3. A square $ABCD$ with side length $a$ is rotated $45^{\circ}$ around vertex $A$, so that vertex $B_{1}$ of the resulting square $A_{1} B_{1} C_{1} D_{1}$ is inside the given square. Let $M$ be the intersection of lines $C_{1} D_{1}$ and $BC$. Determine the area of the square with side $AM$.
Solution. Triangle $M B C_{1}$ is an isosceles right triangle (diagram on the right), so

$$
M B=B C_{1}=A C_{1}-A B=a(\sqrt{2}-1)
$$

From the right triangle $A B M$ we get

$$
\begin{aligned}
A M^{2} & =A B^{2}+M B^{2} \\
& =a^{2}+a^{2}(\sqrt{2}-1)^{2} \\
& =2 a^{2}(2-\sqrt{2})
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_375a647c88a2b0592badg-2.jpg?height=382&width=353&top_left_y=1270&top_left_x=1112)