6.80*. On the sides of a triangle, three squares are constructed externally. What should the angles of the triangle be so that the six vertices of these squares, distinct from the vertices of the triangle, lie on one circle?
6.80. Suppose that on the sides of triangle $A B C$, squares $A B B_{1} A_{1}, B C C_{2} B_{2}, A C C_{3} A_{3}$ are constructed externally, and the vertices $A_{1}, B_{1}, B_{2}, C_{2}, C_{3}, A_{3}$ lie on a single circle $S$. The perpendicular bisectors of segments $A_{1} B_{1}$, $B_{2} C_{2}$, $A_{3} C_{3}$ pass through the center of circle $S$. Clearly, the perpendicular bisectors of segments $A_{1} B_{1}$, $B_{2} C_{2}$, $A_{3} C_{3}$ coincide with the perpendicular bisectors of the sides of triangle $A B C$, so the center of circle $S$ coincides with the circumcenter of triangle $A B C$.

Let the circumcenter of triangle $A B C$ be denoted by $O$. The distance from point $O$ to the line $B_{2} C_{2}$ is $R \cos A + 2 R \sin A$, where $R$ is the radius of the circumcircle of triangle $A B C$. Therefore, $O B_{2}^{2} = (R \sin A)^{2} + (R \cos A + 2 R \sin A)^{2} = R^{2}(3 + 2(\sin 2 A - \cos 2 A)) = R^{2}(3 - 2 \sqrt{2} \cos (45^{\circ} + 2 A))$. Clearly, for the triangle to have the required property, it is necessary and sufficient that $O B_{2}^{2} = O C_{3}^{2} = O A_{1}^{2}$, i.e., $\cos (45^{\circ} + 2 \angle A) = \cos (45^{\circ} + 2 \angle B) = \cos (45^{\circ} + 2 \angle C)$. This equality holds when $\angle A = \angle B = \angle C = 60^{\circ}$. If $\angle A \neq \angle B$, then $(45^{\circ} + 2 \angle A) + (45^{\circ} + 2 \angle B) = 360^{\circ}$, i.e., $\angle A + \angle B = 135^{\circ}$. Then $\angle C = 45^{\circ}$ and $\angle A = \angle C = 45^{\circ}, \angle B = 90^{\circ}$ (or $\angle B = 45^{\circ}, \angle A = 90^{\circ}$). We see that the triangle must be either equilateral or isosceles right.