Problem 1297

Solution. By MiqueL's theorem the circles $(A E F)=\omega_{A},(B F D)=\omega_{B}$ and $(C D E)=\omega_{C}$ have a common point, for arbitrary points $D, E$ and $F$ on $B C, C A$ and $A B$. So $\omega_{A}$ passes through the common point $P \neq D$ of $\omega_{B}$ and $\omega_{C}$.

Let $\omega_{A}, \omega_{B}$ and $\omega_{C}$ meet the bisectors $A I, B I$ and $C I$ at $A \neq A^{\prime}, B \neq B^{\prime}$ and $C \neq C^{\prime}$ respectively. The key observation is that $A^{\prime}, B^{\prime}$ and $C^{\prime}$ do not depend on the particular choice of $D, E$ and $F$, provided that $B D+B F=C A, C D+C E=A B$ and $A E+A F=B C$ hold true (the last equality follows from the other two). For a proof we need the following fact.
Lemma. Given is an angle with vertex $A$ and measure $\alpha$. A circle $\omega$ through $A$ intersects the angle bisector at $L$ and sides of the angle at $X$ and $Y$. Then $A X+A Y=2 A L \cos \frac{\alpha}{2}$.
Proof. Note that $L$ is the midpoint of arc $\widehat{X L Y}$ in $\omega$ and set $X L=Y L=u, X Y=v$. By PtolemY's theorem $A X \cdot Y L+A Y \cdot X L=A L \cdot X Y$, which rewrites as $(A X+A Y) u=A L \cdot v$. Since $\angle L X Y=\frac{\alpha}{2}$ and $\angle X L Y=180^{\circ}-\alpha$, we have $v=2 \cos \frac{\alpha}{2} u$ by the law of sines, and the claim follows.

Apply the lemma to $\angle B A C=\alpha$ and the circle $\omega=\omega_{A}$, which intersects $A I$ at $A^{\prime}$. This gives $2 A A^{\prime} \cos \frac{\alpha}{2}=A E+A F=B C$; by symmetry analogous relations hold for $B B^{\prime}$ and $C C^{\prime}$. It follows that $A^{\prime}, B^{\prime}$ and $C^{\prime}$ are independent of the choice of $D, E$ and $F$, as stated.

We use the lemma two more times with $\angle B A C=\alpha$. Let $\omega$ be the circle with diameter $A I$. Then $X$ and $Y$ are the tangency points of the incircle of $A B C$ with $A B$ and $A C$, and hence $A X=A Y=\frac{1}{2}(A B+A C-B C)$. So the lemma yields $2 A I \cos \frac{\alpha}{2}=A B+A C-B C$. Next, if $\omega$ is the circumcircle of $A B C$ and $A I$ intersects $\omega$ at $M \neq A$ then $\{X, Y\}=\{B, C\}$, and so $2 A M \cos \frac{\alpha}{2}=A B+A C$ by the lemma. To summarize,
$$2 A A^{\prime} \cos \frac{\alpha}{2}=B C, \quad 2 A I \cos \frac{\alpha}{2}=A B+A C-B C, \quad 2 A M \cos \frac{\alpha}{2}=A B+A C .$$

These equalities imply $A A^{\prime}+A I=A M$, hence the segments $A M$ and $I A^{\prime}$ have a common midpoint. It follows that $I$ and $A^{\prime}$ are equidistant from the circumcenter $O$. By symmetry $O I=O A^{\prime}=O B^{\prime}=O C^{\prime}$, so $I, A^{\prime}, B^{\prime}, C^{\prime}$ are on a circle centered at $O$.

To prove $O P=O I$, now it suffices to show that $I, A^{\prime}, B^{\prime}, C^{\prime}$ and $P$ are concyclic. Clearly one can assume $P \neq I, A^{\prime}, B^{\prime}, C^{\prime}$.

We use oriented angles to avoid heavy case distinction. The oriented angle between the lines $l$ and $m$ is denoted by $\angle(l, m)$. We have $\angle(l, m)=-\angle(m, l)$ and $\angle(l, m)+\angle(m, n)=\angle(l, n)$ for arbitrary lines $l, m$ and $n$. Four distinct non-collinear points $U, V, X, Y$ are concyclic if and only if $\angle(U X, V X)=\angle(U Y, V Y)$.

Suppose for the moment that $A^{\prime}, B^{\prime}, P, I$ are distinct and noncollinear; then it is enough to check the equality $\angle\left(A^{\prime} P, B^{\prime} P\right)=\angle\left(A^{\prime} I, B^{\prime} I\right)$. Because $A, F, P, A^{\prime}$ are on the circle $\omega_{A}$, we have $\angle\left(A^{\prime} P, F P\right)=\angle\left(A^{\prime} A, F A\right)=\angle\left(A^{\prime} I, A B\right)$. Likewise $\angle\left(B^{\prime} P, F P\right)=\angle\left(B^{\prime} I, A B\right)$. Therefore
$$\angle\left(A^{\prime} P, B^{\prime} P\right)=\angle\left(A^{\prime} P, F P\right)+\angle\left(F P, B^{\prime} P\right)=\angle\left(A^{\prime} I, A B\right)-\angle\left(B^{\prime} I, A B\right)=\angle\left(A^{\prime} I, B^{\prime} I\right) .$$

Here we assumed that $P \neq F$. If $P=F$ then $P \neq D, E$ and the conclusion follows similarly (use $\angle\left(A^{\prime} F, B^{\prime} F\right)=\angle\left(A^{\prime} F, E F\right)+\angle(E F, D F)+\angle\left(D F, B^{\prime} F\right)$ and inscribed angles in $\left.\omega_{A}, \omega_{B}, \omega_{C}\right)$.

There is no loss of generality in assuming $A^{\prime}, B^{\prime}, P, I$ distinct and noncollinear. If $A B C$ is an equilateral triangle then the equalities (*) imply that $A^{\prime}, B^{\prime}, C^{\prime}, I, O$ and $P$ coincide, so $O P=O I$. Otherwise at most one of $A^{\prime}, B^{\prime}, C^{\prime}$ coincides with $I$. If say $C^{\prime}=I$ then $O I \perp C I$ by the previous reasoning. It follows that $A^{\prime}, B^{\prime} \neq I$ and hence $A^{\prime} \neq B^{\prime}$. Finally $A^{\prime}, B^{\prime}$ and $I$ are noncollinear because $I, A^{\prime}, B^{\prime}, C^{\prime}$ are concyclic.

Comment. The proposer remarks that the locus $\gamma$ of the points $P$ is an arc of the circle $\left(A^{\prime} B^{\prime} C^{\prime} I\right)$. The reflection $I^{\prime}$ of $I$ in $O$ belongs to $\gamma$; it is obtained by choosing $D, E$ and $F$ to be the tangency points of the three excircles with their respective sides. The rest of the circle $\left(A^{\prime} B^{\prime} C^{\prime} I\right)$, except $I$, can be included in $\gamma$ by letting $D, E$ and $F$ vary on the extensions of the sides and assuming signed lengths. For instance if $B$ is between $C$ and $D$ then the length $B D$ must be taken with a negative sign. The incenter $I$ corresponds to the limit case where $D$ tends to infinity.