Here is a very short and elegant solution, which I had found in contest Let $U$ be the foot of $H$ onto $YZ$. Claim 1 : $HQBZ$ and $HRCY$ are cyclic. Proof : Since $\angle ZOB + \angle YOC = 180^{\circ}$, $O$ has an isogonal conjugate in quadrilateral $BCYZ$. Looking at angles $\angle CBZ$ and $\angle BCY$, the isogonal conjugate must be $H$. Hence, $\measuredangle HZB = \measuredangle YZO = \measuredangle UZO = \measuredangle UQO = \measuredangle HQB$. This implies that $HQBZ$ is cyclic, and so is $HRCY$ by symmetry. Claim 1 : $QZ$ and $RY$ are tangents to $(AYZ)$. Proof : $\measuredangle QZY = \measuredangle QZU = 90^{\circ} - \measuredangle UQZ = 90^{\circ} - \measuredangle HQZ = 90^{\circ} - \measuredangle HBZ = \measuredangle BAC = \measuredangle ZAY$. Hence $QZ$ is tangent to $(AYZ)$, and the other follows by symmetry. To finish, observe that $T=QZ \cap RY$. Hence $\measuredangle QTR = \measuredangle ZTY = 2 \measuredangle ZAY = 2 \measuredangle BAC = \measuredangle BOC = \measuredangle QOR$. Hence $Q,R,O,T$ are concyclic, and we are done!

This problem was proposed by Kazi Aryan Amin (me!) and KV Sudharshan. Here's the solution we sent : Define $K$ to be the point on $YZ$ such that $HK \perp YZ$. Let $A'$ be a point on the circumcircle of $\triangle ABC$ such that $AA' \parallel BC$ Lemma : $YZ$ is the perpendicular bisector of $HA'$. Proof: Let $H_B$ denote the reflection of $H$ in $AC$. Then note that $A'H_B \parallel CO$. This is because $\angle (AA',A'H_B) = 90-\angle A = \angle (CO,BC)$. Now $YO \perp CO$ implies $OY\perp A'H_B$. This shows $YA'=YH_B=YH$. Similarly $ZH=ZA'$. Lemma : $OK\perp BC$ Proof: Let $D$ and $D'$ denote the foot of perpendiculars from $H,A'$ onto $BC$ respectively. Then note that if $M$ is the foot of perpendicular from $K$ onto $BC$, then by similarity, $DM=MD'$ as $HK=KA'$. However $BD=D'C$, so $MB=MC$, as desired. Lemma : $RY$ and $QZ$ are tangents to the circumcircle of $\triangle AYZ.$ Proof: Note that $RKYO$ is a cycle quadrilateral. Now we angle chase : $\angle RYK=\angle ROK=\angle (CO, OK)= \angle BAC = \angle YAZ$. Thus, $RY$ is tangent to the circumcircle of $\triangle AYZ$. Similarly for $QZ$. Finally note that $O$ is the miquel point of quadrilateral $QZYR$, as $RKYO$ and $QKZO$ are cyclic. Thus $T= QZ \cap YR$ lies on the circumcircle of $\triangle QOR$, as desired. $\square$

The part $HYZ \sim ABC$ could be done easily with complex bash.

Since $\angle ZOB+\angle YOC=180$ we have that $O$ has an isogonal conjugate in $ZYCB$, in fact it is $H$, therefore $\triangle HYZ \sim \triangle ABC$ and since $\angle ZHQ=90-\angle ACB=\angle ABO$ we get $ZQHB$ cyclic, in a similar way we get $YRHC$ cyclic, note that depending on whether $R,Q$ are on one part of their line or in the other w.r.t. $H$, because in either case the angle chase remains correct, i will show the one in my diagram which is $\angle QBH=\angle ACB-\angle BAC$ so $\angle YZT=\angle BAC=\angle ACB-(\angle ACB-\angle BAC)=\angle YZH-\angle HZQ=\angle YZQ$ therefore $Z,Q,T$ are colinear. In a similar way $Y,R,T$ are colinear as well for either case. And now since $\angle RTQ=2\angle BAC=\angle BOC$ we have $RTQO$ cyclic as desired thus we are done .

