Let $Q$ be the intersection of $XY$ and $ZT$ and $P'$ be the intersection of $YZ$ and $XT$. By Pascal on $XYYZTT$, $Q$, $C$, and $P'$ are collinear. By Pascal on $XXYZZT$, $A$, $Q$, and $P'$ are collinear. Hence, $P'$ is on $AC$. Similarly, we can show that $P'$ is on $BD$, so $P'=P$. Now, invert about $P$ fixing $\Gamma$, and denote by $K'$ the image of point $K$ under this inversion. Since line $ABX$ is tangent to $\Gamma$, the circumcircle of $PA'B'T$ is tangent to $\Gamma$. Because $ABCD$ is cyclic, $A'B'$ is parallel to line $CD$, which is the tangent to $\Gamma$ at $T$. This implies that $TA'=TB'$. If $T^*$ is the antipode of $T$ with respect to $\Gamma$, then clearly the circumcircle of $T^*A'B'$ is also tangent to $\Gamma$ at $T^*$, so $T^*=S'$. But this means that $\angle PST=\angle PXS'=TXT^*=90^\circ$, as desired.

Let $ J $ be the pole of $ SX $ WRT $ \Gamma $ and let $ V $ $ \equiv $ $ AB $ $ \cap $ $ YZ $. It's well-known that $ P $ lies on $ XT, $ $ YZ $ and $ (A,B;V,X) $ $ = $ $ -1 $, so from $ {JX}^2 $ $ = $ $ JA $ $ \cdot $ $ JB $ we get $ J $ is the midpoint of $ VX $, hence $ J $ is the circumcenter of $ \triangle SVX $. Since $ TX, $ $ YZ $ is parallel to the bisector of $ \angle (AB,CD), $ $ \angle (BC,DA) $, respectively, so $ \angle XPV $ $ = $ $ 90^{\circ} $ $ \Longrightarrow $ $ P, $ $ S, $ $ V, $ $ X $ lie on a circle with diameter $ VX $, hence we conclude that $ \measuredangle TSP $ $ = $ $ \measuredangle XSP $ $ - $ $ \measuredangle XST $ $ = $ $ \measuredangle XVP $ $ - $ $ \measuredangle VXP $ $ = $ $ 90^{\circ} $.

Let $O$ and $K$ be the centers of $\odot(ABCD)$ and $\Gamma.$ $E \equiv AD \cap BC$ and $F \equiv AB \cap CD.$ Since $E(F,Y,Z,P)=-1,$ then the pole of $EF$ WRT $\Gamma$ is on $EP$ and similarly it must be on $FP$ $\Longrightarrow$ $P$ is the pole of $EF$ WRT $\Gamma$ $\Longrightarrow$ $X,T,P$ are collinear and $EF \perp PK$ at $Q,$ i.e. $K \in OP.$ Since the polar of $Q$ WRT $(O)$ passes through $P,$ then it follows that $P,Q$ are the limiting points of $\Gamma,(O).$ Thus if the common tangent of $\Gamma,\Omega$ cuts $AB$ at $M,$ we get $MS^2=MX^2=MA \cdot MB$ $\Longrightarrow$ $\odot(M,MS)$ is orthogonal to $(O)$ and $\Gamma$ $\Longrightarrow$ $P \in \odot(M,MS)$ $\Longrightarrow$ $\angle PST=\angle MSP+\angle MST =90^{\circ}-\angle SXT+\angle SXT=90^{\circ}.$

Let $AD\cap BC=E, AB\cap CD=F, XY\cap TZ=G, XZ\cap TY=H, YZ\cap AB=Q$. Note that polar of $H$ wrt $\Gamma$ is $AC$, polar of $G$ is $BD$, hence polar of $GH$ must be $P=XT\cap YZ$. $YZ,XT$ are parallel to the internal angle bisectors of $\angle AEB,\angle AFB$, which are also parallel to the internal bisectors of $\angle APB,\angle APD$ since $ABCD$ is cyclic$\implies PZ\perp PX$, and $PQ$ is the angle bisector of $\angle APB\implies (A,B;Q,X)=-1$. Let the tangent at $S$ to the two circles meet $AB$ at $M$. $MX^2=MS^2=MA\times MB$, so $M$ is the midpoint of $QX$, combining with $MS=MX\implies QS\perp SX$. Hence $Q,P,S,X$ lie on a circle with diameter $QX\implies \angle PSQ=\angle PXQ=\angle TSX\implies \angle PST=\angle QSX= 90^{\circ}$.

