Difficult question.Just observation:The orthocenter of triangles ABC and ARQ are collinear with S.Hope it is true

@above Your claim is true. Another Desargues Involution Theorem's application. Let $OP=\ell$. Let $A', B', C', P', S'$ be reflections of $A, B, C, P, S$ across point $O$. So $\angle ASP=90^{\circ}\implies A', S,P$ are colinear. By symmetry, $A, P', S'$ are also colinear. Let $\{T, T'\}=\ell\cap\Omega$, $Q'=BS'\cap \ell$ and $R'=CS'\cap \ell$. Claim : $Q', R'$ are reflections of $Q, R$ across $O$. Proof : Apply Desargues' Involution Theorem on line $\ell$, intersecting quadrilateral $ABCS'$ which has circumconic $\Omega$. We get an involution on $\ell$ swapping $(P, P'), (T,T'), (Q,Q')$ and $(R, R')$. Since $(P, P'), (T, T')$ are already symmetric across $O$, we also get that $(Q, Q'), (R, R')$ are symmetric across $O$, implying the claim. Now by the claim, $B, Q', S'$ are colinear $\implies B', Q, S$ are colinear $\implies \angle BSQ=90^{\circ}$. So $S\in\odot(BQ)$ and similarly lies on $\odot(CR)$. Hence $S$ lies on Steiner line $\sigma$ of complete quadrilateral $BRQCAP$ (which is radical axis of $\odot(BQ), \odot(CR)$). Let $H_1$ be the orthocenter of $\Delta AQR$. Since $H, H_1\in\sigma$, we get that $H, S, H_1$ are colinear. Since $X \in \odot(AP)\cap\sigma$, $X$ must be second concurrency point of $\odot(AP), \odot(BQ), \odot(CR)$. Since $\angle BYQ = \angle CZR=90^{\circ}$, we get $Y\in\odot(BQ), Z\in\odot(CR)$. This implies that quadrilaterals $BYXS, CZXS$ are concyclic. Hence \begin{align*}\angle YXZ &= 360^{\circ} - \angle AYX - \angle YAZ - \angle AZY \\ &= 360^{\circ} - \angle BSX - \angle BAC - \angle CSX \\ &= 360^{\circ} - \angle BSC - \angle BAC \\ &= 180^{\circ}\end{align*}implying the desired colinearity. EDIT : I have attached link to my article on Desargues' Involution Theorem.

@above, Instead of Desargues' Involution Theorem we can simply use pascal on $B'BCAA'S$ and $C'CBAA'S$

Is there any solution without introducing antipodes?This problem appeared in Second Round.I believe there exists medium solution.

Official Solution: Let $AD,BE,CF$ be altitudes of $\triangle ABC$ and $\Gamma_A,\Gamma_B,\Gamma_C$ be circles with diameter $\overline{AP},\overline{BQ},\overline{CR}$, respectively. Then the powers of $H$ w.r.t. $\Gamma_A,\Gamma_B,\Gamma_C$ are $HA\cdot HD,HB\cdot HE,HC\cdot HF$, and they're the same. Let $A',B'$ be the reflection points of $A,B$ w.r.t. $O$. Let $A'P$ intersects $B'Q$ at $S'$, then because $O,P,Q$ are collinear, we can apply reverse of Pascal's theorem and get that $A,A',S',B',B,C$ lie on a conic. So $S'$ lies on $\Omega$, which means that $S=S'$. Since $\angle BSQ=\angle BSB'=90^\circ$, $S$ lies on $\Gamma_B$, and similarly $S$ lies on $\Gamma_C$. So $\Gamma_A,\Gamma_B,\Gamma_C$ are coaxial with axis $SH$, hence the other intersection of $\Gamma_A,\Gamma_B,\Gamma_C$ is $X$. Because of the fact that $Y,Z$ lie on $\Gamma_B,\Gamma_C$ respectively, it's true that $$\measuredangle YXZ=\measuredangle YXS+\measuredangle SXZ=\measuredangle ABS+\measuredangle SCA=180^\circ$$which means $X,Y,Z$ are collinear.$\quad\blacksquare$

