Problem 3. Let segments $A B$ and $B C$ be mutually perpendicular and $\overline{A B}=\overline{B C}$ $=a$. Let $K$ and $L$ be the midpoints of segments $A B$ and $B C$, and $K^{\prime}$ and $L^{\prime}$ be their axis-symmetric images with respect to the line $A C$ as the axis of symmetry, respectively. What is the area of the hexagon $A K L C L ' K '$.
Solution. Let the image of $\triangle A B C$ under the reflection about the line $A C$ be the triangle $\triangle A C D$. From $\overline{A B}=\overline{B C}=a$ and $A B \perp B C$ it follows that $\triangle A B C$ is an isosceles right triangle, and therefore triangle $\triangle A C D$ is also an isosceles right triangle. Consequently, quadrilateral $A B C D$ is a square.

Since $K$ and $L$ are the midpoints of segments $A B$ and

![](https://cdn.mathpix.com/cropped/2024_06_05_2e9ecab3ff55f9ab2430g-08.jpg?height=365&width=372&top_left_y=1822&top_left_x=1112)
$B C$, it follows that their reflections $K^{\prime}$ and $L^{\prime}$ are the midpoints of segments $A D$ and $C D$, respectively. Clearly, triangles $\triangle K B L$ and $\triangle K^{\prime} L^{\prime} D$ are isosceles and right. From the above, it follows that

$$
P_{\triangle K B L}=\frac{\frac{a}{2} \cdot \frac{a}{2}}{2}=\frac{a^{2}}{8}, P_{\triangle K ' L^{\prime} D}=\frac{\frac{a}{2} \cdot \frac{a}{2}}{2}=\frac{a^{2}}{8}, P_{\square A B C D}=a^{2}
$$

therefore

$$
P_{A K L C L^{\prime} K^{\prime}}=P_{\square A B C D}-P_{\triangle K B L}-P_{\triangle K K^{\prime} L^{\prime} D}=a^{2}-\frac{a^{2}}{8}-\frac{a^{2}}{8}=\frac{3}{4} a^{2}
$$