1. Let $A B C$ be a triangle with $\angle A=60^{\circ}$. Line $\ell$ intersects segments $A B$ and $A C$ and splits triangle $A B C$ into an equilateral triangle and a quadrilateral. Let $X$ and $Y$ be on $\ell$ such that lines $B X$ and $C Y$ are perpendicular to $\ell$. Given that $A B=20$ and $A C=22$, compute $X Y$.
Answer:
21
Solution: Let the intersection points of $\ell$ with $A B$ and $A C$ be $B^{\prime}$ and $C^{\prime}$. Note that $A B^{\prime}+A C^{\prime}=$ $2 B^{\prime} C^{\prime}, B B^{\prime}=2 X B^{\prime}$, and $C C^{\prime}=2 Y C^{\prime}$. Adding gives us
$$
A B+A C=A B^{\prime}+A C^{\prime}+B B^{\prime}+C C^{\prime}=2\left(B^{\prime} C^{\prime}+X B^{\prime}+Y C^{\prime}\right)=2 X Y \text {. }
$$

Thus, $X Y=\frac{20+22}{2}=21$.