,F,K,H$ are collinear, by symmetry $I,E,K,G$ are collinear. Moreover, $$(I,K;E,G)\overset{B}{=}(BC,BD;BB,BG)=-1$$Therefore, by Menelaus theorem, $$\frac{EK}{KG}=\frac{EI}{IG}=\frac{EB}{BF}\frac{FC}{CG}=\frac{BE}{CG}$$$\blacksquare$ [asy][asy] size(10cm); 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 = -27.356605005249566, xmax = 21.147225047890352, ymin = -21.167503371439867, ymax = 20.034674845743506; /* image dimensions */ /* draw figures */draw(circle((-5.095929724659887,2.027115961420098), 2.2041302756648538), linewidth(0.8)); draw(circle((-2.924579145174239,-5.056135709280694), 4.725512814009619), linewidth(0.8)); draw(circle((2.0550590444698473,4.166553169351109), 4.797167263167613), linewidth(0.8)); draw(circle((5.938980623808618,-3.9310012117447393), 3.035012815620033), linewidth(0.8)); draw((-7.9865872209071895,-11.687338507312862)--(9.581370927343654,-5.931370365143414), linewidth(0.8)); draw((-7.9865872209071895,-11.687338507312862)--(-7.245955629908703,3.1671768111928897), linewidth(0.8)); draw((-7.245955629908703,3.1671768111928897)--(4.92312734479103,12.332046685258797), linewidth(0.8)); draw((4.92312734479103,12.332046685258797)--(9.581370927343654,-5.931370365143414), linewidth(0.8)); draw((-7.405529462161146,-0.03332390257078012)--(8.334675708951755,-1.0434948591234776), linewidth(0.8)); draw((-4.659903153931072,5.11480379994105)--(4.819056360177046,-7.4916957427587825), linewidth(0.8)); /* dots and labels */dot((-0.4534949668144329,-0.47948983622964875),dotstyle); label("$O$", (-0.27964611357737224,-0.04486770313699714), NE * labelscalefactor); dot((-5.095929724659887,2.027115961420098),dotstyle); label("$O_1$", (-4.930102937668744,2.475940668800382), NE * labelscalefactor); dot((5.938980623808618,-3.9310012117447393),dotstyle); label("$O_3$", (6.109299242884607,-3.478382554568945), NE * labelscalefactor); dot((-2.924579145174239,-5.056135709280694),dotstyle); label("$O_2$", (-4.712791871122419,-4.999560020393226), NE * labelscalefactor); dot((2.0550590444698473,4.166553169351109),dotstyle); label("$O_4$", (2.806171031380454,3.997118134624663), NE * labelscalefactor); dot((-7.245955629908703,3.1671768111928897),linewidth(4pt) + dotstyle); label("$A$", (-7.059751389822737,3.519033788222746), NE * labelscalefactor); dot((-7.9865872209071895,-11.687338507312862),linewidth(4pt) + dotstyle); label("$B$", (-7.7986090160802455,-11.34504316354594), NE * labelscalefactor); dot((9.581370927343654,-5.931370365143414),linewidth(4pt) + dotstyle); label("$C$", (9.76012516086288,-5.564568793413673), NE * labelscalefactor); dot((4.92312734479103,12.332046685258797),linewidth(4pt) + dotstyle); label("$D$", (5.109668336771508,12.689560796477695), NE * labelscalefactor); dot((-7.405529462161146,-0.03332390257078012),linewidth(4pt) + dotstyle); label("$P$", (-7.233600243059798,0.3028300033371242), NE * labelscalefactor); dot((4.819056360177046,-7.4916957427587825),linewidth(4pt) + dotstyle); label("$Q$", (4.979281696843712,-7.1292084725472185), NE * labelscalefactor); dot((8.334675708951755,-1.0434948591234776),linewidth(4pt) + dotstyle); label("$R$", (8.49972097489419,-0.6968009027759746), NE * labelscalefactor); dot((-4.659903153931072,5.11480379994105),linewidth(4pt) + dotstyle); label("$S$", (-4.495480804576093,5.474833387139679), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] We now return to the original problem. For a point $X$ and a circle $\Gamma$ let $f(X,\omega)$ be the length of tangent from $X$ to $\omega$. Let the incircle of $ASOP$ be $\omega_A$, define $\omega_B$, $\omega_C$ and $\omega_D$ similarly. By Menelaus theorem it suffices to show that $$\frac{AP}{AS}\cdot\frac{BQ}{BP}\cdot\frac{RC}{QC}\cdot\frac{SD}{DR}=1 \hspace{20pt}(1)$$Now by Ceva's theorem and the lemma, $$\frac{AP}{AS}=\frac{\sin\angle AOP}{\sin\angle SOA}\cdot\frac{\sin\angle APO}{\sin\angle ASO}=\frac{f(P,\omega_A)}{f(S,\omega_A)}\cdot\frac{OS}{OP}\cdot\frac{\sin\angle APO}{\sin\angle ASO}$$multiplying similar expressions, it suffices to show $$\prod_{cyc}\frac{f(P,\omega_A)}{f(S,\omega_A)}=1 \hspace{20pt}(2)$$Indeed, extend $AP$ and $SO$ to meet at $V$ then $\omega_A$ is the incircle of $\triangle VPO$ while $\omega_B$ is the $V-$excircle of $\triangle VOQ$, therefore, $$f(P,\omega_A)=f(O,\omega_B)=f(Q,\omega_C)$$By symmetry, $$f(R,\omega_D)=f(O,\omega_C)=f(Q,\omega_B)$$$$f(P,\omega_B)=f(O,\omega_A)=f(S,\omega_D)$$$$f(R,\omega_C)=f(O,\omega_D)=f(S,\omega_A)$$Therefore $(2)$ is true, this completes the proof.

Every time I try to do geo with the intention of getting better at things which are not projective spam, I end up using projective spam. Let $\omega_A$ be the incircle of $APOS$, etc. Let $X_{AB}$ be the exsimilicenter of $\omega_A$ and $\omega_B$, etc. Let $X = X_{AC}$ for convenience. By Monge's theorem, $X = X_{AB}X_{BC} \cap X_{AD}X_{DC}$. Lemma: $X \in AC$ Proof 1 (DDIT): By DDIT on point $X_{AC}$ and quadrilateral $APOS$, there exists an involution swapping $(XA, XO), (XX_{AB}, XX_{AD})$, and the two common external tangents of $\omega_A$ and $\omega_C$. By DDIT on point $X_{AC}$ and quadrilateral $CQOR$, there exists an involution swapping $(XC, XO), (XX_{CD}, XX_{CB})$, and the two common external tangents of $\omega_A$ and $\omega_C$. But because $X = X_{AB}X_{BC} \cap X_{AD}X_{DC}$, these two involutions are the same, and because $(XA, XO)$ and $(XC, XO)$ are pairs of this involution, $XA = XC$, i.e. $X \in AC$. Proof 2 (Moving Points): Let $I_A$ be the center of $\omega_A$, etc. Fix $O, P, Q, A, B, C$ and vary $I_D$ projectively along the internal bisector of $\angle POQ$. Then $X_{AD} = I_AI_D \cap OP$ moves projectively, so line $X_{AD}S$, as the second tangent from $X_{AD}$ to $\omega_A$, has degree 2. Thus as $PB$ is also tangent to $\omega_A$, $A = BP \cap X_{AD}S $ has degree 1 (i.e. moves projectively.) Similarly $C$ has degree 1, so to show that $X, A, C$ are collinear, we only need to check the claim for $(0+1+1)+1 = 3$ choices of $I_D$. But for $I_D$ equal to the reflection of $I_B$ over $O$, it is obvious by symmetry, while for $I_D$ such that $\omega_A, \omega_C, \omega_D$ share a common tangent, it is again obvious. For $I_D = O$, we have $A = OS \cap BP = X_{AB}$, and similarly $C = X_{BC}$. Then the claim follows by Monge. Now, we use Pappus. By Pappus on $PQX_{BC}SRX_{CD}$, $PQ \cap RS, QX_{BC} \cap RX_{CD} = C, X_{BC}S \cap X_{CD}P$ are collinear, so it suffices to show $X_{BC}S \cap X_{CD}P \in AC$. By Pappus on $PX_{AB}X_{BC}SX_{AD}X_{CD}$, $PX_{AB} \cap SX_{AD} = A, X_{AB}X_{BC} \cap X_{AD}X_{CD} = X, X_{BC}S\cap X_{CD}P$ are collinear But we know $X \in AC$, so we are done. $\blacksquare$

