Library / /767 of 1394

, 2015

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $ABCD$ be a quadrilateral with $\angle BAD = \angle ABC = 90^{\circ}$ , and suppose $AB = BC = 1$ , $AD = 2$ . The circumcircle of $ABC$ meets $\overline{AD}$ and $\overline{BD}$ at points $E$ and $F$ , respectively. If lines $AF$ and $CD$ meet at $K$ , compute $EK$ .

Solution

Solution:

Answer: $\frac{\sqrt{2}}{2}$

Assign coordinates such that $B$ is the origin, $A$ is $(0,1)$ , and $C$ is $(1,0)$ . Clearly, $E$ is the point $(1,1)$ . Since the circumcenter of $ABC$ is $\left(\frac{1}{2}, \frac{1}{2}\right)$ , the equation of the circumcircle of $ABC$ is $\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$ . Since line $BD$ is given by $x=2y$ , we find that $F$ is at $\left(\frac{6}{5}, \frac{3}{5}\right)$ . The intersection of $AF$ with $CD$ is therefore at $\left(\frac{3}{2}, \frac{1}{2}\right)$ , so $K$ is the midpoint of $CD$ . As a result, $EK=\frac{\sqrt{2}}{2}$ .

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