Let $O$ be the center of $\Gamma$ and let $U$ be the antipode of $T$ WRT $\Gamma.$ Set $A', B'$ as the midpoints of $\overline{XY}, \overline{XZ}$ respectively. It's enough to show that $U, P, S$ are collinear. Since $A, C, P$ are collinear, their polars WRT $\Gamma$ are concurrent. Hence, the polar of $P$ passes through $XZ \cap YT.$ Similarly, the polar of $P$ passes through $XY \cap ZT.$ Thus, by Brokard's Theorem, $P \equiv XT \cap YZ.$ On the other hand, since $ABCD, AXOY, CZOT$ are cyclic, we have \begin{align*} \angle XOY = 180^{\circ} - \angle XAY = 180^{\circ} - \angle BAD = \angle BCD = \angle ZCT = 180^{\circ} - \angle ZOT = \angle ZOU. \end{align*}It follows that $XYZU$ is an isoceles trapezoid. In particular, $XU \parallel YZ$ and consequently $XT \perp YZ.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 47.56791601481347, xmax = 141.43518998779098, ymin = 67.51898117506758, ymax = 115.02398417689444; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); draw((83.781803123793,104.20364210100978)--(103.21873344876656,73.00856874003924)--(87.64516081544994,78.37493546929944)--cycle, green); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 5., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 5., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((83.781803123793,104.20364210100978)--(103.21873344876656,73.00856874003924), green); draw((103.21873344876656,73.00856874003924)--(87.64516081544994,78.37493546929944), green); draw((87.64516081544994,78.37493546929944)--(83.781803123793,104.20364210100978), green); draw(circle((101.62676058590549,93.66953523799953), 20.722208237201073)); draw(circle((88.37417738907266,86.37032180594338), 5.592453509655573), linetype("2 2") + blue); draw((63.56322515264037,69.9530246987366)--(83.781803123793,104.20364210100978), red); draw((63.56322515264037,69.9530246987366)--(103.21873344876656,73.00856874003924), red); draw((75.23289367676819,89.72164306554374)--(87.64516081544994,78.37493546929944), red); draw((77.7599506260836,71.04691358807975)--(81.08052005822805,96.3643291827659), red); draw((63.56322515264037,69.9530246987366)--(79.67650205911538,85.65950643704794), dotted); draw((75.23289367676819,89.72164306554374)--(77.7599506260836,71.04691358807975), dotted); draw((77.77339544777973,94.02529658127877)--(83.4755880367815,83.67229315241157)); draw((76.23634813721327,82.30620250855763)--(122.17300111358293,90.97474129323315), dotted); draw(circle((77.77339544777973,94.02529658127877), 11.819461930231995), linetype("2 2") + ffxfqq); /* dots and labels */ dot((83.781803123793,104.20364210100978),linewidth(3.pt) + dotstyle); label("$X$", (82.50286667606075,105.31076191363431), NE * labelscalefactor); dot((103.21873344876656,73.00856874003924),linewidth(3.pt) + dotstyle); label("$Y$", (103.23529066661403,71.51854328346882), SE * labelscalefactor); dot((87.64516081544994,78.37493546929944),linewidth(3.pt) + dotstyle); label("$Z$", (86.66567621747105,76.90570857233578), SSW * labelscalefactor); dot((93.50026828627978,88.6061054205245),linewidth(3.pt) + dotstyle); label("$A'$", (93.84856326931629,89.06764233053543), NE * labelscalefactor); dot((85.71348196962147,91.28928878515461),linewidth(3.pt) + dotstyle); label("$B'$", (84.05371728952733,91.27148267598102), N * labelscalefactor); dot((77.77339544777973,94.02529658127877),linewidth(3.pt) + dotstyle); label("$K$", (76.46271165519089,94.53643133590039), NNW * labelscalefactor); dot((83.4755880367815,83.67229315241157),linewidth(3.pt) + dotstyle); label("$S$", (81.0131243770713,82.84372447466444), SE * labelscalefactor); dot((122.17300111358293,90.97474129323315),linewidth(3.pt) + dotstyle); label("$U$", (122.98823005918842,90.61849294399714), NE * labelscalefactor); dot((81.08052005822805,96.3643291827659),linewidth(3.pt) + dotstyle); label("$T$", (79.80928403161879,96.16890566586007), S * labelscalefactor); dot((76.23634813721327,82.30620250855763),linewidth(3.pt) + dotstyle); label("$P$", (75.36485077371023,80.49715209824708), S * labelscalefactor); dot((63.56322515264037,69.9530246987366),linewidth(3.pt) + dotstyle); label("$A$", (62.58667985048987,68.66171320603937), SW * labelscalefactor); dot((79.67650205911538,85.65950643704794),linewidth(3.pt) + dotstyle); label("$C$", (78.4665520006434,86.78217826859188), NNE * labelscalefactor); dot((75.23289367676819,89.72164306554374),linewidth(3.pt) + dotstyle); label("$B$", (73.60588157775244,89.88387949551529), NW * labelscalefactor); dot((77.7599506260836,71.04691358807975),linewidth(3.pt) + dotstyle); label("$D$", (77.76869111916275,69.55957408751719), S * labelscalefactor); dot((71.76498777176647,83.84695106154776),linewidth(3.pt) + dotstyle); label("$J$", (70.09606176832807,83.35398217567653), SW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Meanwhile, inversion in $\Gamma$ sends $\odot(ABS) \mapsto \odot(A'B'S).$ Therefore, $\odot(A'B'S)$ is tangent to $\Gamma$ as well. Hence, $K \equiv XX \cap SS \cap A'B'$ is the radical center of $\Gamma, \odot(A'B'S), \odot(A'B'X).$ Clearly $KS = KX$ by equal tangents; also $KX = KP$ because $A'B'$ is the perpendicular bisector of $\overline{XP}.$ Therefore, $K$ is the circumcenter of $\triangle SXP.$ Thus, if $J$ is the reflection of $X$ in $K$, we have $(JP \parallel XU) \perp XP.$ By the converse of Reim's Theorem for $XXJ$ and $UPS$, it follows that $U, P, S$ are collinear, as desired. _________________________________________________________________________________________________ Remark: This problem is essentially 2011 ISL G4, with $\triangle XYZ$ as the reference triangle.

