A square with side $a$ is divided into two equal right-angled isosceles triangles by one of its diagonals. What is the perimeter and area of the quadrilateral whose vertices are: the endpoints of the diagonal and the centers of the circles inscribed in the two right-angled isosceles triangles? (Without trigonometric tables).
The figure shows the labeling. The quadrilateral $A E C F$ is a rhombus because its diagonals are perpendicular to each other and the quadrilateral is symmetric about the intersection of the diagonals.

![](https://cdn.mathpix.com/cropped/2024_05_02_e784abca3ac2fb89aa2cg-1.jpg?height=458&width=418&top_left_y=232&top_left_x=840)

![](https://cdn.mathpix.com/cropped/2024_05_02_e784abca3ac2fb89aa2cg-1.jpg?height=49&width=1171&top_left_y=713&top_left_x=214)

$$
\begin{gathered}
K E=E L=L B=\varrho \\
A L=A K=\frac{a \sqrt{2}}{2}
\end{gathered}
$$

and thus

$$
\varrho=L B=A B-A L=a-\frac{a \sqrt{2}}{2}=a\left(1-\frac{\sqrt{2}}{2}\right)
$$

therefore

$$
\varrho^{2}=a^{2}\left(1-\sqrt{2}+\frac{1}{2}\right)=a^{2}\left(\frac{3}{2}-\sqrt{2}\right)
$$

The side of the rhombus, according to the Pythagorean theorem, is

$$
A E=\sqrt{A L^{2}+L E^{2}}=\sqrt{\left(\frac{a \sqrt{2}}{2}\right)^{2}+\varrho^{2}}=\sqrt{\frac{a^{2}}{2}+\frac{3 a^{2}}{2}-a^{2} \sqrt{2}}=a \sqrt{2-\sqrt{2}}
$$

Thus, the perimeter we are looking for is

$$
k=4 \cdot A E=4 a \sqrt{2-\sqrt{2}}
$$

and the area we are looking for is

$$
t=A C \cdot K E=a \sqrt{2} \cdot \varrho=a \sqrt{2} \cdot a\left(1-\frac{\sqrt{2}}{2}\right)=a^{2}(\sqrt{2}-1)
$$

Kornélia Jójárt (Esztergom, Dobó Katalin Secondary School, 1st grade)