A projective type solution: As seen in many of the above solutions, we have that quad $ABCD$ has an incircle. Let $W = \overline{AB}\cap \overline{QS}$, $X = \overline{BC}\cap \overline{PR}$, $Y = \overline{CD}\cap\overline{QS}$, $Z = \overline{DA}\cap\overline{PR}$. Then by Monge's theorem, we obtain that lines $AC,WX,YZ$ concur (at the exsimilicenter of incircles of quad $APOS,CROQ$), in particular it suffices to prove the following lemma: Lemma: Let $A,B,C,D$ be any four points and $\ell_1,\ell_2$ be any two lines. Let $\ell_1$ and $\ell_2$ intersect lines $AB,BC,CD,DA$ at points $P,X,R,Z$ and $W,Q,Y,S$, respectively. Then $$\overline{AC},\overline{PQ},\overline{RS} \text{ concur} ~~ \iff ~~ \overline{AC},\overline{WX},\overline{YZ} \text{ concur}$$ proof: Because of symmetry, WLOG assume $\overline{AC},\overline{WX},\overline{YZ} \text{ concur}$. Consider the following argument: $\overline{AC},\overline{WX},\overline{YZ} \text{ concur}$. Then $\triangle AWY, \triangle CXZ$ are perspective. Points $B = \overline{AW} \cap \overline{CX} ~,~ \ell_1 \cap \ell_2 = \overline{WY} \cap \overline{XZ} ~,~ \overline{AY} \cap \overline{CZ}~$ lie on some line $\ell$. Pappus on $\{A,S,Z\},\{C,R,Y\}$ gives points $\overline{AR} \cap \overline{CS} ~,~ \ell_1 \cap \ell_2 = \overline{SY} \cap \overline{RZ} ~,~ \overline{AY} \cap \overline{CZ}$ are collinear. All these three points must lie on $\ell$ as the last two points lie on $\ell$. So points $B = \overline{AP} \cap \overline{CQ} ~,~ \ell_1 \cap \ell_2 ~,~ \overline{AR} \cap \overline{CS}~$ are collinear (as they all lie on $\ell$). Then $\triangle APR, \triangle CQS$ are perspective. Consequently, $ \overline{AC},\overline{PQ},\overline{RS} \text{ concur}$. This completes the proof of the lemma. $\blacksquare$

Let \(I_A\), \(I_B\), \(I_C\), \(I_D\) be the incenters of \(APOS\), \(BQOP\), \(CROQ\), \(DSOR\). We begin with this estimate: Claim: We have \[\frac{SA}{AP}\cdot\frac{SO}{OP}=\left(\frac{SI_A}{PI_A}\right)^2\]and cyclic variations. Proof. This is immediate upon noting by law of sines that \[ \frac{AS}{\sin\frac{A+S}2}=\frac{SI_A}{\sin\frac A2},\quad \frac{AR}{\sin\frac{A+R}2}=\frac{RI_A}{\sin\frac A2},\quad \frac{OS}{\sin\frac{A+R}2}=\frac{SI_A}{\sin\frac O2},\quad \frac{RO}{\sin\frac{A+S}2}=\frac{RI_A}{\sin\frac O2}. \]\(\blacksquare\) Now we may observe that \begin{align*} \frac{AP}{PB}\cdot\frac{BQ}{QC}\cdot\frac{CR}{RD}\cdot\frac{DS}{SA} &=\prod_\mathrm{cyc}\left(\frac{SA}{AP}\cdot\frac{SO}{OP}\right) =\prod_\mathrm{cyc}\left(\frac{SI_A}{PI_A}\right)^2 =\prod_\mathrm{cyc}\left(\frac{PI_B}{PI_A}\right)^2\\ &=\prod_\mathrm{cyc}\left(\frac{\sin\angle POI_B}{\sin\angle POI_A}\right)^2 =\prod_\mathrm{cyc}\left(\frac{\sin\angle SOI_A}{\sin\angle POI_A}\right)^2 =1, \end{align*}so by Menelaus on \(\triangle ABC\) and \(\triangle ADC\) we have that \(\overline{PQ}\) and \(\overline{RS}\) intersect \(\overline{AC}\) at the same point.

