HMMT February, 2022

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

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 a quadrilateral. Let $X$ and $Y$ be on $\ell$ such that lines $BX$ and $CY$ are perpendicular to $\ell$ . Given that $AB = 20$ and $AC = 22$ , compute $XY$ .

Solution

Solution:

Let the intersection points of $\ell$ with $AB$ and $AC$ be $B'$ and $C'$ . Note that $AB' + AC' = 2B'C'$ , $BB' = 2XB'$ , and $CC' = 2YC'$ . Adding gives us
$AB + AC = AB' + AC' + BB' + CC' = 2\left(B'C' + XB' + YC'\right) = 2XY$
Thus, $XY = \frac{20 + 22}{2} = 21$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.