Problem 2. (3 points) A segment of length $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is given. Is it possible to construct a segment of length 1 using only a straightedge and compass?
Solution: We will prove that it is possible to construct a segment of length 1. From a segment of length $a$, we can construct $A B=a(\sqrt{2}+\sqrt{3}+\sqrt{5})$ (by constructing a segment of length $a \sqrt{2}$ as the diagonal of a square with side $a$, then $a \sqrt{3}$ as the diagonal of a rectangle $a \times a \sqrt{2}$, and then $a \sqrt{5}$ as the diagonal of a rectangle $a \sqrt{2} \times a \sqrt{3}$).

Let $A C=\sqrt{2}+\sqrt{3}+\sqrt{5}$, and $C$ does not lie on $A B$. On the segment $A B$, we choose a point $D$ such that $A D=a$. We construct a line $D E \| B C$ ($E$ lies on $A C$). From the similarity of triangles $A B C$ and $A D E$, we have that $\frac{A E}{A D}=\frac{A C}{A B}$, i.e., $A E=\frac{A C . A D}{A B}=1$.