A1. The point $S\left(\frac{5-\sqrt{5}}{2}, \frac{5-3 \sqrt{5}}{2}\right)$ is the midpoint of the line segment $A B$. One endpoint of the line segment is the point $A(3-2 \sqrt{5}, 4-\sqrt{5})$. Which point is the other endpoint of this line segment?
(A) $B(-2-\sqrt{5}, 1+2 \sqrt{5})$
(B) $B(1-7 \sqrt{5}, 3+\sqrt{5})$
(C) $B(2+\sqrt{5}, 1-2 \sqrt{5})$
(D) $B\left(\frac{\sqrt{5}}{2}, \frac{-\sqrt{5}}{2}\right)$
(E) $B(1-2 \sqrt{5}, 2+\sqrt{5})$
A1. We use the formula for calculating the center or midpoint of a segment $S\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$. We write the relationship $\frac{3-2 \sqrt{5}+x_{1}}{2}=\frac{5-\sqrt{5}}{2}$ and $\frac{4-\sqrt{5}+y_{1}}{2}=\frac{3-3 \sqrt{5}}{2}$. Solve the resulting equations and we get the solutions $x_{1}=2+\sqrt{5}$ and $y_{1}=1-2 \sqrt{5}$.