Projective KO: 1. Lemma: $P$ is $YZ$ intersect $TX$. Proof. 1a. With respect to $\Gamma$, the pole of $P$ is the line through the polars of $BD$ and $AC$. 1b. But the polar of $BD$ is the intersection of the poles of $B$ and $D$, or $XY$ and $TZ$. 1c. Similarly $AC$ polar is intersection of $XZ$ and $YT$. 1d. Thus pole of $P$ is line through aforementioned intersections. But by Brokard's theorem the polar of this line is also the intersection of $YZ$ and $TX$. 2. $YZ$ intersect $AB$ at $M$ such that $A, B; M, X$ harmonic. Proof. 2a. $BY$ and $BX$ are tangents, so $ZA, BY, ZB, ZX$ form harmonic bundle. 2b. Thus $ZA, ZM, ZB, ZX$ are harmonic, so $A, B; M, X$ harmonic. 3. $\angle ZPX = 90^\circ$ by easy angle chasing because $ABCD$ is cyclic, so $\angle A + \angle D = 180^\circ$. 4. Because the two circles are tangent at $S$, easy angle chasing show that $SX$ is external angle bisector of $ABS$. 5. Thus by famous lemma $\angle MSX = 90^\circ$. 6. $MPSX$ is cyclic. 7. $PST$ and $MSX$ are spirally similar, so $\angle PST = \angle MSX = 90^\circ$.