Li4 wrote: Let $ABC$ be a triangle with circumcircle $\Omega$, circumcenter $O$ and orthocenter $H$. Let $S$ lie on $\Omega$ and $P$ lie on $BC$ such that $\angle ASP=90^\circ$, line $SH$ intersects the circumcircle of $\triangle APS$ at $X\neq S$. Suppose $OP$ intersects $CA,AB$ at $Q,R$, respectively, $QY,RZ$ are the altitude of $\triangle AQR$. Prove that $X,Y,Z$ are collinear. Proposed by Shuang-Yen Lee Let $\overline{AD}$ be an altitude in $\triangle ABC$. Then pentagon $APSDX$ is cyclic hence $\overline{HA} \cdot \overline{HD}=\overline{HX} \cdot \overline{HS}$. Now fix $\triangle ABC$ and move point $S$ on $\Omega$. Then $S \mapsto P$ is a projectivity $\pi: \Omega \mapsto \overline{BC}$ and $\overline{OP}$ is a pencil with same cross-ratio as $S$. Now consider the map $\Gamma$ that dilates at $A$ with factor $\cos A$ and reflects in the internal bisector of angle $A$. Then $\Gamma: \overline{QR} \mapsto \overline{YZ}$ hence $\overline{YZ}$ is a pencil of same cross-ratio emanating from the midpoint $N$ of $\overline{AH}$. Define $X_1$ as the intersection of $\overline{YZ}$ with the nine-point circle $\gamma$ of $\triangle ABC$. Then $S \mapsto X_1$ is a projectivity $\Omega \mapsto \gamma$. However $S \mapsto X$ is also a projectivity $\Omega \mapsto \gamma$. Thus, in order to show $X=X_1$ it is enough to prove the claim for any three choices of point $S$. Note that the result is trivial when $S$ coincides with the vertices of $\triangle ABC$, thus completing the proof! $\blacksquare$

here's a solution with moving points fix $\triangle ABC$ and move $P$ on the line $BC$ claim(1): $P \rightarrow S$ is projective proof: let $A'$ be the antipode of $A$ wrt $(ABC)$ so $A',P,S$ are collinear $P \rightarrow A'P \rightarrow S$ is projective $\blacksquare$ claim(2): $S \rightarrow X$ is projective proof: let $D=AH \cap (ABC)$ and $X'=SX \cap (ABC)$ since $AH.AD=HS.HX \implies AH.HD=SH.HH' \implies X$ is the midpoint of $HH'$ so $S \rightarrow SH \rightarrow H' \rightarrow X $ $\blacksquare$ now it's easy to see that $P \rightarrow PO \rightarrow R \rightarrow RZ \rightarrow Z$ is projective so $deg(Z)=deg(Y)=deg(X)=deg(P)=1$ so it suffice to consider four position for $P$ on $BC$ 1) $P=AO \cap BC$ so $Z=Y$ 2)$P=B$ (trivial) 3)$P=C$ (trivial) 4) $P=\infty_{BC}$ : so $S=D$ and so $X$ is the midpoint of $AH$ and it's easy to see that the result is true by doing a homothety around $A$ with facrot $2$ and we win

Ali3085 wrote: Let $D=AH \cap (ABC)$ and $X'=SX \cap (ABC)$ I think you meant $H'$ instead of $X'$. An issue about your proof is that the following statement : Quote: $deg(X)=1$ so it suffice to consider four position for $P$ on $BC$ Is false. What you proved is $S\overset {H}{\mapsto}H'\mapsto X$ is a homography, so actually $X$ moves on a circle which transforms to $\Omega$ by homothey at $H$ with ratio $2$ (ie, the nine point circle of $(ABC)$) hence actually $\deg X =2$. So the assertion of $X,Y,Z$ collinear is of degree atmost $(2+1+1)$; so we need to check $(2+1+1)+1$ cases. Another easy case is take $S=(AH)\cap (ABC)$, which means $P$ is the midpoint of $BC$. To deal with this case, we need to prove that if $E$ is foot from $A$ to $QR$, then $(QR;PE)=-1$. This is easy, note: $-1= (B,C; AE\cap (ABC) , A') \overset {A}{=} (QR;EP)$

