2001 AMC 12 Problems/Problem 22

Problem

In rectangle $ABCD$ , points $F$ and $G$ lie on $AB$ so that $AF=FG=GB$ and $E$ is the midpoint of $\overline{DC}$ . Also, $\overline{AC}$ intersects $\overline{EF}$ at $H$ and $\overline{EG}$ at $J$ . The area of the rectangle $ABCD$ is $70$ . Find the area of triangle $EHJ$ .

Let $[X]$ denote the area of the polygon $X$ .

Note that the triangles $AFH$ and $CEH$ are similar, as they have the same angles. Hence $\frac {AH}{HC} = \frac{AF}{EC} = \frac 23$ .

We can now compute $[EHJ]$ as $[ACD]-[AHD]-[DEH]-[EJC]$ . We have:

$[ACD]=\frac{[ABCD]}2 = 35$ .
$[AHD]$ is $2/5$ of $[ACD]$ , as these two triangles have the same base $AD$ , and $AH$ is $2/5$ of $AC$ , therefore also the height from $H$ onto $AD$ is $2/5$ of the height from $C$ . Hence $[AHD]=14$ .
$[HED]$ is $3/10$ of $[ACD]$ , as the base $ED$ is $1/2$ of the base $CD$ , and the height from $H$ is $3/5$ of the height from $A$ . Hence $[HED]=\frac {21}2$ .
$[JEC]$ is $3/14$ of $[ACD]$ for similar reasons, hence $[JEC]=\frac{15}2$ .

As in the previous solution, we note the similar triangles and prove that $H$ is in $2/5$ and $J$ in $4/7$ of $AC$ .