Solution with Danielle Wang: Ignore $ABCD$ cyclic for now, and focus entirely on $\Gamma$. Let $Q$ be the inverse of $P$ with respect to $\Gamma$. Since $P = AC \cap BD$, it follows $P$ lies on the polars of $\overline{TY} \cap \overline{XZ}$ and $\overline{YZ} \cap \overline{TX}$. By Brokard's Theorem, this implies $P = \overline{YZ} \cap \overline{XT}$. Therefore $Q$ is the Miquel point of cyclic quadrilateral $YTXZ$. Next let $\gamma$ be the circumcircle of $\triangle PQX$, and $M$ its center. Thus $\gamma$ is orthogonal to $\Gamma$. So if $W$ is the second intersection of $\Gamma$ and $\gamma$, then $\overline{OW}$ and $\overline{OX}$ are tangents to $\gamma$. Angle chasing, \[ \angle WPT = \frac12 \angle WMX = 90^{\circ} - \angle WOX = \angle WTX - 90^{\circ} \implies \angle PWT = 90^{\circ}. \]Thus we have shown that a circle centered at $M \in AB$ passes through $P$, $W$, $Q$, $X$ with $\angle PWT = 90^{\circ}$. Now suppose $ABCD$ is cyclic, centered at $N$ with circumcircle $\omega$. If $E = AB \cap CD$, $F = BC \cap DA$, then $Q \in EF$, $PQ \perp EF$, so $Q$ is the Miquel point of cyclic quadrilateral $ABCD$. Consequently, $N$, $P$, $Q$, $O$ collinear, and $P$ and $Q$ are inverses with respect to both $\omega$ and $\Gamma$. Now the circle with diameter $PQ$ is orthogonal to both $\omega$ and $\Gamma$, thus the midpoint $H$ of $PQ$ is on the radical axis of $\omega$ and $\Gamma$. Thus from $HM \perp PQ$, $M$ lies on this radical axis as well. Then $MA \cdot MB = MW^2$, so $W = S$ and we're done. [asy][asy]size(16cm); pair O = origin; pair X = dir(-90); pair T = dir(210); pair Y = dir(165); pair Z = T/Y*(-X); pair A = conj(2/(X+Z)); pair B = conj(2/(Y+X)); pair C = conj(2/(Y+T)); pair D = conj(2/(T+Z)); pair E = conj(2/(T+X)); pair F = conj(2/(Y+Z)); draw(unitcircle, blue); pair P = extension(A, C, B, D); pair Q = extension(P, O, E, F); draw(A--B--C--D--cycle, orange); draw(A--C, orange); draw(B--D, orange); draw(F--C--E, orange); draw(D--Z, orange); draw(B--X, orange); draw(circumcircle(A, B, C), orange+dashed); draw(Y--Z--T--X--cycle, lightblue+1.5); draw(Y--P--T, lightblue+1.5); pair N = circumcenter(A, B, C); draw(N--O, heavygreen); pair H = midpoint(P--Q); draw(CP(H, P), deepgreen+dashed); pair M = circumcenter(P, Q, X); draw(H--M, heavygreen); draw(CP(M, X), lightolive); pair L = -T; pair W = foot(T, P, L); draw(P--W--T, mediumgreen); dot("$O$", O, dir(45)); dot("$X$", X, dir(-45)); dot("$T$", T, dir(T)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$A$", A, dir(225)); dot("$B$", B, dir(-45)); dot("$C$", C, dir(170)); dot("$D$", D, dir(90)); dot("$E$", E, dir(-90)); dot("$F$", F, dir(F)); dot("$P$", P, dir(135)); dot("$Q$", Q, dir(Q)); dot("$N$", N, dir(N)); dot("$H$", H, dir(225)); dot("$M$", M, dir(-90)); dot("$W$", W, dir(45)); /* Source generated by TSQ */ [/asy][/asy]

