2002 AMC 10A Problem 20

Sketch $A$ through $F$ on a horizontal line one unit apart, with $G$ above, and draw $GD$ and $GF$.

Segment $HC$ is parallel to $AG$ and joins a point of $GD$ to a point of $AD$, so it cuts triangle $DAG$ into a smaller similar triangle $DCH$ with the ratio of similarity
$$\frac{HC}{AG} = \frac{DC}{DA} = \frac{1}{3}.$$

Likewise $JE \parallel AG$ cuts triangle $FAG$ into the similar triangle $FEJ$, with
$$\frac{JE}{AG} = \frac{FE}{FA} = \frac{1}{5}.$$

Dividing the two relations eliminates $AG$:
$$\frac{HC}{JE} = \frac{1/3}{1/5} = \frac{5}{3}.$$

The answer is $\boxed{\textbf{(D)}\ 5/3}$.

A segment drawn parallel to one side of a triangle creates a similar triangle sharing the opposite vertex, and the scale factor is read off along the base. Both unknown segments are compared to the same reference $AG$, so their ratio is a ratio of scale factors. Note the distinct similarity centers, $D$ for $HC$ and $F$ for $JE$; each ratio is measured from its own center.

Assign a convenient length, say $AG = 15$. Then $HC = 15 \cdot \frac{1}{3} = 5$ and $JE = 15 \cdot \frac{1}{5} = 3$, so the ratio is $\frac{5}{3}$. Coordinates also work: with $A = (0,0)$, $G = (0, 1)$, the line $GD$ is $y = 1 - x/3$, giving $HC = 1/3$ at $x = 2$, and $GF$ is $y = 1 - x/5$, giving $JE = 1/5$ at $x = 4$.

• Using the wrong vertex as the similarity center (measuring from A instead of from D and F), which gives ratios like 2/3 or 4/5.
• Inverting the final division and answering $3/5$, which is not offered, then guessing among the choices.