solution with Psyduck909, islander7 To begin with, draw in the circles $\omega_1, \omega_2, \omega_3$ with diameters $AP$, $BQ$, and $CR$, respectively. The main claim is as follows: Claim: Points $S, X$ lie on $\omega_1, \omega_2, \omega_3$. We know that $S, X \in \omega_1$ just by definition. To prove the other two, add in $A', B', C'$ which are the antipodes of $A, B, C$. Observe that $A' \in \overline{SP}$ by the right angle and $A' \in \overline{AO}$ by definition. Consider the following application of Pascal on $C'SA'ABC$: we get that $\overline{C'S} \cap \overline{AB}$, $\overline{SA'} \cap \overline{BC} = P$, and $\overline{AA'} \cap \overline{CC'} = O$ are collinear. However, since $R$ lies on $\overline{PO}$ as well as $\overline{AB}$, we know that $R$ lies on $\overline{C'S}$. As a result, we have $\measuredangle RSC = \measuredangle C'SC = 90^\circ$, proving the fact that $S$ lies on $\omega_3$. By symmetry $S$ also lies on $\omega_2$. In order to get $X$ on all three circles, we notice that $H$ has equal power to all three circles too since $HA \cdot HD = HB \cdot HE = HC \cdot HF$, where $D, E, F$ are the feet of the altitudes. Consequently $\omega_1, \omega_2, \omega_3$ are coaxial, so $X$ precisely describes the second concurrency point of the three circle, as desired. $\square$ The finish from here is simply angle chasing: observe that \begin{align*} \measuredangle ZXS = \measuredangle ZCS = \measuredangle ACS = \measuredangle ABS = \measuredangle YBS = \measuredangle YXS, \end{align*}as desired.