Let the incircles of $APOS,BQOP,CROQ,$ and $DSOR$ be $\omega_1,\omega_2,\omega_3,$ and $\omega_4,$ respectively. Let $\omega_1$ and $\omega_2$ touch $\overline{AB}$ at $A_1$ and $A_2.$ Let $\omega_1$ touch $\overline{PR}$ and $\overline{QS}$ at $P_1$ and $Q_1.$ Similarly define $B_1,B_2,$ etc and $P_2,Q_2,$ etc. WLOG let $A_1$ be closer to $P$ than $A_2$ and let $A_3$ be closer to $Q$ than $A_4.$ Notice $$2PP_1+P_1P_2=A_1A_2=Q_1Q_2=OP_2+OP_1=2OP_2+P_1P_2$$so $PP_1=OP_2.$ Hence, $$PR-QS=(PA_1+A_1A_3+RA_3)-(QQ_3+Q_3Q_1+SQ_1)=(PA_1-QQ_3)+(RA_3-SQ_1)=0.$$By Pitot, $DR=OR+SD-OS$ etc, so $$AB+CD=2(OR+OP-OS-OQ)+SD+CQ+BQ+AS=BC+AD+2(PR-QS)=BC+AD.$$Therefore, $ABCD$ has an incircle, $\omega.$ Let $W=\overline{AB}\cap\overline{QS},$ $X=\overline{BC}\cap\overline{PR},$ $Y=\overline{CD}\cap\overline{QS},$ $Z=\overline{DA}\cap\overline{PR},$ and let $T$ be the exsimilicenter of $\omega_1$ and $\omega_3.$ By Monge $\omega_1,\omega_3,$ and $\omega,$ we see $T,A,$ and $C$ are collinear. By Monge on $\omega_1,\omega_2,$ and $\omega_3,$ we see $T$ lies on $\overline{WX}$ and similarly $\overline{YZ}.$ Let $U=\overline{PQ}\cap\overline{RS}$ and $V=\overline{SX}\cap\overline{PY}.$ By Pappus on $\{X,P,Z\}$ and $\{W,Y,S\},$ we see $V,A,$ and $T$ are collinear. By Pappus on $\{P,R,X\}$ and $\{Y,S,Q\},$ we see $V,U$ and $C$ are collinear. Hence, we ascertain that $U,A,$ and $C$ are collinear. $\square$

We easily see, that $ABCD$ has incircle, so exsimilicenter $K$ of incircles $APOS,CROQ$ lies on $AC$ by Monge. Let $QS$ meet $AB,CD$ at $W,X$ and $PR$ meet $AD,BC$ at $Y,Z$ respectively. By Monge $K=XZ\cap WY.$ Pappus on $(SWX),(PYZ)$ gives $SZ\cap PX\in AC,$ analogously $QZ\cap RX\in AC.$ Pappus on $(SQX),(PRZ)$ gives $PQ\cap RS\in \overline{(SZ\cap PX)(QZ\cap RX)}= AC,$ so done.

