$\angle HXP= \angle AQB = 90^0 + \angle PAC $ Now , $\angle PAC = \angle PXC $ So , $\angle HXC =90^0$ So, $X$ is the intersection of circle $ABC$ and circle with the diameter $HC$ .Hence $X$ is a fixed point and unique.

Thnks a lot u relieved me..I too did the same but proved with radical axis thm

but how does it prove that the circumcircle of HPQ intersect the circumcircle of ABC at point X.

My solution: Let $ C' $ be the antipode of $ C $ and $ T=C'H \cap (ABC) $ . ( Easy to see $ T $ is a fixed point on $ (ABC) $ ) Since $ AC' \parallel HQ $ , so from Reim theorem we get $ T \in (HPQ) $ . Q.E.D

There is a general version of the problem which is true : $ABC$ be a triangle and $Y$ is a fixed point inside the triangle . Let$P$ be any point on $ \odot ABC$ . $ BY \cap AP = Q $ .Pt the circumcircle of $YPQ$ passes through a fixed point $X$ on $ \odot ABC$

BBAI wrote: There is a general version of the problem which is true : $ABC$ be a triangle and $Y$ is a fixed point inside the triangle . Let$P$ be any point on $ \odot ABC$ . $ BY \cap AP = Q $ .Pt the circumcircle of $YPQ$ passes through a fixed point $X$ on $ \odot ABC$ My solution: Let $ R\in (ABC) $ s.t. $ AR \parallel BY $ and $ T=RY \cap (ABC) $ . Easy to see $ T $ is a fixed point on $ (ABC) $ . Since $ AR \parallel QY $ , so from Reim theorem we get $ T \in (PQY) $ . Q.E.D

My solution. First we take any point P in the circle and construct a circumcircle of ΔHQP and determine X. Now we need to prove that for any other point P', X must lie on circumcircle of ΔHQ'P', where Q' is the intersection of AP' with BH. From cyclic quad. PQHX, θ + φ = 180° [ θ = ∠QHX and φ = ∠QPX ] .... (1). Also P' and Q' are independant for the reader to select, thus it is sufficient to prove that X also lies on the circumcircle of ΔHQ'P' , or to prove that HQ'P'X is a cyclic quadrilateral. As ∠ suspended by AX is same at P and P', thus ∠AP'X = φ. From (1), in HQ'P'X, θ + φ = 180°. Thus HQ'P'X is cyclic. Hence proved for any point P'.

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Telvcohl ,u r right . The solution of the general problem is same as the previous of mine too . Except saying that $X$ lies on circle with diameter $HC$ ,Here it will be that $XYFC$ is cyclic where $BY \cap AC =F$.

My Solution. Let $AD,BE$ be the altitudes of $\triangle ABC.$ Let $AD,BE$ produced meet $\odot ABC$ again at $D_1,E_1$ respectively. Let $X$ be the $2^{nd}$ intersection point of $\odot HQP$ and $\odot ABC.$ Then the quadrilaterals $HQPX,APCX$ will be cyclic. Let $\angle PXH=\theta.$ Then $\angle HQP = 180^{\circ}-\theta.\rightarrow \angle PQE_1=\angle AQE=\theta.$ Then, $\angle QEA=90^{\circ}.$ Which means $\angle EAQ=\angle CAP=90^{\circ}-\theta.$ As $APCX$ is cyclic, we have $\angle CXP=\angle CAP=90^{\circ}-\theta.$ Or, $\angle CXH=\theta+(90^{\circ}-\theta)=90^{\circ}.$ Let $\Omega$ be the circle with diameter $CH.$ Then $X\in \Omega.$ Now, $X,C\in \Omega,\odot ABC.$ Which means $\Omega,\odot ABC$ intersect at $X,C.$ Clearly, $\Omega$ and $\odot ABC$ are unique as $\triangle ABC$ and $H$ are unique. Which means $X$ is also unique. $\Square$

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BBAI wrote: There is a general version of the problem which is true : $ABC$ be a triangle and $Y$ is a fixed point inside the triangle . Let$P$ be any point on $ \odot ABC$ . $ BY \cap AP = Q $ .Pt the circumcircle of $YPQ$ passes through a fixed point $X$ on $ \odot ABC$ Which theorem did you use ?

There are multiple configurations to this problem, so either all the configurations should be taken into configuration or directed angles have to be used.

Vrangr wrote: There are multiple configurations to this problem, so either all the configurations should be taken into configuration or directed angles have to be used. exactly (Though, the official solution has considered all the possible configurations)