We proceed with pure complex numbers, because why not. Define $(ABC)$ as the unit circle, $h = a + b + c$, and $S = s$. We have: \[p + bc\overline{p} = b + c\]\[\frac{p-s}{a-s} = \frac{\overline{p} - \frac{1}{s}}{\frac{1}{a} - \frac{1}{s}}\]\[\implies p - s = as\overline{p} - a \implies bc(p + a - s) = as(b + c - p)\]\[p(bc + as) = asb + asc +bcs - abc \implies p = \frac{asb + asc + bsc - abc}{bc + as}\]Now, observe \[\frac{r}{p} = \frac{\overline{r}}{\overline{p}}, r + ab\overline{r} = a + b\]\[\implies r + ab \frac{\overline{p}}{p} r = a + b \implies r(1 + ab\frac{\overline{p}}{p}) = a + b\]\[r(1 + ab\cdot \frac{a + b + c- s}{asb + asc + bsc - abc}) = a + b\]\[r\left(\frac{(a + b)(ab + sc)}{asb + asc + bcs - abc}\right) + a + b\implies r = \frac{asb + asc + bsc - abc}{ab + sc}\]Similarly, $q = \frac{asb + asc + bcs - abc}{ac + sb}$. Now, to solve for $y$, we have \[y = \frac12 (a + b + q - ab\overline{q}) = \frac12 \left(\frac{a^2c + 2asb + b^2s + bsc + asc - ab(a+b+c-s)}{ac + sb}\right)\]\[= \frac12 \left(\frac{3asb + a^2c + b^2s + bcs + acs - a^2 - ab^2 - abc}{ac + sb}\right)\]Similarly, $z = \frac12(\frac{3asc + a^2b + c^2s + bcs + abs - a^2c _ ac^2 - abc}{ab + sc})$. Now, we calculate $x$. Since $X\in (ASP)$, we have $\angle AXP = 90$.This means we have the following equation: \[\frac{x-a}{x-p} = - \frac{\overline{x} - \frac{1}{a}}{\overline{x} - \overline{p}}\]\[\implies \overline{x}x - \overline{p}x - a\overline{x} + a\overline{p} = -x\overline{x} + \frac{x}{a} + p\overline{x} - \frac{p}{a}\]\[\implies 2x\overline{x} - x(\overline{p} + \frac{1}{a}) - \overline{x}(a + p) + (a\overline{p} + \frac{p}{a}) = 0\]Since $X$ lies on $SH$, we also ahve \[\frac{x-h}{h-s} = \frac{\overline{x} - \overline{h}}{\overline{h} - \frac{1}{s}}\implies x\overline{h} - \frac{x}{s} + \frac{h}{s} = \overline{x}h - \overline{x}s + \overline{h}s \]\[\implies x(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{1}{s}) - s(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) + \frac{a + b + c}{s} = \overline{x}(a + b + c-s)\]Now, we can substitute this formula for $\overline{x}$ into our other equation to get a quadratic in $x$: \[2x\left(\frac{x(abs + acs + bcs - abc) + abc(a + b + c) - s^2(ab + ac + bc)}{abcs (a+b+c-s)}\right)\]\[ - x\left(\frac{a + b + c - s}{bc + as} + \frac1a\right) - \left(\frac{x(abs + acs + bcs - abc) + abc(a+b+c) - s^2(ab + ac + bc)}{abcs(a+b+c_s)}\right)\left(\frac{a^2s + abs + acs + bcs}{bc + as}\right)\]\[ + \frac{a(a+b+c-s) + \frac1a(-abc + abs + bcs + acs)}{as + bc} = 0\]Observe that $x = s$ is a solution to this (since $S\in HS$ and $\angle ASP = 90$). Therefore, dividing the constant term by the coefficient of $x^2$ and $s$ gives us the other solution. Thus, \[x = \frac{\frac{abcs (a + b + c - s)(a(a+b+c-s) + \frac1a(-abc + abs + bcs + acs) - (abc(a+b+c) - s^2(ab + ac + bc))(s)(a+b)(a+c)}{abcs(a+b+c-s)(bc+as)}}{\frac{2(abs + acs + bcs - abc)s}{abcs(a+b+c-s)}}\]\[= \frac{(a + b + c-s)bc((a + b + c-s)a^2 + (abs + bcs + acs - abc)) - (abc(a+b+c) - s^2(ab + ac + bc))(a+b)(a+c)}{2(bc + as)(abs + acs + bcs - abc)}\]We first calculate the numerator: \[s^2(bca^2 - bc(ab + bc + ac) + (ab + ac + bc)(a+b)(a+c)) + s(bca^2(-2)(a+b+c) + (a+b+c)(bc)(ab + ac + bc) + bc(abc))\]\[ + (a+b+c)^2bca^2 + (a+b+c)(bc)(-abc) - abc(a+b+c)(a+b)(a+c) \]\[= s^2a(abc + (ab + bc + ac)(a+b+c)) + s(bc)((a + b + c)(ab + ac + bc - 2a^2) + abc) + (a + b + c)(abc)(-2bc)\]\[= (as + bc)(s(abc + (ab + ac + bc)(a+b+c)) - 2a(a + b + c)bc)\]Therefore, \[x = \frac{s(abc + (ab + ac + bc)(a + b + c)) - 2abc(a+b+c)}{2(abs + acs + bcs - abc}\]Finally, we calculate: \[(x-y)(abs + acs + bcs - abc)(ac + sb)\]\[= (abs + acs + bcs - abc)(3asb + a^2c + b^2s + bcs + acs - a^2b - ab^2 - abc) - (ac + sb)(s(abc + (ab + ac + bc)(a+b+c)) - 2a(a + b + c)bc)\]\[= s^2((ab + ac + bc)(3ab + b^2 + bc + ac) - b(abc + (ab + ac + bc)(a+b+c))\]\[+ s((ab + ac + bc)(-1)(a^2b + ab^2 + abc) - (abc)(3ab + b^2 + bc + ac) + a^2c(ab + ac + bc) + 2ab(a+b+c)(bc) - ac(abc + (ab + ac + bc)(a + b + c)))\]\[+ (abc)(a^2b + ab^2 + abc) + 2a^2c (a+b+c)(bc) - abc(a^2c)\]\[= s^2((ab + ac + bc)(ac + 2ab) - ab^2c) + sa((ab + ac + bc)(a+b+c)(b + c) + (-ab^2c - abc^2 + b^3c + b^2c^2 + a^2c^2 + a^2bc))\]\[ + (abc)(ab(a+b+c) + 2ac(a+b+c) - a^2c)\]\[= s^2(a)((ab + ac + bc)(c +2b) - b^2c) + sa((ab + ac + bc)(a + b + c)(c + b) - ab^2c - abc^2 + b^3c + b^2c^2 + a^2c^2 + a^2bc) + ((abc)(a)(a+b+c)(2b + c) - a^3bc^2)\]Now, we can calculate \[\overline{(x-y)(abs + acs + bcs - abc)(ac + sb)\cdot (ab + sc)\cdot ((abc)^3\cdot a \cdot s^2)}\]\[= ((a + b + c)(1)(1)(2c + b)(abc(a) - a^3bc^2) + s^2(a(ab + ac + bc)(c + 2b) - b^2c)\]\[ + s((1)(a + b + c)(ab + ac + bc)(c + b)a - abc^2 - ab^2c + a^2c^2 + a^2bc + b^3c + b^2c^2)\]\[= x - y\]Therefore, \[\frac{x-y}{x-z} = \overline{\frac{x-y}{x-z}}\]so we conclude that $X,Y,Z$ are collinear. The end.

