## Task A-1.3.

Two congruent squares with side lengths of $1+\sqrt{2}$ have the same center, and their intersection is a regular octagon. What is the area of this octagon?
## Solution.

Let's mark the vertices of the square as in the figure and the center as $S$. Let $K$ and $L$ be the intersections of the sides $\overline{A^{\prime} D^{\prime}}$ and $\overline{A^{\prime} B^{\prime}}$ with the side $\overline{A B}$, respectively, and let $M$ be the intersection of the segments $\overline{S A^{\prime}}$ and $\overline{A B}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_21f39afeb1dfc245d841g-03.jpg?height=600&width=614&top_left_y=405&top_left_x=652)

Notice that the area $P$ of the octagon is equal to the area of the square $A B C D$ minus the areas of four congruent triangles congruent to triangle $A^{\prime} L K$.

That is, we have $P = P(A B C D) - 4 P(A^{\prime} L K)$.

1 point

The segment $\overline{S A^{\prime}}$ connects the center of the square $S$ with the vertex of the square $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, so its length is half the length of the diagonal of that square:

$$
\left|S A^{\prime}\right| = \frac{\sqrt{2}}{2}(1 + \sqrt{2}) = 1 + \frac{\sqrt{2}}{2}
$$

The segment $\overline{S M}$ connects the center of the square with the midpoint of the side of the square $A B C D$, so its length is half the length of the side of the square:

$$
|S M| = \frac{1}{2}(1 + \sqrt{2}) = \frac{1}{2} + \frac{\sqrt{2}}{2}
$$

Triangle $A^{\prime} L K$ is an isosceles right triangle. The length of the altitude to the hypotenuse is $\left|A^{\prime} M\right| = \left|S A^{\prime}\right| - |S M| = \frac{1}{2}$.

The length of the hypotenuse is $|K L| = 2 \left|A^{\prime} M\right| = 1$, so the area of triangle $A^{\prime} L K$ is

$$
P(A^{\prime} K L) = \frac{1}{2} \cdot \left|A^{\prime} M\right| \cdot |K L| = \frac{1}{4}
$$

Therefore, the area of the octagon is

$$
P = P(A B C D) - 4 P(A^{\prime} L K) = (1 + \sqrt{2})^2 - 4 \cdot \frac{1}{4} = 2 + 2 \sqrt{2}
$$

Note: The desired area can also be calculated as eight times the area of triangle $S K L$, using the same segment lengths as in the given solution.