Taiwan IMO Selection Camp

Below we use $(XYZ)$ to denote the circumcircle of $\triangle XYZ$. Let $D, E, F$ be three points on $BC, CA, AB$ respectively. By Miquel's theorem, the three circumcircles $(AEF) = \omega_A$, $(BFD) = \omega_B$, $(CDE) = \omega_C$ have a common point $P \neq D$.
Let $\omega_A, \omega_B, \omega_C$ meet $AI, BI, CI$ respectively at points $A \neq A', B \neq B', C \neq C'$. The key point here is that the three points $A', B', C'$ do not depend on the points $D, E, F$, as long as they satisfy $BD + BF = CA, CD + CE = AB, AE + AF = BC$ (the last equation can be derived from the first two). For this we first prove a lemma.
Lemma. Given $\angle A = \alpha$. A circle $\omega$ passes through point $A$, and intersects the angle bisector of $\angle A$ at $L$, and intersects the two sides of $\angle A$ at points $X, Y$ respectively. Then $AX + AY = 2AL \cos \frac{\alpha}{2}$.
Proof. Note that point $L$ is the midpoint of arc $\text{XY}$ on $\omega$, so we may let $XL = YL = u, XY = v$. By Ptolemy's theorem, $AX \cdot YL + AY \cdot XL = AL \cdot XY$, which can be rewritten as $(AX + AY)u = AL \cdot v$. Since $\angle LXY = \frac{\alpha}{2}$ and $\angle XLY = 180^\circ - \alpha$, by the law of cosines we get $v = 2u \cos \frac{\alpha}{2}$, and thus the lemma is proved.

Applying this lemma to $\angle BAC = \alpha$ and the circle $\omega = \omega_A$, where $\omega_A$ intersects $AI$ at point $A'$, we obtain $2AA' \cos \frac{\alpha}{2} = AE + AF = BC$. Similarly we can derive the equations satisfied by $BB', CC'$. From this it follows that the positions of the points $A', B', C'$ do not depend on the choice of the points $D, E, F$.
We again apply this lemma twice to $\angle BAC = \alpha$. Let $\omega$ be the circle with $AI$ as diameter. In this case, the two points $X, Y$ are respectively the points where the incircle of $\triangle ABC$ touches sides $AB, AC$, so $AX = AY = \frac{1}{2}(AB+AC-BC)$. By the lemma we get $2AI \cos \frac{\alpha}{2} = AB+AC-BC$. Now let $\omega$ be the circumcircle of $\triangle ABC$, and let $AI$ intersect $\omega$ at point $M \neq A$. In this case $\{B, C\} = \{X, Y\}$, so by the lemma $2AM \cos \frac{\alpha}{2} = AB + AC$. We summarize the results obtained so far as follows:
$$\begin{aligned} 2 AA' \cos \frac{\alpha}{2} &= BC, \\ 2 AI \cos \frac{\alpha}{2} &= AB + AC - BC, \\ 2 AM \cos \frac{\alpha}{2} &= AB + AC. \end{aligned} \qquad (1)$$

From the above equations we can derive $AA' + AI = AM$, and hence segment $AM$ and $IA'$ have the same midpoint.
From this we can conclude that point $I$ and point $A'$ are equidistant from the circumcenter $O$. By symmetry we know $OI = OA' = OB' = OC'$, so $I, A', B', C'$ are concyclic, with center $O$.
To prove $OI = OP$, it now suffices to prove that $I, A', B', C', P$ are concyclic. If $P$ equals one of $I, A', B', C'$, then there is nothing to prove; so suppose $P \neq I, A', B', C'$.

In the argument below we use directed angles to avoid distinguishing signs. Denote the directed angle between lines $l, m$ by $\angle(l, m)$. For any lines $l, m, n$ we naturally have $\angle(l, m) = -\angle(m, l)$

and $\angle(l, m) + \angle(m, n) = \angle(l, n)$. The necessary and sufficient condition for four distinct points $U, V, X, Y$ to be concyclic
is $\angle(UX, VX) = \angle(UY, VY)$.
Let us temporarily assume that the four points $I, P, A', B'$ are all distinct and not collinear; then it suffices to verify the equation $\angle(A'P, B'P) = \angle(A'I, B'I)$. Since $A, F, P, A'$ lie on circle $\omega_A$, we get $\angle(A'P, FP) = \angle(A'A, FA) = \angle(A'I, AB)$. By similar reasoning we get $\angle(B'P, FP) = \angle(B'I, AB)$. Thus we have
$$\begin{aligned} \angle(A'P, B'P) &= \angle(A'P, FP) + \angle(FP, B'P) \\ &= \angle(A'I, AB) - \angle(B'I, AB) = \angle(A'I, B'I). \end{aligned}$$
In the above computation we assumed $P \neq F$. If $P = F$, then $P \neq D, E$, and the conclusion can similarly be shown to hold (using $\angle(A'F, B'F) = \angle(A'F, EF) + \angle(EF, DF) + \angle(DF, B'F)$ together with the inscribed angles in the circles $\omega_A, \omega_B, \omega_C$).
Assuming that $A', B', P, I$ are distinct and not collinear does not lose generality. If $ABC$ is an equilateral triangle, then equation (??) implies that the six points $A', B', C', I, O, P$ coincide, in which case $OP = OI$. Otherwise, at most one of the three points $A', B', C'$ can coincide with $I$, for the following reason: suppose $C' = I$, then by the previous argument we know $OI \perp CI$; so $A', B' \neq I$, hence $A' \neq B'$. Finally, since $I, A', B', C'$ are concyclic, $A', B', I$ cannot be collinear.
This completes the proof.