Problem 3. Given segments a and $b(a>b)$. Construct a segment of length $\frac{a^{2}+b^{2}}{a-b}$ using a compass and a straightedge.
Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_f7a16de7ae2a30fbf517g-2.jpg?height=571&width=1055&top_left_y=1368&top_left_x=635)

Construct an equilateral triangle $\mathrm{ABC}$ with sides $\mathrm{AB}=\mathrm{AC}=\mathrm{BC}=a$. On the line $\mathrm{AB}$, mark segments $\mathrm{BD}=\mathrm{DE}=b$. Construct a line $\mathrm{EF}$ parallel to $\mathrm{DC}$, where $\mathrm{F}$ is the intersection of $\mathrm{AC}$ and $\mathrm{EF}$. On the line AF in the direction of point A, mark the segment $\mathrm{FG}=b$. The segment $\mathrm{AG}$ is the desired one. Proof.

1). Transform the given expression:

$$
\frac{a^{2}+b^{2}}{a-b}=\frac{a^{2}+b^{2}-2 a b+2 a b}{a-b}=\frac{(a-b)^{2}}{a-b}+\frac{2 b}{a-b}=a-b+\frac{2 b}{a-b}
$$

2). $\triangle A D C \sim \triangle A E F \Rightarrow \frac{A E}{A D}=\frac{A F}{A C} \Rightarrow \frac{A D+D E}{A D}=\frac{A C+C F}{A C} \Rightarrow \frac{D E}{A D}=\frac{C F}{A C} \Rightarrow C F=\frac{2 a b}{a-b}$.

$$
A G=A F-G F=a-b+\frac{2 a b}{a-b}
$$