B3. With a broad brush, the diagonals of a square tile are painted, as shown in the figure. Exactly half of the tile's surface is painted. The width of the paintbrush is 1, as indicated. Calculate the length of the side of the tile.

![](https://cdn.mathpix.com/cropped/2024_04_17_8dd363468b37c79bb3bag-2.jpg?height=284&width=330&top_left_y=823&top_left_x=1662)
B3. With a broad brush, the diagonals of a square tile are painted, see the figure. Exactly half of the tile's surface is painted. The width of the paintbrush is 1, as indicated. Calculate the length of the side of the tile. $2+2 \sqrt{2}$

Solution We can suffice by looking at a quarter of the tile: $\triangle A B C$. The area of $\triangle P Q R$ is half that of $\triangle A B C$. The triangles are similar, so corresponding sides

![](https://cdn.mathpix.com/cropped/2024_04_17_cb8708f0b7aaf218de84g-2.jpg?height=290&width=307&top_left_y=1449&top_left_x=1668)
relate as $1: \sqrt{2}$, so $|Q R|:|B C|=1: \sqrt{2}$. Now we calculate $|B Q|$ using Pythagoras in $\triangle B Q Q^{\prime}$: $2|B Q|^{2}=\left|B Q^{\prime}\right|^{2}+|B Q|^{2}=1^{2}$, so $|B Q|=\sqrt{\frac{1}{2}}=\frac{1}{2} \sqrt{2}$. With $x=|Q R|$ we then find $x+\sqrt{2}=\sqrt{2} \cdot x$, so $x(\sqrt{2}-1)=\sqrt{2}$ or $x=\frac{\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}(\sqrt{2}+1)}{2-1}=2+\sqrt{2}$. Then $|B C|=x+\sqrt{2}=2+2 \sqrt{2}$ (or $\left.|B C|=\sqrt{2} \cdot x=\sqrt{2} \cdot(2+\sqrt{2})=2 \sqrt{2}+2\right)$.