$(1)$: $XT\cap YZ=P$. proof: let $TZ\cap XY=R,XZ\cap TY=W$. apply pascal respectively to the hexagons: $(TTXYYZ);(XXTZZY);(YYZXXT);(TTXZZY)$ to see that $XT\cap YZ\in AC\cap BD=P$. next let $G$ the intersection of the tangent at $P$ to $(ABP)$ and $AB$. $(2)$: $S\in \Gamma(G,r=GP)$. proof: on the one hand: $XPG=XPB-GPB=180-AXT-XBD-BAC=ABD-BAC+180-AXT$ on the other hand $2AXT=180-BAD-ADC$, equating the latters yellds $XPG=GXP$, hence $GX^2=GP^2=GB.GA\rightarrow$ $G$ is on the radical axis of $(XYZ);(\Omega)$ i.e lays on the tangent of $(XYZ)$ at $S$. $(3)$: $TSP=90$ proof: $TSP=PSG+GST=90-1/2PGS+TXS=90-PXS+PXS=90$ which ends the proof.

First by Pascal on $XXTYYZ$ and $TTXZZY$, we get that $A$, $C$, and $XT\cap ZY$ are collinear. Applying Pascal again on $XXTZZY$ and $YYXTTZ$ gives that $B$, $D$, and $XT\cap ZY$ are collinear, so we conclude that $P\equiv XT\cap ZY$. Then $$\angle ZYX=\frac{\angle ZOX}{2}=\frac{180^\circ - \angle ZBX}{2}=\frac{\angle ABC}{2}=\frac{180^\circ-\angle ADC}{2}=\frac{\angle TDY}{2}=\frac{180^{\circ}-\angle TOY}{2}=90^\circ-\angle TXY$$so that $\angle YPZ=90^\circ$. If $F\equiv YZ\cap AB$, then $$(A,B;F,X)\stackrel{Y}{=}(D,B;P, XY\cap BD)\stackrel{X}{=}(DX\cap \odot(XYZ),X;T,Y)=-1$$which, combined with $\angle FPX=90^{\circ}$, gives that $PF$ and $PX$ are the internal and external angle bisectors of $\angle APB$. Then $\odot(FPX)$ is the $P$-Apollonius circle in $\triangle APB$, so its center $M$ lies on the common internal tangent of $\odot(ASB)$ and $\odot(XYZ)$ by the radical axis theorem, hence $S$ lies on $\odot(FPX)$ as well. Finally, $$\angle SPT=\angle SPX=\angle SFX$$and $$\angle STP=180^{\circ}-\angle STX=\angle SYX=\angle SXF$$whence $\angle PST=\angle FSX=90^\circ$, as desired.

Ignore cyclicity of $ABCD$, and move $T$. then $BD\cap YZ\rightarrow D\rightarrow T\rightarrow C\rightarrow AC\cap YZ$ are homography, which is identity on $T=X,Y,Z$, hence is identity for all $T\in \Gamma$. So $P\in YZ$, similarly $P\in XT$. \Define $M$:= $YZ\cap AB$, $U$:= $AD\cap BC$, $J$:= $SS\cap AB$ \end Define; Since $ABU$ and $YZX$ are perspective, by FT of projective geometry $(A,B;M,X)=-1$. So $JX^2=JS^2=JA\cdot JB$ implies that $J$ is midpoint of $MX$ and $\angle MSX=\pi/2$. Also $YZ\perp TX$, hence $\angle MPX=\pi/2$, which means $\angle PSM=\angle PXM=\angle TSX$. Therefore $\angle PST=\pi/2$, which means $PS\perp ST$, as desired. \return;

