Problem 1. A regular right prism $A B C A^{\prime} B^{\prime} C^{\prime}$, with $A B=a$, has the property that there exists a unique point $M \in\left(B B^{\prime}\right)$ such that $m\left(\angle A M C^{\prime}\right)=90^{\circ}$.

Determine the measure of the angle formed by the line $A M$ with the plane $\left(A C C^{\prime}\right)$.
Solution. We will first show that $M$ is the midpoint of the edge $\left[B B^{\prime}\right]$. Assuming the contrary, let $M^{\prime} \in \left(B B^{\prime}\right)$ be the symmetric point of $M$ with respect to the midpoint of the edge $\left[B B^{\prime}\right]$; then $M^{\prime} \neq M$. Since $\triangle M A B \equiv \Delta M^{\prime} C^{\prime} B^{\prime}$ and $\Delta M^{\prime} A B \equiv \Delta M C^{\prime} B^{\prime}$, it follows that $[M A] \equiv \left[M^{\prime} C^{\prime}\right]$ and $\left[M^{\prime} A\right] \equiv \left[M C^{\prime}\right]$. Therefore, $\triangle M A C^{\prime} \equiv \Delta M^{\prime} C^{\prime} A$, so $m\left(\varangle A M C^{\prime}\right) = m\left(\varangle A M^{\prime} C^{\prime}\right) = 90^{\circ}$, which contradicts the uniqueness of the choice of $M$.

3 points

Let $B B^{\prime} = h$, and calculate $A M$, $M C^{\prime}$, and $A C^{\prime}$, then apply the Pythagorean theorem in the triangle

![](https://cdn.mathpix.com/cropped/2024_06_07_537b9f14cff679319578g-3.jpg?height=60&width=1797&top_left_y=1263&top_left_x=175)

If $O$ is the center of the face $\left(A C C^{\prime} A^{\prime}\right)$, then $M O \perp \left(A C C^{\prime}\right)$, so the angle formed by the line $A M$ with the plane $\left(A C C^{\prime}\right)$ is the angle $M A O$.

Since $M O = O A = \frac{a \sqrt{3}}{2}$, triangle $M O A$ is a right isosceles triangle, so $m(\varangle M A O) = 45^{\circ}$. 2 points