[ Rectangles and squares. Properties and characteristics ] Area of a triangle.

On the sides of a right triangle with legs $a$ and $b$, squares are constructed lying outside the triangle. Find the area of the triangle with vertices at the centers of the squares.
Let $O_1, O_2$ and $O_3$ be the centers of the squares constructed on the legs $BC = a$, $AC = b$, and the hypotenuse $AB$ of the right triangle $ABC$. Since $CO_1$ and $CO_2$ are the bisectors of the angles of the squares, the points $O_1, C$, and $O_2$ lie on the same line. From the vertex $K$ of the square $ABKL$ constructed on the hypotenuse $AB$, drop the perpendicular $KM$ to the extension of the leg $BC$, and from the vertex $L$ drop the perpendicular $LQ$ to the extension of the leg $AC$. Let this perpendicular intersect the line $KM$ at point $P$. Then $CMPQ$ is a square with side $a + b$, and its center coincides with the center $O_3$ of the square $ABKL$ (if the vertices of one parallelogram are located one on each side of another parallelogram, then the centers of the parallelograms coincide). The segment $CO_3$ is half the diagonal of the square $CMPQ$, so $CO_3 = \frac{a + b}{\sqrt{2}}$ and

$$
\angle O_1CO_3 = \angle O_1CB + \angle O_3CB = 45^\circ + 45^\circ = 90^\circ
$$

Thus, $O_3C$ is the altitude of the triangle $O_1O_2O_3$. Therefore,

$$
S_{\triangle O_1O_2O_3} = \frac{1}{2} O_1O_2 \cdot CO_3 = \frac{1}{2} \left( \frac{a}{\sqrt{2}} + \frac{b}{\sqrt{2}} \right) \cdot \frac{a + b}{\sqrt{2}} = \frac{(a + b)^2}{4}
$$

## Answer

$\frac{(a + b)^2}{4}$