Taiwan IMO Selection Camp, 2018

Let $AD$, $BE$, $CF$ be the altitudes of $\triangle ABC$, and let $\Gamma_A$, $\Gamma_B$, $\Gamma_C$ be the circles with diameters $\overline{AP}$, $\overline{BQ}$, $\overline{CR}$ respectively. Then the power of $H$ with respect to $\Gamma_A$, $\Gamma_B$, $\Gamma_C$ are $HA \cdot HD$, $HB \cdot HE$, $HC \cdot HF$ respectively, so the powers of $H$ with respect to the three circles are all equal. Let $A'$, $B'$ be the reflections of $A$, $B$ over $O$ respectively, and let $S'$ be the intersection of $A'P$ and $B'Q$. Then since $O$, $P$, $Q$ are collinear, by the converse of Pascal's theorem, $A$, $A'$, $S'$, $B'$, $B$, $C$ lie on a conic (considering the broken line $AA'S'B'BC$), so $S'$ lies on $\Omega$, i.e. $S = S'$. From $\angle BSQ = \angle BSB' = 90^\circ$ we know $S$ lies on $\Gamma_B$, and similarly $S$ lies on $\Gamma_C$. Hence $\Gamma_A$, $\Gamma_B$, $\Gamma_C$ are coaxial, with $SH$ being the radical axis, so the second intersection point of $\Gamma_A$, $\Gamma_B$, $\Gamma_C$ is $X$. Therefore, since $Y$, $Z$ lie on $\Gamma_B$, $\Gamma_C$ respectively, we obtain
$$\angle YXZ = \angle YXS + \angle SXZ = \angle ABS + \angle SCA = 180^\circ$$
that is, $X$, $Y$, $Z$ are collinear. This completes the proof.

### Alternative Solution

In fact, we can prove a stronger result. Let the midpoint of $AH$ be $N$, and let the tangent to the circumcircle at $A$ meet the circle with diameter $AP$ at $F$. Then $F$, $N$, $X$, $Y$, $Z$ are five collinear points, and this line is antiparallel to $OP$ with respect to $AB$, $AC$. We prove this in several steps below:

1. $N$, $Y$, $Z$ are collinear and antiparallel to $OP$ with respect to $AB$, $AC$.
Consider the reflective homothety centered at $A$, with the angle bisector of $\angle BAC$ as the axis of symmetry, and ratio $\cos A$. Since $QR \perp RA$ and $\angle QAR = \angle BAC$, we know that the image of $Q$ under this reflective homothety is $Y$. Similarly, the image of $R$ is $Z$. Therefore, the image of $QR$ is $YZ$. Note that $AN$, $AO$ are isogonal with respect to $BC$, and
$$AN = \frac{1}{2}AH = \frac{1}{2}R \cos A = AO \cos A,$$
so the image of $O$ is $N$. Note that $O$, $P$, $Q$, $R$ are collinear, so via this reflective homothety we can obtain that $N$, $Y$, $Z$ are collinear, and $NYZ$ is antiparallel to $OPQR$ with respect to $AB$, $AC$.

2. Let $A'$ be the reflection of $A$ over $F$. Then $A$, $A'$, $S$, $H$ are concyclic.
Let $AH$ meet the circumcircle of $ABC$ at $H'$, and meet $BC$ at $D$. Then $HD = H'D$. Consider the inversion centered at $A$, and suppose the image of any point $T$ under this inversion is $T^*$. By the angle-preserving property of inversion, we know:
$$\angle AP^*S^* = \angle ASP = 90^\circ, \angle AP^*D^* = \angle ADP = 90^\circ.$$
Moreover,
$$\angle AP^*A'^* = \angle AA'P = \angle A'AP = \angle A'^*AP^*,$$
so $A'^*A = A'^*P^*$. Since the circle $ASH'$ is tangent to $AA'$, $S^*H'^*$ is parallel to $AA'^*$. Since the midpoint of $HH'$ is $D$, $H^*H'^*$ is harmonically divided by $AD^*$. To prove that $A$, $A'$, $S$, $H$ are concyclic, it suffices to prove that $A'^*$, $S^*$, $H^*$ are collinear, or that $S^*(A'^*, H'^*; A, D^*)$ is a harmonic pencil. Note that $S^*H'^*$ meets $AA'^*$ at infinity, so, if $S^*D^*$ meets $AA'^*$ at $K$, then
$$S^*(A'^*, H'^*; A, D^*) = (A'^*, \infty; A, K).$$

Note that $AP^*K$ is a right triangle with the right angle at $P$, and $A'^*$ lies on the hypotenuse satisfying $A'^*A = A'^*P^*$, so $A'^*$ is the circumcenter of $AP^*K$. In particular, $A'^*$ is the midpoint of $AK$, so
$$S^*(A'^*, H'^*; A, D^*) = (A'^*, \infty; A, K) = -1,$$
which completes the proof.

3. $F$, $N$, $X$ are collinear.
By (2), we have
$$\angle AFN = \angle AA'H = \angle ASH = \angle ASX = \angle AFX.$$
Hence $F$, $N$, $X$ are collinear.

Since $AO$, $AN$ are isogonal with respect to $AB$, $AC$, it suffices to prove that $FN$, $OP$ are antiparallel with respect to $AN$, $AO$, that is,
$$\angle ANF + \angle POA = 180^\circ \text{ suffices.}$$
Let the midpoint of $AS$ be $L$. Then by (2) applying a homothety of ratio $1/2$ centered at $A$, we know $A$, $F$, $L$, $N$ are concyclic. Therefore
$$\angle ANF = \angle ALF.$$
Let $U$ be the reflection of $A$ over $O$, and let $V$ be the projection of $S$ onto $AF$. Then by the tangent-chord angle,
$$\angle VAS = \angle SUA.$$
Also, since
$$\angle SVA = \angle ASU = 90^\circ,$$
we have $\triangle SVA \sim \triangle ASU$. Since $P$ lies on $SU$ and $A$, $F$, $V$ are the projections of $U$, $P$, $S$ respectively, we know
$$\frac{AF}{FV} = \frac{UP}{PS}.$$
Also, since $L$, $O$ are the midpoints of $SA$, $AU$ respectively, $\triangle ALF \sim \triangle VOP$. Therefore
$$\angle ANF + \angle POA = \angle ALF + \angle POA = \angle VOP + \angle POA = 180^\circ.$$
By (1) and (4), we know that $F$, $Y$, $Z$ all lie on the line through $N$ that is antiparallel to $OP$ with respect to $AB$, $AC$, so $F$, $N$, $Y$, $Z$ are collinear. And by (3), $X$ lies on $FN$, so $F$, $N$, $X$, $Y$, $Z$ are five collinear points.