Nice problem

Hard problem,especially if you are in a contest. Take $M$ to be the Miquel point of $BZCY$. We have the following claims: Claim 1: $AM\parallel BC$ Proof:Observe that $MCY\sim MBZ$ so $\frac{MC}{MB}=\frac{CY}{BZ}$. Take $L$ to be the midpoint of $AB$.Then $BO^2=BL\cdot BZ=\frac{BA\cdot BZ}{2}$ so $BZ=\frac{2R^2}{BA}$,similary $CY=\frac{2R^2}{AC}$ so $\frac{MC}{MB}=\frac{AB}{AC}$ from which it follows easily that $AM\parallel BC$ Claim 2:$M$ and $H$ are symmetric wrt $YZ$ Proof:Employ complex coordinates.WLOG $(ABC)$ unit circle so we will have that $z=\frac{b^2+ba}{b-a};y=\frac{c^2+ca}{c-a},h=a+b+c , m=\frac{bc}{a}$ and we want to prove that: $$\frac{\overline m-\overline z}{\overline y-\overline z}=\frac{h-z}{y-z}$$Plug in and after a 5 min calculation simplication you can see that they are both equal to $\frac{c-a}{c-b}$ so we are done. Claim 3: Let $K=HM\cap YZ$,then $K$ is on the perpendicular bisector of $BC$ Proof:$K$ is the midpoint of $HM$ and since $\angle MAH=90$ we have $K$ is on the perpendicular bisector of $AM$.Since $AMCB$ is an isosceles trapezoid,they have the same perpendicular bisector,as desired ! Claim 4:$QZ$ is tangent to $(AYZ)$,similary $RY$ meaning $T=QZ\cap RY$ Proof:We have that $Q\in MK$ so $\angle ZOQ=\angle ZKQ=90$ meaning $ZOKQ$ cyclic so $\angle QZK=180-\angle QOK=A$ so $QZ$ is tangent what was nedeed. Claim 5:$ QROT $cyclic. Proof:$\angle QOR=180-2A=\angle ZTY$ so we are done. PS:this is my first hybrid solution on aops !

Let $K$ be the projection of $H$ to $ZY$. $\angle ZOB+\angle YOC=180$ implies that $O$ has a isogonal conjugate wrt $BZYC$, which can be easily verifies to be $H$ so $HYZ \sim ABC$ which in turn by e $\angle ZHQ=90-\angle ACB=\angle ABO$ means that $ZHQB$ is cyclic, analogously $YRHC$ is cyclic. Now $\angle QZK = 90 - \angle ZQH = 90 - \angle A$ which implies that $ZQ$ is tangent to $AZY$ and analogously $RY$ becomes tangent to it as well. So $\angle ZTY = \angle BOR = 180 - 2 \angle A$ finishing the proof.

Straightforward complex bash also works. Let $(ABC)$ be a unit circle, $A$ has coordinate $a$ and so on. Then $h=a+b+c$. Let $K$ be points of intersection of tangent lines of points $B,C$ to $(ABC)$. Then $k=2bc/(b+c)$. Let's find $y$. We have \[ \begin{array}{lr} y+ac\overline{y}=a+c\\ c\overline{y}+\overline{c}y=0 \end{array} ,\]thus \[y = \frac{c(a+c)}{c-a}\]and similalrly \[z=\frac{b(a+b)}{b-a}. \]Now we can find \[y-z=\frac{(b-c)(-bc+ca+ab+a^2)}{(c-a)(b-a)} \]and \[ \overline{y-z}=\frac{(c-b)(bc+ca+ab-a^2)}{(c-a)(b-a)bc}. \]We have \[ \begin{array}{lr} \bar{r}=r/c^2 \text{ (since $R$ lies on $CO$)}\\ (h-r)(\overline{y-z})+(\overline{h-r})(y-z)=0 \text{ (since $RH \perp YZ$)} \end{array} ,\]so after some manipulations we can find \[r=\frac{c(a+c)(a^2-bc)}{a(b+c)(a-c)} \]and similarly \[ q=\frac{b(a+b)(a^2-bc)}{a(b+c)(a-b)}.\]Now note that $\Delta YTZ \sim \Delta CKB$, because $\angle KBC=\angle KCB=\angle A=\angle TYZ=\angle TZY$, so we have \[ \frac{t-y}{t-z}=\frac{k-c}{k-b}=-c/b \Rightarrow t=\frac{2bc(bc-a^2)}{(b+c)(a-c)(a-b)}.\]Finally we are ready to solve the problem. It is known that $T \in (QOR) \iff z=\frac{(t-r)q}{(t-q)r} \in \mathbb{R}$, and using formulas above we can find \[z=\frac{(a+b)(a-c)}{(a+c)(a-b)}, \]which is clearly real, so we solved problem! $\square$

