Problem 1146

Solution:

Let $\rho$ be the $90^{\circ}$ rotation about the center of square $A B D E$, counterclockwise (orienting $\triangle A B C$ to have its vertices in counterclockwise order). Note that segments $C A$ and $Z E$ are congruent and perpendicular (thanks to square $C A H I$ and parallelogram $H A E Z$), so $\rho(C) = Z$. Likewise, segments $B C$ and $D X$ are congruent and perpendicular, implying $\rho(X) = C$. Now $\rho(X C) = C Z$ which implies $\angle Z C X = 90^{\circ}$. Likewise $\angle X A Y = \angle Y B Z = 90^{\circ}$. With three of the angles of the reentrant hexagon $X A Y B Z C$ known, the sum of the other three is readily computed:

$$\begin{aligned} & \angle A Y B + \angle B Z C + \angle C X A \\ & = \left(180^{\circ} - \angle B A Y - Y B A\right) + \left(180^{\circ} - \angle C B Z - \angle Z C B\right) + \left(180^{\circ} - \angle A C X - \angle X A C\right) \\ & = 540^{\circ} - (\angle B A Y + \angle X A C) - (\angle C B Z + \angle Y B A) - (\angle A C X + \angle Z C B) \\ & = 540^{\circ} - (\angle X A Y + \angle B A C) - (\angle Y B Z + \angle C B A) - (\angle Z C X + \angle A C B) \\ & = 540^{\circ} - 3 \cdot 90^{\circ} - (\angle B A C + \angle C B A + \angle A C B) \\ & = 540^{\circ} - 270^{\circ} - 180^{\circ} = 90^{\circ}. \end{aligned}$$