Amazing problem !! Claim 1:-$S-H-H'-X$ are collinear.

Now take the reflection of $A,B,C$ around $O$ and name it $A',B',C'$ Claim 2:- $S-P-A',C'-S-R,S-Q-B'$ are collinear.

Claim 3:- $R-P-O$ collinear.

Claim 4:- $S,X \in (ASP),(BSQ),(RXC)$

The rest is just angle chasing, $$\angle ZXS=\angle ZCS=\angle ACS=\angle ABS=\angle YBS=\angle YXS$$which concludes our problem.

hard

Here's a non-projective angle-chasing solution: Let $A'$ be the antipode of $A$ in $(ABC)$. Let $D,E,F$ be the feet of altitudes of $ABC$. Let the midpoints of $BC,AH,SA'$ be $M,N,L$. Finally, let $H'$ be the reflection of $H$ in $BC$. Note that as $\angle ASP = 90$, we have that $S,P,A'$ are collinear. Let $\mathcal{T}$ denote the transformation composed of taking a dilation with ratio $\cos(\angle A)$ at center $A$ and then reflecting in the angle bisector of $\angle BAC$. Observe that $\mathcal{T}$ takes $ABC$ to $AEF$ and hence takes $O$, the circumcenter of $ABC$ to $N$ the circumcenter of $AEF$. Also, $\mathcal{T}$ takes $Q,R$ to $Y,Z$, so it takes $\overleftrightarrow{QR}$ to $\overleftrightarrow{YZ}$ Hence, $QR$ passes through $N$ and $\measuredangle (AN, QR) = \measuredangle POA$ So, it suffices to show that $\measuredangle ANX = \measuredangle POA$ or $\measuredangle ANX = \measuredangle POA'$ Now, observe that as $(AP)$ i.e. circle with diameter $AP$. passes thorough $D$. $HS\cdot HX = HA\cdot HD = HN\cdot HN' \implies N,X,H',S$ are concyclic $\implies \measuredangle H'SH = \measuredangle HNX = \measuredangle ANX$. So, we need to show that $\measuredangle H'SH = \measuredangle POA'$. Observe that $O,P,L,M$ are concyclic on $(OP)$ Now we finish with an angle chase: \begin{align*} \measuredangle H'SH &= \measuredangle H'SA' + \measuredangle A'SH\\ &= \measuredangle H'AA' + \measuredangle A'LM\\ &= \measuredangle MOA' + \measuredangle POM\\ &= \measuredangle POA' \end{align*}

