3. Given a circle with radius 1 and a square, such that the circle touches one side of the square and moreover two vertices of the square lie on the circle.

How long are the sides of the square?
![](https://cdn.mathpix.com/cropped/2024_04_17_37097645698d56aa28c8g-3.jpg?height=804&width=514&top_left_y=1418&top_left_x=1431)
B3. In the picture next to it, we have given names to the different distances. So $a$ is the distance from the center of the circle to the top side of the square, and $b$ is half the length of the side of the square. All line segments with length 1 are the radius of the circle.

Using the Pythagorean theorem, we find that $a^{2}+b^{2}=1$. Furthermore, $a+1$ is exactly the length of the side of the square, so $a+1=2 b$, or $a=2 b-1$. If we substitute this into the first equation, we get

![](https://cdn.mathpix.com/cropped/2024_04_17_b71e27e72afbe21d81ceg-5.jpg?height=306&width=308&top_left_y=201&top_left_x=1528)
$(2 b-1)^{2}+b^{2}=1$. We can simplify this to $5 b^{2}-4 b=0$. Since we know that $b \neq 0$, we can divide the equation by $b$ and get $5 b-4=0$, or $b=\frac{4}{5}$. The length of the sides of the square is therefore $2 \cdot \frac{4}{5}=\frac{8}{5}$.