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The center of the circle passing through the midpoints of sides of isosceles triangle $ABC$ lies on the circumcircle of triangle $ABC$. If largest angle of the triangle is $x$ and smallest is $y$. find $x-y$.

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The center of any circle with $D$ and $E$ on it must pass through the (potentially extended) bisector of $\angle A$. For this center to be on the circumcircle of $\triangle ABC$, the only possibility is for the center to be $A$ itself.

$AF=AD$ since they are both radii of the same circle. $AD=DB$ since $D$ is the midpoint of $\overline{AB}$. $\overline{AF}\perp \overline{BC}$, since $\triangle ABC$ is isosceles. Therefore, since $AB=2AF$, $\angle B=30^\circ$. That makes $\angle A=120^\circ$, so the difference between them is $90^\circ$.

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