3. Let $A B C$ be a right-angled triangle at $C$ that is not isosceles with legs $b > a$.

(i) Find the side length of the square $A X Y Z$ that circumscribes the triangle $A B C$ (vertices $B$ and $C$ must be on different sides of the square).

(ii) Explain step by step how to construct the square $A X Y Z$ with a ruler and compass.
Solution. (i) Let $l$ be the side length of the square and $x$ the length of the segment $X C$. The right triangles $A X C$ and $B Y C$ are similar (since $\angle B C Y=\pi / 2-\angle A C X=\angle C A X$), so $l / b=(l-x) / a$, which means $x / l=(b-a) / b$. Then, applying the Pythagorean Theorem,

$$
b^{2}=l^{2}+x^{2}=l^{2}\left(1+\left(\frac{x}{l}\right)^{2}\right)=l^{2}\left(1+\frac{(b-a)^{2}}{b^{2}}\right)
$$

from which $l=b^{2} / \sqrt{(b-a)^{2}+b^{2}}$.

![](https://cdn.mathpix.com/cropped/2024_04_26_8d8943c65e3c25ed875fg-3.jpg?height=551&width=525&top_left_y=478&top_left_x=777)

(ii) To construct the square, we observe that, from (i), the tangent of the angle $\alpha=\angle C A X$ is $x / l=(b-a) / b$. We extend the side $B C$ of the triangle to a point $B^{\prime}$ such that $B B^{\prime}$ measures $b$ units. Thus, $C B^{\prime}$ measures $b-a$ and the angle $\angle C A B^{\prime}$ has a tangent of $(b-a) / b$. The vertex $X$ of the square we are looking for must then lie on the line containing $A$ and $B^{\prime}$. On the other hand, the angle $\angle A X C$ must be $90^{\circ}$, so $X$ must lie on the circle with diameter $A C$. Therefore, it suffices to draw this circle and its intersection with the line through $A$ and $B^{\prime}$ will be the point $X$ we are looking for. The points $Y$ and $Z$ that complete the square can now be easily obtained.