HMMT February, 2024

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $ABCD$ be a square, and let $\ell$ be a line passing through the midpoint of segment $\overline{AB}$ that intersects segment $\overline{BC}$ . Given that the distances from $A$ and $C$ to $\ell$ are $4$ and $7$ , respectively, compute the area of $ABCD$ .

Solution

Solution:

Consider the line $\ell'$ through $B$ parallel to $\ell$ , and drop perpendiculars from $A$ to $\ell'$ and $C$ to $\ell'$ . Note that because $\ell$ passes through the midpoint of segment $AB$ , the distance from $B$ to $\ell$ is $4$ . Thus, the distances from $A$ to $\ell'$ and from $C$ to $\ell'$ are $4+4=8$ and $4+7=11$ , respectively. Let $P$ be the foot from $A$ to $\ell'$ . Rotating the square $90^\circ$ from $B$ to $A$ sends the altitude from $C$ to $\ell'$ to the segment along $\ell'$ between $B$ and the foot from $A$ to $\ell'$ ; hence $BP=11$ . So the side length of the square is $\sqrt{AP^2+BP^2}=\sqrt{8^2+11^2}$ , which means the area of the square is $8^2+11^2=185$ .

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