Consider inversion wrt $\Gamma$. Vertices of $ABCD$ go to midpoints of $XZ,XY,TY,ZT$ and still form a cyclic quadrilateral; since it's also a parallelogram, it must be a rectangle. Then $YZ \perp XT$. Let $M,N$ be the midpoints of $XY,XZ$. Then $(MSN)$ and $\Gamma$ are tangent. $AC$ is a pole of $XZ \cap YT$ and $DB$ is a pole of $XY \cap ZT$. Then by Brocard's thm their intersection, $P$, also is an intersection of $XT$ and $YZ$. Introduce points $U$ and $V$ where the tangent from $X$ to $\Gamma$ meets $MN$ and $YZ$. Then $US$ is also tangent to $\Gamma$ and $U$ is the midpoint of $XV$. Now $UV=UX=US$, so $\angle XSV=90^{\circ}=\angle XPV$, so $XSPV$ is cyclic and $\angle PSV=\angle PXV=\angle XST$. Finally, $\angle PST=\angle PSV+\angle VST=\angle XST+\angle VST=\angle VSX=90^{\circ}$.

Let $Q$ be the inverse of $P$ wrt $\Gamma$, let $O$ be the center of $\Gamma$, and let $R_1=YT\cap XZ$, $R_2=YX\cap TZ$. Claim. $Q$ is the Miquel point of quadrilateral $YTZX$. Proof. Because $YT$ is the polar of $C$ and $XZ$ is the polar of $A$, we know that $AC$ is the polar of $YT\cap XZ=R_1$. Similarly $BD$ is the polar of $R_2$. Thus $P=AC\cap BD$ is the pole of $R_1R_2$. This means that $Q$ lies on $R_1R_2$ and $OQ\perp R_1R_2$, which implies that $Q$ is the Miquel point of cyclic quadrilateral $YTZX$. Consider the circumcircle of $PQX$, and redefine $S$ as the other intersection between $\omega=(PQX)$ and $\Gamma$. Claim. $\angle PST=90^{\circ}.$ Proof. \begin{align*} \angle PST &= \angle PSX+\angle XST\\ &= 90^{\circ} -\angle XQR_2 +\angle XZT\\ &= 90^{\circ} -\angle XZR_2 +\angle XZT\\ &= 90^{\circ}. \end{align*}Next, observe that $\omega$ is orthogonal to $\Gamma$. Let $O'$ be the center of $\omega$. Then $O'$ lies on $AX$ and $O'S$ is tangent to $\Gamma$ because $\omega$ and $\Gamma$ are orthogonal. Hence, to prove that $(ABS)$ is tangent to $\Gamma$, we only need $O'P^2=O'S^2=OA\cdot OB$, which is equivalent to $\angle BPO'=\angle CAB$. Claim. $PO'\parallel CD$, which implies $\angle BPO'=\angle CAB$. Proof. \begin{align*} \angle AO'P &= 180^{\circ} - \angle XO'P\\ &= 180^{\circ} - 2\angle XSP\\ &= 2\angle XST\\ &= \angle XOT\\ &= \measuredangle(AX, DT). \end{align*}

Can't believe I get to say this, but not that hard for China TST #6! Let $DA \cap CB \equiv M$, and $BA \cap CD \equiv L$. Let $O$ be the center of $\Omega$. We claim that $P \equiv TX \cap YZ$, and $TX \perp YZ$. Assume WLOG that $M$ is closer to $AB$, and $L$ is closer to $AD$. Then, $\angle TYZ = \angle TOZ/2 = \angle CDA/2$. Similarly, $\angle XTY = \angle ABC/2$. Thus $TX \perp YZ$. Now, wrt $\Omega$, $B$ is pole of $YX$, $D$ is pole of $ZT$, so $BD$ is polar of $YX \cap ZT$. Similarly, $AC$ is polar of $YT \cap ZX$, so, by Brokard's theorem, $P \equiv BD \cap AC$ satisfies $P \equiv YZ \cap XT$. Invert about $\Omega$. $A^*$ and $B^*$ are midpoints of $ZX, YX$ respectively, and $S^* \equiv S$. As $X \neq S$, and $S^*A^*B^*$ is tangent to $\Omega$, $S$ is the $2011 G4$ point, wrt $XZY$, with $X$ being the vertex at top. Note $P$ is foot of altitude from $X$ to $YZ$. Let $PS$ meet $\Omega$ again in $X'$. Then, $XX' \parallel ZY$. Then $\angle TXX' = \angle TPY = 90$, so $\angle TSX' = 90$, so $ST \perp SP$, and we're done.

