Taiwan IMO Selection Camp

Let $\Omega$ denote the circumcircle of $\triangle ABC$. Recall that the reflection of $H$ across $\overline{BC}$, which we denote by $X$, is a point of $\Omega$. Define $Y, Z$ similarly. Thus the circumcircle $\omega_A$ of $\triangle HYZ$ is centered at $A$. Define $\omega_B, \omega_C$ similarly.

Then, if we let
$$r = AH + AK = BH + BK = CH + CK.$$
then it follows $K$ has the following description: it is the center of the (unique) circle tangent externally to all three of $\omega_A, \omega_B, \omega_C$, and has radius $r$.

Proof. First, note that quadrilateral $PYDB$ is cyclic, since
$$\angle DPY = \angle HZY = \angle CZY = \angle CBY = \angle DBY.$$
In particular $HB \cdot HY = HP \cdot HD$, and repeating this gives
$$HA \cdot HX = HB \cdot HY = HC \cdot HZ = HD \cdot HP = HE \cdot HQ = HF \cdot HR. \quad (1)$$

In particular PEDB and EFQR are cyclic too by Eq. (1). Now if $\ell$ is the tangent line to $\omega_A$ at $P$, then working with directed angles modulo $180^\circ$ gives
$$\begin{align*} \angle(\ell, PQ) &= \angle(\ell, PH) + \angle(PD, PQ) = \angle(PY, BY) + \angle(DE, EQ) \\ &= \angle(PD, DB) + \angle(DE, EQ) = \angle(PH, EQ) + \angle(DE, DB) \\ &= \angle(PD, EQ) + \angle(FE, DF) = \angle(PD, DF) + \angle(FE, EQ) \\ &= \angle(PR, RH) + \angle(RF, FQ) = \angle(PR, RQ). \end{align*}$$

Finally, if we define $P'$, $Q'$, $R'$ as the second intersection of $\overline{HP}$, $\overline{HQ}$, $\overline{HR}$ with the incircle, then it follows from Eq. (1) and
$$HD \cdot HP' = HE \cdot HQ' = HF \cdot HR'$$
that $\triangle P'Q'R'$ is homothetic to $\triangle PQR$ through $H$. As $I$ is the circumcenter of $\triangle P'Q'R'$ this implies $K, I, H$ are collinear.