The hypotenuse of a right triangle serves as a side of a square located outside the triangle. Find the distance between the vertex of the right angle of the triangle and the center of the square, if the sum of the legs of the triangle is $d$.

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Construct on the other sides of the square, using them as hypotenuses, three right triangles equal to the given one.

## Solution

Consider a square with side $d$ and place four triangles equal to the given one at its corners so that their hypotenuses form a square (see figure). The required segment is equal to half the diagonal of the larger square, that is, $\frac{d \sqrt{2}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-07.jpg?height=389&width=372&top_left_y=2212&top_left_x=846)

## Answer

$\frac{d \sqrt{2}}{2}$.

In an isosceles triangle, the height drawn to the base is bisected by the point of intersection of the altitudes. Find the angles of this triangle.

## Solution

Let $B P$ be the height of the isosceles triangle $A B C$ drawn to the base $A C$, and $H$ be the point of intersection of the altitudes of the triangle. Denote

$\angle A B P=\angle C B P=\alpha, A C=2 a$. Then $\angle H C P=\alpha, B P=a \operatorname{ctg} \alpha, H P=a \operatorname{tg} \alpha$.

By the condition $B P=2 H P$, or $a \operatorname{tg} \alpha=2 a \operatorname{tg} \alpha$, from which $\operatorname{tg} \alpha=\frac{1}{\sqrt{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-08.jpg?height=355&width=546&top_left_y=580&top_left_x=756)

## Answer

$2 \operatorname{arctg} \frac{1}{\sqrt{2}}, 90^{\circ}-\operatorname{arctg} \frac{1}{\sqrt{2}}, 90^{\circ}-\operatorname{arctg} \frac{1}{\sqrt{2}}$.