Consider the following problem (found here along with the solutions):
Let $ABC$ be a triangle with $\angle A=60^\circ$. Line $\ell$ intersects segments $AB$ and $AC$ and splits triangle $ABC$ into an equilateral triangle and quadrilateral. Let $X$ and $Y$ be on $\ell$ such that $BX$ and $CY$ are perpendicular to $\ell$. Given that $AB=20$ and $AC=22$ compute $XY$
Please post other solutions to this question. Here's what I did:
Note that the segments $BX$ and $CY$ are variable. So we can set $BX=0$, then $CY>0$ because $AC>AB$. The large(st) equilateral triangle formed has sides of $20$ since $AB$ is one of its sides. Let the intersection of $\ell$ with $AC$ be denoted as $C'$. Then, $BC'=AB$. Also, $CC'=2$. We now need to find $C'Y$. Since $\angle AC'B'=60^\circ$, $\angle CC'Y=60^\circ$, so $C'Y=1$. So the answer is $XY=BC'+C'Y=20+1=21$
Here is the original solution made by the competition organizers:
Let the intersection points of $ℓ$ with $AB$ and $AC$ be $B′$ and $C ′$. Note that $AB′ + AC′ = 2B′C'$, $BB′ = 2XB′$, and $CC′ = 2YC′$. Adding gives us $AB + AC = AB′ + AC′ + BB′ + CC′ = 2(B′C′ + XB′ + Y C′) = 2XY$. Thus, $XY =\frac{20+22}2 = 21$.