18. Collinear points $A, B$, and $C$ are given in the Cartesian plane such that $A=(a, 0)$ lies along the $x$-axis, $B$ lies along the line $y=x, C$ lies along the line $y=2 x$, and $A B / B C=2$. If $D=(a, a)$, the circumcircle of triangle $A D C$ intersects $y=x$ again at $E$, and ray $A E$ intersects $y=2 x$ at $F$, evaluate $A E / E F$.
Answer: $\square$
Let points $O, P$, and $Q$ be located at $(0,0),(a, 2 a)$, and $(0,2 a)$, respectively. Note that $B C / A B=1 / 2$ implies $[O C D] /[O A D]=1 / 2$, so since $[O P D]=[O A D],[O C D] /[O P D]=1 / 2$. It follows that $[O C D]=[O P D]$. Hence $O C=C P$. We may conclude that triangles $O C Q$ and $P C A$ are congruent, so $C=(a / 2, a)$.
It follows that $\angle A D C$ is right, so the circumcircle of triangle $A D C$ is the midpoint of $A C$, which is located at $(3 a / 4, a / 2)$. Let $(3 a / 4, a / 2)=H$, and let $E=(b, b)$. Then the power of the point $O$ with respect to the circumcircle of $A D C$ is $O D \cdot O E=2 a b$, but it may also be computed as $O H^{2}-H A^{2}=13 a / 16-5 a / 16=a / 2$. It follows that $b=a / 4$, so $E=(a / 4, a / 4)$.
We may conclude that line $A E$ is $x+3 y=a$, which intersects $y=2 x$ at an $x$-coordinate of $a / 7$. Therefore, $A E / E F=(a-a / 4) /(a / 4-a / 7)=(3 a / 4) /(3 a / 28)=7$.
Remark: The problem may be solved more quickly if one notes from the beginning that lines $O A$, $O D, O P$, and $O Q$ form a harmonic pencil because $D$ is the midpoint of $A P$ and lines $O Q$ and $A P$ are parallel.