Let the intersection of tangents to $(AYZ)$ at $Y,Z$ be $W$ since we will use $T$ for another point. Claim: If $AT\parallel BC$ and $T\in (ABC)$, then $A,T,Y,Z$ are concyclic and $T$ is the reflection of $H$ over $YZ$. Proof: Invert around $(ABC)$. Then perform $\sqrt{bc}$ inversion and reflect over the angle bisector of $\measuredangle CAB$. Let $B',C'$ be the antipodes of $B,C$ and $P,Q\in (ABC)$ such that $BA=BP,CA=CQ$. We want to show that $AA\cap BC=D,AC'\cap PC,AB'\cap QB$ are concurrent. Pascal at $AAC'BCP$ gives $D,AC'\cap CP,C'B\cap AP$ are collinear and pascal at $AAB'CBQ$ implies $D,AB'\cap QB,B'C\cap AQ$ are collinear. Let's prove the collinearity of $D,C'B\cap AP,B'C\cap AQ$. Set $AP\cap BC'=M,AQ\cap B'C=N$. By angle chasing we see that $\measuredangle AMB=90-\measuredangle A$ and $BAM=180-\measuredangle C$. Similarily $\measuredangle CNA=90-\measuredangle A$ and $\measuredangle NAC=180-\measuredangle B$. Thus, \[\frac{MB}{NC}=\frac{\sin C.\frac{AB}{\cos A}}{\sin B.\frac{AC}{\cos A}}=\frac{AB^2}{AC^2}=\frac{DB}{DC}\]Which yields $D,M,N$ are collinear. We have proved that $A,T,Y,Z$ are concyclic. $T$ is the miquel point of $BCYZ$. Since $\measuredangle BOZ+\measuredangle YOC=180$, isogonal conjugate of $O$ exists in quadrilateral $BCYZ$ and that must be $H$. Hence $\measuredangle HYZ=\measuredangle CYO=\measuredangle B$ and $\measuredangle YZH=\measuredangle OZB=\measuredangle C$ which implies $HYZ\sim ABC\sim TYZ$. We observe that $T$ is the reflection of $H$ onto $YZ$.$\square$ Claim: $H,R,Y,C$ and $H,Q,Z,B$ are concyclic. Proof: $\measuredangle RCY=90-\measuredangle B=\measuredangle YTH=\measuredangle RHY$ and $\measuredangle ZBQ=90-\measuredangle C=\measuredangle QTZ=\measuredangle ZQH$ which prove the results.$\square$ We see that $\measuredangle RYZ=90-\measuredangle TRY=90-\measuredangle HCA=\measuredangle A$ hence $RY$ is tangent to $(AYZ)$. Similarily $\measuredangle YZQ=90-\measuredangle ZQH=90-\measuredangle ABH=\measuredangle A$ so $QZ$ is tangent to $(AYZ)$. These yield $W,Q,Z$ and $W,R,Y$ are collinear. Thus, $\measuredangle ZWY=2\measuredangle A=\measuredangle QOR$ which gives the desired result.$\blacksquare$

No new points added lol. Claim $O$ and $H$ are isogonal conjugates in $BCYZ$. Proof See that $\angle ZOB+\angle YOC$ means $O$ has an isogonal conjugate as it is well known that this is equivalent of $O$ having a pedal circle. Simple angle chase shows that is is $H$. $\square$ Claim $(BQHZ)$ is cyclic and similarly so is $(HRYC)$. Proof $\angle ZBQ=90^{\circ}-\angle BZO=90^{\circ}-\angle YZH=\angle ZHR=180^{\circ}-\angle ZHQ$. $\square$ Claim $T= \overline{QZ} \cap \overline{RY}$. Proof Just see that \begin{align*} \angle QZY=\angle QZH+\angle HZY=\angle OBH+\angle OZB &= 90^{\circ}-\angle ABO+\angle OBH \\ &=\angle ACB+\angle ABC-2\angle ABH \\ &=\angle BAC=\angle ZAY \end{align*}So $ \overline{QZ}$ is tangent to $(AZY)$ and similarly so is $ \overline{RY}$. $\square$ Now just see that \[\angle QOR=180^{\circ}-2\angle BAC=180^{\circ}-\angle YZT-\angle ZYT=\angle YTZ=180^{\circ}-\angle QTR\]And problem khatam. Remark: Bruh why is P1>>P2>>P3 in the geo problems for ind tst 2024 lols