, 2024

Solution:

Because the circles have equal radii, $\angle PDA=\angle ABP$, so if $(PDA)$ intersects line $AB$ again at a point $B'$, then we have $\angle PB'B=\angle PBB'$, which means $PB=PB'$, similarly for the second intersection of $(PCB)$ with $AB$, $A'$; thus, $(PDA)$ and $(PCB)$ are congruent mirror images across the $P$-altitude, as they are $\left(PAB'\right)$ and $\left(PBA'\right)$, respectively.

Consider $C'$, the reflection of $C$ across the $P$-altitude. We want to prove that $C'$ lies on $(XCD)$, as then the circumcenter of $(XCD)$ will lie on the perpendicular bisector of $CC'$. Because of our earlier observation, $C', D, P, A$ are concyclic.

We present two approaches to finishing the angle chase from here:

- Add point $C''$, the intersection of $CC'$ with $(XDA)$. Because $AB$ is parallel to the line between the centers, and so is $C''C$, then $ABCC''$ is a parallelogram; thus,
$$\angle C'DA=\angle C'C''A=\angle C'CB=\angle C'CX$$

- Add point $B'$, the reflection of $B$ over the $P$-altitude. Note that $B'$ lies on $(XDA)$; in particular, $C'CB B'$ is an isosceles trapezoid, as $C'B'$ is the reflection of $CB$ over the $P$-altitude of $\triangle PAB$. Thus,
$$\angle C'DA=180-\angle C'B'A=180-\angle C'B'B=\angle C'CB=\angle C'CX$$

Solution:

Invert about $P$. Because the circles $(PDA),(PAB),(PBC)$ all have equal radii and pass through $P$, the resulting lines $D'A', A'B', B'C'$ are equal distances away from $P$; letting $H'$ be the intersection of lines $D'A'$ and $B'C'$, it follows that $P$ is an incenter or excenter of $\triangle A'B'H'$. Also, in the original diagram, the $P$-altitude of $\triangle PAB$ includes the second intersection of the circles $(PDA)$ and $(PCB)$ (as in the first solution); thus this $P$-altitude inverts to line $PH'$. Finally, $X'$ is the intersection of $\left(PD'A'\right)$ and $\left(PB'C'\right)$.

Note that the center of $(XCD)$ lies on the $P$-altitude of $PAB$ iff the inverse of the center of $\left(X'C'D'\right)$ does. Thus we want to show that the center of $\left(X'C'D'\right)$ lies on $H'P$. Let $D''$ be the second intersection of $X'C'D'$ with $B'H'$. Then
$$\begin{aligned} \measuredangle\left(D'D'', H'P\right) & =\measuredangle D'D''H'+\measuredangle D''H'P \\ & =\measuredangle D'X'C'+\measuredangle B'H'P \\ & =\measuredangle D'X'P+\measuredangle PX'C'+\measuredangle PH'A' \\ & =\measuredangle D'A'P+\measuredangle PB'C'+\measuredangle PH'A' \\ & =\measuredangle PA'B'+\measuredangle PB'H'+\measuredangle PH'A' \\ & =90^\circ \end{aligned}$$
(where the last step follows from the fact that $P$ is an incenter or excenter of $\triangle A'B'H'$ ), and
$$\begin{aligned} \measuredangle H'D'D'' & =180^\circ-\measuredangle D''H'D'-\measuredangle D'D''H' \\ & =180^\circ-\measuredangle D''H'P-\left(\measuredangle D''H'P+\measuredangle D'D''H'\right) \\ & =90^\circ-\measuredangle D''H'P \\ & =\measuredangle D'D''H', \end{aligned}$$
so $\triangle H'D'D''$ is isosceles, and thus $H'P$ is the perpendicular bisector of $D'D''$. Thus the center of $\left(X'C'D'D''\right)$ lies on $H'P$, which means we're done.

Solution:

Let $A'$ be the other intersection of line $PA$ with $(PDX)$ and $B'$ be the other intersection of $PB$ with $(PCX)$. Consider circles $\left(PA'B'\right)$ and $(XCD)$. Note that $A$ and $B$ have equal power with respect to both circles, because of $\left(PDA'X\right)$ and $\left(PCB'X\right)$. Thus, $AB$ is the radical axis of the two circles. However,
$$\measuredangle PA'X=\measuredangle PDX=\measuredangle PDA=\measuredangle ABP$$
and
$$\measuredangle XB'P=\measuredangle XCP=\measuredangle BCP=\measuredangle PAB$$
where the last step follows from the fact that the circles have equal radii. Because $\measuredangle B'PA'=\measuredangle BPA$, it follows that $A', X, B'$ are collinear, and in fact $\triangle PAB \sim \triangle PB'A'$. In particular, this means that the $P$-altitude of $\triangle PAB$ passes through the circumcenter of $\triangle PB'A'$, as the circumcenter and orthocenter are isogonal conjugates. Thus, as the circumcenter of $PA'B'$ lies on the $P$-altitude, and the line between the centers of $\left(PA'B'\right)$ and $(XCD)$ must be perpendicular to their radical axis $AB$, then the circumcenter of $(XCD)$ must lie on the $P$-altitude as well, completing the proof.