66. An isosceles triangle \(ABC\) with a perimeter of 12 is inscribed in a circle \(\omega\). Points \(P\) and \(Q\) are the midpoints of the arcs \(ABC\) and \(ACB\) respectively. The tangent to the circle \(\omega\) at point \(A\) intersects the ray \(PQ\) at point \(R\). It turns out that the midpoint of segment \(AR\) lies on the line \(BC\). Find the length of segment \(BC\).

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\text{(A. Kuznetsov)}
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66. Answer: 4. Let $I_{A}, I_{B}, I_{C}$ be the centers of the excircles of triangle $ABC$, touching sides $BC, CA$, and $AB$ respectively. Then the lines $A I_{A}, B I_{B}, C I_{C}$ will be the angle bisectors of triangle $ABC$, and the lines $I_{B} I_{C}, I_{C} I_{A}, I_{A} I_{B}$ will be its external angle bisectors. Therefore, points $A, B, C$ will be the feet of the altitudes of triangle $I_{A} I_{B} I_{C}$, and circle $\omega$ will be its nine-point circle. Then point $P$ is the point of intersection of $I_{A} I_{C}$ with $\omega$ other than $B$. Therefore, $P$ is the midpoint of $I_{A} I_{C}$. Similarly, $Q$ is the midpoint of $I_{A} I_{B}$. Thus, $PQ$ is the midline of triangle $I_{A} I_{B} I_{C}$.

Let $K$ and $L$ be the feet of the external and internal angle bisectors of angle $A$ of triangle $ABC$, and let $M$ be the point of intersection of lines $AR$ and $BC$. By the problem's condition, we know that $AM = MR$. We will prove that $AM = ML = MK$. Indeed, $\angle MAL = \angle MAC + \angle CAL = \angle ABC + \angle LAB = \angle ALM$ (point $M$ lies on ray $BC$ since $R$ lies on $PQ$). Therefore, $AM = ML$, and since triangle $AKL$ is right-angled, $AM = MK$.

Thus, $AM = ML = MK = MR$. Therefore, $ALRK$ is a rectangle and $LR \parallel I_{B} I_{C}$. We obtain that lines $PQ$ and $LR$ are parallel to $I_{B} I_{C}$ and have a common point $R$. Then these lines coincide. This means that point $L$ lies on the midline of triangle $I_{A} I_{B} I_{C}$ and, therefore, divides segment $A I_{A}$ in half. Further, applying the property of the external angle bisector to triangles $ABL$ and $ACL$, we get $AB / BL = AC / CL = AI_{A} / I_{A} L = 2$. Then $AB + AC = 2BC$ and, therefore, $BC = 4$.