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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Let $ABCD$ be a unit square (that is, the labels $A$ , $B$ , $C$ , $D$ appear in that order around the square). Let $X$ be a point outside of the square such that the distance from $X$ to $AC$ is equal to the distance from $X$ to $BD$ , and also that $AX = \frac{\sqrt{2}}{2}$ . Determine the value of $CX^{2}$ .

Solution

Solution:

Since $X$ is equidistant from $AC$ and $BD$ , it must lie on either the perpendicular bisector of $AB$ or the perpendicular bisector of $AD$ . It turns that the two cases yield the same answer, so we will just assume the first case. Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $CD$ . Then, $XM$ is perpendicular to $AB$ , so $XM = \frac{1}{2}$ and thus $XN = \frac{3}{2}$ , $NC = \frac{1}{2}$ . By the Pythagorean Theorem we find $XC = \frac{\sqrt{10}}{2}$ and the answer follows.

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