Firstly, we prove that $ABCD$ has an incircle $\omega$. Let $X= AB \cap SQ, Y= BC \cap PR, Z= CD \cap SQ, W= AD \cap SR$. By Pitot Theorem and some length bash, we have that $ABCD$ has an incircle if and only if $PR=QS$. Let $P_A,P_B$ be the touching point of $P$ WRT $\omega_A, \omega_B$, respectively, where $\omega_A$ is the incircle of $APOS$ and $\omega_B, \omega_C, \omega_D$ are defined similarly. Observe that $OP_A=OS_A, OR_C=OQ_C$, so $PR=QS \iff PP_A+RR_C=SS_A+QQ_C$. However, observe that $\omega_B, \omega_D$ are the incircle and $Y-$excircle of $YOQ$, respectively, so $QQ_C=OQ_B$. Similarly, $SS_A=OS_D, RR_C= OR_D, PP_A= OP_B$, but $OP_B=OQ_B, OS_D=OR_D$, so we have the desired equality. Therefore, $ABCD$ has an incircle $\omega$. Let $U$ be the exsimilicenter of $\omega_A, \omega_C$. By Monge's Theorem on $\omega, \omega_A, \omega_C$, we know that $U$ lies on $AC$. Furthermore, by Monge's Theorem on $\omega_A, \omega_B, \omega_C$ and $\omega_C, \omega_D, \omega_A$ we know that $U= XY \cap ZW$. Now, by Pappus Theorem on $\{WYR\}, \{XZQ\}$, we have that $WZ \cap XY=U, WQ \cap XR, YQ \cap ZR=C$ are collinear. Moreover, since $U \in AC$, $WQ \cap XR \in AC$. Again, by Pappus Theorem on $\{WPR\}, \{XSQ\}$, we have that $WS \cap XP= A, WQ \cap XR, PQ \cap SR$ are collinear. But since $WQ \cap XR$ lies on $AC$, we have that $PQ \cap SR$ lies on $AC$. Therefore, $AC, PQ, SR$ are concurrent. $\blacksquare$

Surprisingly, combining spartacle's start with the standard DDIT finish to this problem gives a solution that uses only DDIT and obvious applications of Monge's theorem, which is written up below. Let $\omega_A$, $\omega_B$, $\omega_C$, $\omega_D$ be the incircles of $\square ASOP$, $\square BPOQ$, $\square CQOR$, and $\square DROS$, respectively. Let $T_{AB} = AB\cap QS$, $T_{CD}=CD\cap QS$, $T_{DA}=AD\cap PR$, and $T_{BC}=BC\cap PR$, so that $T_{AB}$ is the exsimilicenter of the incircles of $\omega_A$ and $\omega_B$, and similarly for three other points. Thus, Monge's theorem implies that $X = T_{AB}T_{DA}\cap T_{BC}T_{CD}$ is the exsimilicenter of $\omega_A$ and $\omega_C$. Let $XU$ and $XV$ be the common external tangents of $\omega_A$ and $\omega_C$. Then, we apply DDIT at $X$ twice: On $APOS$, we get an involution $\Phi_1$ swapping $\{XA, XO\}$, $\{XP, XS\}$, $\{XT_{AB}, XT_{DA}\}$, and $\{XU, XV\}$. On $CQOR$, we get an involution $\Phi_2$ swapping $\{XC, XO\}$, $\{XQ, XR\}$, $\{XT_{BC}, XT_{CD}\}$, and $\{XU, XV\}$. From the last two pairs, we get that $\Phi_1=\Phi_2$. Thus, the first pair gives $X\in AC$. Finally, if $Z=PR\cap QS$, then by applying DDIT on $PQSR$, we get an involution $\Phi_3$ swapping $\{XP, XS\}$, $\{XQ, XR\}$, and $\{XO, XZ\}$. From the first two pairs, we get $\Phi_1=\Phi_2=\Phi_3$, so the last pair gives $Z\in AC$, as desired.

