Track / Stage 4 / 269 of 340 #529 of 1964

Problem 529

AMC 12 late, AIME early

Geometry Difficulty 4.9 Find the answer

7. Given that the midpoint of line segment $A B$ is $C$ , a circle is drawn with point $A$ as the center and the length of $A B$ as the radius. On the extension of line segment $A B$ , take point $D$ such that $B D = A C$ ; then, with point $D$ as the center and the length of $D A$ as the radius, draw another circle, which intersects $\odot A$ at points $F$ and $G$ . Connect $F G$ to intersect $A B$ at point $H$ . Then the value of $\frac{A H}{A B}$ is $\qquad$

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

7. $\frac{1}{3}$ .

As shown in Figure 7, extend $A D$ to intersect $\odot D$ at point $E$ , and connect $A F, E F$ .
From the given conditions, we have
$\begin{array}{l} A C=\frac{1}{3} A D, \\ A B=\frac{1}{3} A E . \end{array}$

In $\triangle F H A$ and $\triangle E F A$ ,
$\begin{array}{l} \angle E F A=\angle F H A=90^{\circ}, \\ \angle F A H=\angle E A F, \end{array}$

thus, Rt $\triangle F H A \backsim \mathrm{Rt} \triangle E F A \Rightarrow \frac{A H}{A F}=\frac{A F}{A E}$ . Since $A F=A B$ , we have $\frac{A H}{A B}=\frac{1}{3}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.