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Let $P$ be point inside triangle $ABC$, let $A_1$, $B_1$, $C_1$ be the vertices of pedal triangle of point P. Let $X$, $Y$, $Z$ be the incenters of triangles $AB_1C_1$, $BA_1C_1$ and $CA_1B_1$. Is it true that if $P$ is circumcenter of $XYZ$, then $P$ is the incenter of $ABC$?

I've tried to use trigonometry and recieved following system $$ \begin{cases} \frac{\sin^2 \beta_1}{\sin^2 \alpha_2}=\frac{1-\cos \beta_1 \cos \beta_2}{1-\cos \alpha_1 \cos \alpha_2} \\ \frac{\sin^2 \beta_2}{\sin^2 \gamma_1}=\frac{1-\cos \beta_1 \cos \beta_2}{1-\cos \gamma_1 \cos \gamma_2} \\ \frac{\sin^2 \alpha_1}{\sin^2 \gamma_2}=\frac{1-\cos \alpha_1 \cos \alpha_2}{1-\cos \gamma_1 \cos \gamma_2} \end{cases},$$ where $\alpha_1 = \angle BCP$, $\alpha_2 = \angle PAC$ and so on. But i don't know how to prove that there isn't any solutions except $\alpha_1=\alpha_2, ...$

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Let $LMN$ be the triangle formed by reflecting $P$ over the sides of $ABC$. It is well known that the circumcenter of this triangle is the isogonal conjugate of $P$ with respect to triangle $ABC$-this can be proved by showing each perpendicular bisector of $LMN$ corresponds to an isogonal of $AP,BP,$ or $CP$. Then by homothety the circumcenter of $XYZ$ is the midpoint of $P$ and its isogonal conjugate, so $P$ is its own isogonal conjugate. This is possible iff $P$ is the incenter or one of the three excenters of $ABC$.

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