Just spam pitot and monge

Claim 1: $ABCD$ has an incircle

Now we spam monge's to every triple of circles Let $\omega_1,\omega_2,\omega_3,\omega_4,\omega$ denote the incircles of $APOS,BPOQ,CQOR,DSOR$ and $ABCD$ respectively Let $G = SQ \cap CD$, $E = SQ \cap AB$, $F = PR \cap BC$, $H= PR \cap AD$, $X$ and $Y$ be the exsimilicenters of $\omega_1,\omega_3$ and $\omega_2,\omega_4$ respectively. By Monge's we get the following collinearities: $X-H-G$,$E-H-Y$,$F-Y-G$,$Y-B-D$,$X-A-C$,$F-X-E$ Now we apply Pappus's theorem to $FPR$ and $GSQ$ to get $FS \cap PG,PQ \cap RS,C$ collinear $HPF$ and $ESG$ to get $FS \cap PG, X, A$ collinear. Hence we're done.

Absolute menace of a problem. Denote the incircles by $\omega_i$ in order $(APOS)$, $(BQOP)$, $(CROQ)$, and $(DSOR)$. 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dot((-207.04159900674418,181.64822758767713),linewidth(4pt) + dotstyle); label("$T$", (-212.09888743744003,186.4567612395372), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: $ABCD$ has an incircle. Proof. Let the tangents from $\omega_1$ and $\omega_2$ meet $\overline{AB}$ at $W_1$ and $W_2$. Consider clockwise and label the tangents from $\overline{BC}$, $\overline{CD}$, $\overline{DA}$ by $X_i$, $Y_i$ and $Z_i$. Similarly denote the tangency points of $\omega_1$ and $\omega_2$ to $\overline{PR}$ by $T_1$ and $T_2$. Continuing, label the tangents at $\overline{QS}$ from $\omega_2$ and $\omega_3$ by $T_3$ and $T_4$, and so on. Then observe that, \begin{align*} AB + CD - AD - BC &= W_1W_2 + Y_1Y_2 - X_1X_2 - Z_1Z_2\\ &= \frac{1}{2} \left(2W_1W_2 + 2Y_1Y_2 - 2X_1X_2 - 2Z_1Z_2 \right) \end{align*}Note then that $W_1W_2 = T_3T_7$ due to the fact that they are both common external tangents to $\omega_1$ and $\omega_2$. Then we have, \begin{align*} \frac{1}{2} \left(2W_1W_2 + 2Y_1Y_2 - 2X_1X_2 - 2Z_1Z_2 \right) &= \frac{1}{2} \left(2T_3T_8 + 2T_4T_7 - 2T_2T_5 - 2T_6T_1 \right)\\ &= \frac{1}{2}(OT_1 + OT_2 + OT_5 + OT_6 - OT_3 - OT_4 - OT_7 - OT_8)\\ &= 0 \end{align*}whence the last step follows by equal tangents. Hence $AB + CD - AD - CB = 0 \iff AB + CD = AD + CB$ and hence $ABCD$ is tangential. $\square$ Let $\omega$ be the incircle of $ABCD$. Now define the following points. $E = \overline{AB} \cap \overline{QS}$. Similarly define the points $F$, $G$, and $H$ with regards to $\overline{BC}$, $\overline{CD}$ and $\overline{DA}$. $T = \overline{PQ} \cap \overline{RS}$. $K$ as the exsimilicenter of $\omega_1$ and $\omega_3$. $M = \overline{FS} \cap \overline{GP}$ and similarly $N = \overline{ER} \cap \overline{HQ}$. Claim: The points $K$, $A$ and $C$ are collinear. Proof. Note from Monge on $\omega_1$, $\omega_3$ and $\omega$ our claim follows. $\square$ Claim: The points $F$, $E$ and $K$ are collinear. Similarly the points $G$, $H$, and $K$ are collinear. Proof. From Monge on $\omega_1$, $\omega_2$ and $\omega_3$ our claim follows. $\square$ Now to finish it off. Claim: The points $K$, $M$, $N$, $A$, $C$ and $T$ are all collinear. Proof. From Pappus Theorem on $\overline{FPH}$ and $\overline{GSE}$ we find $\overline{MKA}$ collinear. Similarly by