2002 AMC 10A Problem 20

Draw it: $A, B, C, D, E, F$ on a horizontal line, one unit apart, and $G$ somewhere above. Draw $GA$, $GD$ and $GF$. Both unknown segments are parallel to $AG$, so compare each of them to $AG$.

1. Segment $HC$. It goes from $C$ (on $AD$) up to $H$ (on $GD$), parallel to $AG$. So it cuts the big triangle $DAG$ into a small triangle $DCH$ at the corner $D$.

1. Why the small triangle is a shrunken copy. Triangles $DCH$ and $DAG$ share the angle at $D$, and because $CH$ is parallel to $AG$, the angles at $C$ and $A$ are equal. Same angles means same shape, so every side of $DCH$ is the same fraction of the matching side of $DAG$.

3. Which fraction? Measure along the base from $D$: $DC = 1$ and $DA = 3$. So
$$HC = \frac{1}{3} \times AG.$$

4. Segment $JE$. In the same way, $JE$ is parallel to $AG$ and cuts triangle $FAG$ into the small triangle $FEJ$ at the corner $F$. Along the base from $F$: $FE = 1$ and $FA = 5$. So
$$JE = \frac{1}{5} \times AG.$$

5. Compare. To make it concrete, say $AG = 15$. Then $HC = 5$ and $JE = 3$, so
$$\frac{HC}{JE} = \frac{5}{3}.$$
Any other length for $AG$ gives the same ratio, because both segments are fractions of the same $AG$.

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

A segment drawn parallel to one side of a triangle cuts off a smaller copy of that triangle, and the shrink factor can be read off along the side you know. Here both unknown segments are compared with the same segment $AG$, so their ratio is just the ratio of the two shrink factors. Each factor must be measured from its own corner: $D$ for $HC$ and $F$ for $JE$.

Use coordinates in a special case. The ratio cannot depend on where $G$ is, so put $A$ at $(0, 0)$, $F$ at $(5, 0)$ and $G$ at $(0, 1)$, which makes $AG$ vertical. Then $HC$ and $JE$ are vertical too. The line from $G$ down to $D = (3, 0)$ has height $1 - \frac{x}{3}$ at position $x$, so at $C$ ($x = 2$) the height is $HC = \frac13$. The line from $G$ to $F = (5, 0)$ has height $1 - \frac{x}{5}$, so at $E$ ($x = 4$) the height is $JE = \frac15$. The ratio is $\frac{1/3}{1/5} = \frac{5}{3}$.

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