Claim: $P$ lies on lines $XT$ and $YZ$. Proof: Take a homography fixing $\Gamma$ and sending $XZYT$ to a rectangle; the claim becomes trivial (we don't need $ABCD$ cyclic). Let the image of $ABCD$ after inversion about $\Gamma$ be $A'B'C'D'$. Note this is the Varignon parallelogram of $XYTZ$, and since it is cyclic, it must be a rectangle. This gives $$YZ \parallel C'D' \perp C'B' \parallel XT$$so in fact $P$ is the foot from $X$ to $Z$. Since $(A'B'S)$ is tangent to $\Gamma$ by 2011 G4 we have that if $PS$ intersects $\Gamma$ again at $X'$, then $XX' \parallel YZ$. Thus $$90^\circ = \angle X'XT = \angle X'ST = \angle PST$$as desired.

This is a very long and messy solution but this was what I really thought. Let $\omega$ be the circumcircle of $\square ABCD$. We fix $\omega$ and $\Gamma$. By Poncelet porism, we can vary $D$ along $\omega$ and the conditions still hold. Claim $P$ is a fixed point. Let $\ell$ be one of the external tangents of $\Gamma$ and $\omega$ and touch $\omega$ at $X$. Another tangent line from $X$ touches $\Gamma$ and intersect $\omega$ at $Z'$ and $Y$ respectively. By Poncelet porism, Another tangent line from $Y$ to $\Gamma$ is also tangent to $\omega$ and intersect $\ell$ at $Z$. By DDIT, the pairs $(\overline{DA},\overline{DC}),(\overline{DX},\overline{DY}),(\overline{DZ},\overline{DZ'})$. Hence, $AC,XY,Z_0Z_0'$ are concurrent ($Z_0=AZ \cap \omega,Z_0'=AZ' \cap \omega$). Let $P' = XY \cap Z_0Z_0'$. $(X,Y;P',Z') \overset{Z_0'}{=} (X,Y;Z_0,D)=-1$. Therefore, $P'$ is fixed and $AC$ always passes through $P'$. Similarly, $BD$ also passes though $P'$. So, $P'=P$. By symmetry, $P$ lies on the line $\ell_0$ passing through the centers of $\Gamma,\omega$. But $(X,Y;P,Z')=-1$. Polar of $P$ , $\ell_P$, wrt. $\Gamma$ passes though $Z'$ and $Z$ and perpendicular to $\ell_0$. Let $N$ be the midpoint of $Z'P$. $NZ^2=NY \cdot NX$. So radical axis of $\Gamma$ and $\omega$ is the midline of $\ell_P$ and $P$ [asy][asy] import geometry; size(300); pair X,Y,Z,M,Ix,Z1,A,Z0,Z01; X=dir(-40); Y=dir(220);Z=dir(90);M=dir(270); triangle t=triangle(X,Y,Z); Ix=excenter(t.BC); Z1=foot(Ix,X,Y); path omega = circle(M,length(M-X)); pair CCC = 2/3*X+1/3*Y; pair CCCC = 2*CCC-Z; A = IP(CCC--CCCC,omega); Z0=IP(Z--CCC,omega); pair[] Z01 = intersectionpoints(A--Z1,omega); clipdraw(omega,deepmagenta); draw(line(A,Z1),fuchsia); drawline(t,blue); draw(line(A,Z),fuchsia); clipdraw(excircle(t.BC),deepmagenta); dot("$D$",A,dir(-60)); dot("$X$",X,dir(60)); dot("$Y$",Y,dir(120)); dot("$Z$",Z,dir(0)); dot("$Z'$",Z1,dir(270)); dot("$Z_0'$",Z01[0],dir(240)); dot("$Z_0$",Z0,dir(60)); pair P = IP(Z0--Z01[0],X--Y); dot("$P$",P,dir(280)); draw(Z0--Z01[0],purple+dashed); draw(box((-2.3,-2), (2,1.5)), invisible); draw(line(Z1,Z),deepcyan); [/asy][/asy] Assume that $S'$ lies on $\Gamma$ and $\angle PS'T = 90^{\circ}$. Then, we'll show that $\Omega$ is tangent to $\Gamma$ at $S'$. Since, $AB \cap CD = R$ lies on $\ell_P$. So, the polar of $R$, $TX$, passes through $P$. Let $T'$ be $T$ antipode wrt. $\Gamma$. $X,S'$ are the foot from $T',T$ to $TP,T'P$ respectively. By Brokard theorem, $H$, the orthocenter of $\triangle PT'Y$ lies on $\ell_P$. Since $AB$ intersects the tangent line to $\Gamma$ at $S'$ is the intersection of tangent line at $S',X$ to $\Gamma$ is $M$, the midpoint of $PH$ (angle chasing after we know that $H$ is orthocenter) lies on radical axis of $\omega,\Gamma$. So, $MS'^2=MA \cdot MB$ and $\Omega$ is tangent to $\Gamma$ at $S'$. $S'=S$.