Pappus Theorem on $\overline{EQG}$ and $\overline{HRF}$ we find $\overline{NKC}$ collinear. Also recall the points $\overline{KAC}$ are collinear. Finally by Pappus Theorem on $\overline{PRF}$ and $\overline{SQG}$ we find $\overline{TMC}$ collinear. Thus as of right now we have, $\overline{MKA}$ $\overline{NKC}$ $\overline{KAC}$ $\overline{TMC}$ Combining the first and third we have $\overline{MKAC}$ collinear. Combining this with the second we have $\overline{MKANC}$ collinear. Finally combining this with the fourth we have $\overline{TMKANC}$ all collinear as desired. Specifically, $T$, $A$ and $C$ are collinear, so we are done. $\square$

We will show $AC$, $PQ$, and $RS$ are concurrent in the projective plane. Claim: $PR=QS$ Proof: Let $O_A$ be the incenter of $APOS$. Define $O_B$, $O_C$, and $O_D$ similarly. Let $O_C'$ and $O_D'$ be the reflection of $O_C$ about $O_BO_D$ and the reflection of $O_D$ about $O_AO_C$, respectively. Let $Q'$, $R'$, and $S'$ be the reflection of $Q$ about $O_BO_D$, the reflection of $R$ about $O$, and the reflection of $S$ about $O_AO_C$. Then, $P$, $Q'$, $R'$, and $S'$ are the intersections of $OP$ with $(OO_AO_B)$, $(OO_BO_C')$, $(OO_C'O_D')$, and $(OO_D'O_A)$, respectively. Let $*$ denote the image of a point under inversion about $O$. Then, $P*$, $Q'*$, $R'*$, and $S'*$ are the intersections of $OP$ with $O_A*O_B*$, $O_B*O_C'*$, $O_C'*O_D'*$, and $O_D'*O_A*$. By DIT, an involution on line $OP$ swaps $P*$ with $R'*$, $Q'*$ with $S'*$, and $O$ with itself. This is an inversion. Assume it has radius $1$. Then, for some $r$, $OP*=1+r$ and $OR'*=1+\frac1r$. Thus, $PR=\frac1{OP*}+\frac1{OR'*}=1$. Similarly, $QS=1$. Then, using Pitot's theorem and adding, we get $AB+CD-AD-BC = 2PR-2QS=0$, so $ABCD$ is tangential. Let $X$ be the exsimilicenter of $\omega_A$ and $\omega_C$. By Monge's theorem on $\omega_A$, $\omega_C$, and the incircle of $ABCD$, $X$ must lie on $AC$. Let $\ell_P$ be the line tangent to $\omega_C$ that passes through $P$ and doesn't pass through $O$. Define $\ell_S$ similarly. Let $Y$ be the intersection of $\ell_P$ and $\ell_S$. By Pitot's theorem, $YP-YS=OP-OS=AP-AS$. Thus, $APYS$ has an incircle. By Monge's theorem on the incircle of $APYS$, $Y$ lies on $AX$. Thus, $Y$ lies on $AC$. Note that $PYSQCR$ has incircle $\omega_C$. Thus, by Brianchon's theorem, $PQ$, $RS$, and $CY$ are concurrent. Line $CY$ is line $AC$, so $PQ$, $RS$, and $AC$ are concurrent.

2015 G7 We start off by showing the following claim. Claim: $ABCD$ is tangential quadrilateral. Proof: Apply pitot theorem multiple times to show that $AB+CD= AD+CB$. $\blacksquare$ Now we introduce the following points. let $H$ be the exsimilicenter of $\omega_A$ and $\omega_C$ and $G$ be the exsimilicenter of $\omega_B$ and $\omega_D$. $Z= \overline{DA}\cap\overline{PR}$, $Y = \overline{AB}\cap \overline{QS}$, $X = \overline{BC}\cap \overline{PR}$, and $W= \overline{CD}\cap\overline{QS}$. Now by Monge’s theorem we have that; $HAC, GBD, YHX, ZWH, YZG, XGW$ are collinear. Hence we can finish the problem by applying Pappus theorem on; $SQW, PRX$ and $XZP, YWS$.

