On the sides of a right angle with vertex $O$ lie the ends of a segment $A B$ of fixed length $a$. In what position of the segment will the area of triangle $AOB$ be the largest?
Let's display the segment $A B$: first with respect to the line $O A$, then with respect to the line $O B$, and then again with respect to the line $OA$. We will obtain four segments forming a rhombus with side $a$. Its area is maximal when it is a square. In this case,

$$
S=a^{2} \sin \alpha \leqslant a^{2}, \alpha=90^{\circ}
$$

The problem can be easily solved algebraically. The problem can be generalized to an arbitrary angle.

![](https://cdn.mathpix.com/cropped/2024_05_06_ae1b0d3bbbf9aa5c38e8g-18.jpg?height=334&width=692&top_left_y=2200&top_left_x=686)

## Answer

When the triangle $A O B$ is isosceles.

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