This is actualy a nice thing but u have to focus on $\Gamma$ to win!. Claim 1: $AC,BD,TX,YZ$ are concurrent at $P$ Proof: Take all the polars here w.r.t. $\Gamma$ (projective is back baby!) and now since $A,P,C$ are colinear we have $\mathcal P_A,\mathcal P_P, \mathcal P_C$ concurrent which means that $YT \cap ZX \in \mathcal P_P$ and since $B,P,D$ are colinear we have $\mathcal P_B, \mathcal P_P, \mathcal P_D$ concurrent which means that $ZT \cap XY \in \mathcal P_P$ but by Brokard the only point satisfying this conditions is $ZY \cap TX$ hence done!. Finishing: The next move was made because of the cyclic quadrilaterals we get. We make an inversion with center $P$ and radius $\sqrt{PT \cdot PX}$ which means that $\Gamma$ is fixed after this transformation now let $A',B'$ be the inverses of $,B$ respectivily and since the line $ABX$ is tangent to $\Gamma$ we have $(PA'B'T)$ cyclic and tangent to $\Gamma$ and also $ABA'B'$ cyclic, now by angle chase: $$\angle BDC=\angle BAC=\angle A'B'B \implies DC \parallel A'B' \implies \angle TA'B'=\angle DTA'=\angle TB'A' \implies TA'=TB'$$Now let $U$ be the antipode of $T$ w.r.t. $\Gamma$, since $TU \perp CD$ we have $TU \perp A'B'$ hence $TU$ is perpendicular bisector of $A'B'$ meaning that $UA'=UB'$ and also that the tangent from $U$ to $\Gamma$ is parallel to $A'B'$ meaning that $(UA'B')$ is tangent to $\Gamma$ BUT remember that $(ABS)$ was tangent to $\Gamma$ so by the inversion we have letting $S'$ the inverse of $S$ that $(A'B'S')$ tangent to $\Gamma$ so $S'=U$ and this means that $\angle PST=\angle UXT=90$ thus we are done

Sketch, with v4913 and crazyeyemoody907. By Brianchon's on $AXBCTD$ and $AZDCYB$, $\overline{XY}$ and $\overline{YZ}$ intersect at $P$. By angle chasing, $\overline{XT}$ and $\overline{YZ}$ are perpendicular. Let $M$ and $N$ be the midpoints of $\overline{XZ}$ and $\overline{XY}$, respectively, let $T'$ be the antipode of $T$ with respect to $\Gamma$, and redefine $S$ as the second intersection of $\overline{PT'}$ with $\Gamma$. By inversion about $\Gamma$, it suffices to show that $\Gamma$ is tangent to the circumcircle of $SMN$ at $S$. By angle chasing, $XYZT'$ is a cyclic isosceles trapezoid, so we are done by ISL 2011 G4.

I'm disappointed that no complex bash has been posted yet.