Solved with ohiorizzler1434. This is going to be pretty similar to other solutions, and indeed this problem is an absolute projective menace . Also wow why does this feel like ISL 2022/G8. Work over the projective plane for this question. We first start off by showing $ABCD$ has an incircle $\omega$. However this is just ELMO 2011/1. So let $\omega_1$, $\omega_2$, $\omega_3$, $\omega_4$ be incircles of $APOS$, $BQOP$, $CROQ$ and $DSOR$ resp. We now constuct $K_1=\overline{PR}\cap\overline{AD}$ (the exsimilicentre of $ \omega_1$ and $\omega_4$), $K_2=\overline{PR}\cap\overline{BC}$ (the exsimilicentre of $ \omega_2$ and $\omega_3$), $L_1=\overline{CD}\cap\overline{QS}$ (the exsimilicentre of $ \omega_3$ and $\omega_4$), $L_2=\overline{AB}\cap\overline{QS}$ (the exsimilicentre of $ \omega_1$ and $\omega_2$), $M_1$ the exsimilicentre of $\omega_1$ and $\omega_3$, $M_2$ the exsimilicentre of $\omega_2$ and $\omega_4$. By repeated applications of Monge we have that the following are collinear: $K_1$, $L_2$, $M_2$ ($\omega_1$, $\omega_2$,$\omega_4$) $K_2$, $L_1$, $M_2$ ($\omega_3$, $\omega_2$,$\omega_4$) $K_1$, $L_1$, $M_1$ ($\omega_1$, $\omega_4$,$\omega_3$) $K_2$, $L_2$, $M_1$ ($\omega_1$, $\omega_2$,$\omega_3$) $A$, $C$, $M_1$ ($\omega$, $\omega_1$,$\omega_3$) $B$, $D$, $M_2$ ($\omega$, $\omega_2$,$\omega_4$) So now delete the incircles. We are now left with the following problem: Incircle-less ISL 2015/G7 wrote: We are given a quadrilateral $ABCD$ with points $P$, $Q$, $R$, $S$ on $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DA}$ such that the following are collinear: $K_1$, $L_2$, $M_2$ $K_2$, $L_1$, $M_2$ $K_1$, $L_1$, $M_1$ $K_2$, $L_2$, $M_1$ $A$, $C$, $M_1$ $B$, $D$, $M_2$ where $K_1=\overline{PR}\cap\overline{AD}$, $K_2=\overline{PR}\cap\overline{BC}$, $L_1=\overline{CD}\cap\overline{QS}$, $L_2=\overline{AB}\cap\overline{QS}$, $M_1=\overline{AC}\cap\overline{K_1L_1}$ and $M_2=\overline{BD}\cap\overline{K_1L_2}$. Show that $\overline{AC}$, $\overline{PQ}$, $\overline{RS}$ concur. By Pappus on $\overline{K_1PK_2}$ and $\overline{L_2SL_1}$, $A=\overline{L_2P}\cap\overline{K_1S}$, $M_1=\overline{K_1L_1}\cap\overline{K_2L_2}$, and $U=\overline{L_1P}\cap\overline{K_2S}$ are collinear. Similarly, using Pappus on $\overline{K_1RK_2}$ and $\overline{L_2SL_1}$, $C=\overline{L_1R}\cap\overline{K_2Q}$, $M_1=\overline{K_1L_1}\cap\overline{K_2L_2}$, and $V=\overline{L_2R}\cap\overline{K_1Q}$ are collinear. Thus both $U$ and $V$ lie on $\overline{AC}$, and so $\overline{UV}=\overline{AC}$. Now we finish off by using Pappus on $\overline{K_1PQ}$ and $\overline{L_2SR}$; then $A=\overline{L_2P}\cap\overline{K_1S}$, $V=\overline{L_2R}\cap\overline{K_1Q}$ and $X=\overline{PQ}\cap\overline{RS}$ are collinear. Since $\overline{AV}$ is just $\overline{AC}$ however, we have $\overline{PQ}\cap\overline{RS}\cap\overline{AC}=X$ which completes the proof.