4. The normal cross-section of a triangular prism $A_{1} B_{1} C_{1} A_{2} B_{2} C_{2}$ is an equilateral triangle $A B C$ with side $a$. Two arbitrary points $P$ and $Q$ are taken on the lateral edges $B_{1} B_{2}$ and $C_{1} C_{2}$. Let $\overline{B P}=x$ and $\overline{C Q}=y$.

a) Find the relationship between $x$ and $y$ for triangle $A P Q$ to be a right triangle with a right angle at vertex $P$.

b) Determine $x$ and $y$ such that triangle $A P Q$ is an isosceles right triangle with a right angle at vertex $P$.
Solution. a) From the right triangles $A B P$ and $A C Q$ (see the diagram), we get $\overline{A P}^{2}=x^{2}+y^{2}$ and $\overline{A Q}=a^{2}+y^{2}$. Let $M$ be a point on the line $C_{1} C_{2}$ such that $P M \| B C$; then the triangle $P M Q$ is a right triangle, so we have $\overline{P Q}^{2}=a^{2}+(y-x)^{2}, y>x$. If the triangle $A P Q$ is a right triangle with a right angle at vertex $P$, then:

$$
\overline{A P}^{2}+\overline{P Q}^{2}=\overline{A Q}^{2}
$$

i.e.

$$
2 x^{2}-2 x y+a^{2}=0
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_969e4efbfe816ce80b6fg-7.jpg?height=584&width=538&top_left_y=1534&top_left_x=942)

$A_{1}$

b) According to the condition of the problem, for the triangle $A P Q$ the following relations hold:

$$
\overline{A P}=\overline{P Q}_{\text {and }} \overline{A P}^{2}+\overline{P Q}^{2}=\overline{A Q}^{2}
$$

i.e.

$$
y^{2}-2 x y=0 \text { and } 2 x^{2}-2 x y+a^{2}=0 \text {. }
$$

Under the assumption $y \neq 0$, we get $x=\frac{a \sqrt{2}}{2}, y=a \sqrt{2}$.