Baltic Way 2019

$(\Rightarrow )$ is trivial. For $(\Leftarrow )$ let $AP\cap BC=D$. Now AD is a bisector of angle XDY, for this we can argue in two ways

Let $XY\cap BC=S$. Then $D(S,D;B,C)=1$ and $\angle ADB=90^{\circ }$, so statement.

Let $\ell$ be a parallel line to BC passing through A. Let $DY\cap \ell =K$ and $DX\cap \ell =L$. Then using Ceva and Tales theorem we see that A is a midpoint of KL.

Since ABC is a scalene we see that in triangle XDY bisector DA and perpendicular bisector of XY intersects at A, so AXDY is inscribed in circle \omega. Now let Z be the second intersection of \omega with BC. Then AXZY is a kite, so circle with centre Z and radius ZX satisfies problem statement.

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Alternative solution.

The implication $\Leftarrow$ is trivial. For the proof of $\Rightarrow$, let $AP\cap BC=D$. We prove that DA bisects angle XDY. We can argue in two ways:

Let $S=XY\cap BC$ and $XY\cap AD=T$. It is well-known that the quadruple $S,T,X,Y$ is harmonic. Since $\angle SDT=\frac{\pi }{2}$, it follows that DT bisects angle YDX.

Let $\ell$ be the line parallel to BC passing through A. Let $DY\cap \ell =K$ and $DX\cap \ell =L$. By Ceva and similarities we obtain $1=\frac{BD}{DC}\cdot \frac{CX}{XA}\cdot \frac{AY}{YB}=\frac{BD}{DC}\cdot \frac{CD}{AL}\cdot \frac{AK}{DB}=\frac{AK}{AL}$. Hence A is the midpoint of KL and since $\ell \perp AD$, it follows that DA bisects the angle XDY.

Since ABC is scalene, DA is not the perpendicular bisector of XY. So, since $AX=AY$ and DA bisects XDY, it follows that AXDY is cyclic. Let \omega be the circumcircle of AXDY. Let $Z\ne D$ be the second intersection of \omega with BC. Then AXZY is a kite, so the circle with centre Z and radius ZX satisfies the problem condition.