## Task 2 - 100932

In a square $ABCD$ with side length $a$, the midpoints of the sides $AB, BC, CD, DA$ are denoted by $E, F, G, H$.

In the broken line $A F D E C H B G A$, the intersection points that appear are denoted by $K, L, M, N, O, P, R$ such that AKELBMFNCOGPDQHR forms a (non-convex) hexadecagon, on whose sides no further intersection points of the above-mentioned broken line with itself lie (see figure).

Calculate the area of this hexadecagon!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0941.jpg?height=497&width=508&top_left_y=1242&top_left_x=1231)
The area of the hexadecagon is obtained as the difference between the area of the square and the areas of the eight (due to symmetry - reflection across the diagonals or the mid-parallel lines of the square, or combinations thereof, transform each pair of such congruent triangles into each other) triangles $A K L$, $E L B$, $B M F$, $F N C$, $C O G$, $G P D$, $D Q H$, and $H R A$.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0941.jpg?height=385&width=394&top_left_y=1989&top_left_x=840)

The area of triangle $A E D$ is $\frac{1}{2} \cdot a \cdot \frac{a}{2} = \frac{a^{2}}{4}$. This can be decomposed into triangle $A K E$ and triangle $D A K$. Due to the parallelism of the lines $A K = A F$ and $H Q = H C$, triangle $D A K$ is a dilation of triangle $D H Q$ with center $D$ and factor 2, since $|D A| = 2|D H|$.

Let $x$ be the area of triangle $A K E$. Then, triangle $D H Q$ also has an area of $x$, and triangle $D A K$ has an area of $2^{2} \cdot x = 4x$.

According to the preliminary consideration, the area of triangle $A E D$ is $4x + x = \frac{1}{4} a^{2}$, so each of the considered "small triangles" has an area of $x = \frac{1}{20} a^{2}$.

The area of the hexadecagon is therefore $a^{2} - 8 \cdot \frac{1}{20} a^{2} = \frac{3}{5} a^{2}$.