Failed to find a single projective fact. Let $S'$ be the miquel point of $ABCPQR$. Claim: $S'$ is reflection of $S$ across $OP$. Proof. Redefine $S$ as the reflection of $S'$ over $OP$, which must lie on $\Omega$. Let $A'$ be the reflection of $A$ over $OP$. Then \[ \measuredangle ASP = \measuredangle A'S'P = \measuredangle A'S'R + \measuredangle RS'P = \measuredangle A'AR + \measuredangle RBP = 90^\circ \]as desired. $\blacksquare$ Let $K$ be the orthocenter of $\triangle ARQ$. Let $AD, BE, CF$ be altitudes of $\triangle ABC$ and let $AW, BY, CZ$ be altitudes of $\triangle AQR$. Claim: $S, H, K$ are collinear. Proof. This is just the Steiner line of $S'$. $\blacksquare$ Claim: $SXCFRZ$ and $SXBEQY$ are cyclic. Proof. Note that $SH \cdot HX = AH \cdot HD = FH \cdot HC$ and that $SK \cdot KX = AK \cdot KW = RK \cdot KZ$, giving the result by symmetry. $\blacksquare$ As such, \[ \measuredangle YXS = \measuredangle YBS = \measuredangle ABS = \measuredangle ACS = \measuredangle ZCS = \measuredangle ZXS. \]as desired.

Let $S'$ be the antipode of $S.$ Then let $R'=\overline{CS'} \cap \overline{OP}, Q'=\overline{BS'} \cap \overline{OP}.$ Claim: $R'$ is the reflection of $R$ over $O$ and $Q'$ is the reflection of $Q$ over $O.$ Proof: Use DIT on $AS'BC,$ going through circumconic $\Omega$ and line $\overline{OP}.$ Note that as $\angle PSA=\angle SAS'=90^{\circ} ,$ then $\overline{AS'}\|\overline{SP}.$ As $O$ lies on the perpendicular bisector of $\overline{AS},$ we have that the reflection of $P$ over $O$ lies on $\overline{AS'}.$ Thus, the rest of the pair of points that $DIT$ gives an involution between must also be reflections through $O$ by Descargues Assistant Theorem. Let $A',B',C',$ be the antipodes of $A,B,C,$ respectively. Note that by the above we have that $S'R'C$ and $S'Q'B$ are collinear. Reflecting these $2$ lines about the origin gives us that $SRC'$ and $SQB'$ are collinear. Thus, $\angle CSC'=\angle CSR=90^{\circ}=\angle BSQ.$ Let the orthocenter of $\triangle RAQ$ be $T.$ Using Gauss-Bodenmiller on complete quadrilateral $BRQCAP,$ we get that the Steiner line $(AP)\cap(CR)\cap(BQ)$ is $\overline{TH}.$ As $S$ is one of the interection points of all $3$ circles as we just proved with the right angles, and as $\overline{SH}\cap (AP)=X,$ we have that $TXHS$ is collinear. Note that $(BQ)$ and $(CR)$ must also go through $X$ as well. \newline Note that we have that $Y$ is on $(BQ)$ as $\angle BYQ=90^{\circ}.$ By the same logic $Z$ is on $(CR).$ We finish with an angle chase. \begin{align*} \measuredangle YXZ &=\measuredangle XYA +\measuredangle YAZ+\measuredangle AZX \\ &=\measuredangle XSB +\measuredangle BAC + \measuredangle CSX \\ &=\measuredangle CSB +\measuredangle BAC \\ &=0.\blacksquare \end{align*}

We use moving points. Vary $P$ with degree $1$, so line $\ell=OP$ has degree $1$, then $Q,R$ move with degree $1$ so $Y,Z$ do as well, and $S$ moves with degree $2$ so $T=SH\cap\Omega$ moves with degree $2$, and $X$ is the midpoint of $HT$ since $H$ has twice the power to $\Omega$ as $(APS)$ by considering the $A$-altitude Thus $X$ has degree $2$. Thus we need to check $5$ cases: When $\ell=OA$ we have $Y=Z=A$. When $\ell=OB$ we have $X=Y$ is the foot from $B$ to $AC$. When $\ell=OC$ we have $X=Z$ is the foot from $C$ to $AB$. When $\ell$ is the perpendicular bisector of $AB$ we have $X=Z$ is the midpoint of $AB$. When $\ell$ is the perpendicular bisector of $AC$ we have $X=Y$ is the midpoint of $AC$.