2016 AMC 10B Problem 19

A rectangle with many given lengths is a good place for coordinates.

1. Put $D = (0,0)$, $C = (5,0)$, $B = (5,4)$ and $A = (0,4)$.
2. $E$ is $1$ left of $B$: $E = (4, 4)$. $G$ is $1$ above $C$: $G = (5, 1)$. $F$ is $2$ right of $D$: $F = (2, 0)$.
3. Line $EF$. From $F$ to $E$, $x$ goes up $2$ and $y$ goes up $4$, so the slope is $2$. Its equation is $y = 2x - 4$ (check: $x = 2$ gives $0$, $x = 4$ gives $4$).
4. Line $AC$. It starts at height $4$ and drops $4$ over $5$ units, so $y = 4 - \frac45 x$. Where it meets $EF$:
$$2x - 4 = 4 - \tfrac45 x \;\Longrightarrow\; \tfrac{14}{5} x = 8 \;\Longrightarrow\; x = \tfrac{20}{7}.$$
This is point $P$.
5. Line $AG$. It starts at height $4$ and drops $3$ over $5$ units, so $y = 4 - \frac35 x$. Where it meets $EF$:
$$2x - 4 = 4 - \tfrac35 x \;\Longrightarrow\; \tfrac{13}{5} x = 8 \;\Longrightarrow\; x = \tfrac{40}{13}.$$
This is point $Q$.
6. Compare lengths without square roots. $P$, $Q$, $E$, $F$ all lie on the same straight line. Along one line, lengths are in the same ratio as their horizontal widths. The width of $PQ$ is
$$\tfrac{40}{13} - \tfrac{20}{7} = \tfrac{280 - 260}{91} = \tfrac{20}{91},$$
and the width of $EF$ is $4 - 2 = 2$. So
$$\frac{PQ}{EF} = \frac{20/91}{2} = \frac{10}{91}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{10}{91}}$.

With a rectangle and lots of whole-number lengths, coordinates turn every point into a pair of numbers and every line into a simple equation. For points on one line, you can compare lengths using only their $x$-coordinates, which avoids square roots completely.

Measure everything as a fraction of $EF$, starting from $F$ (where $x = 2$) and ending at $E$ (where $x = 4$). Point $P$ is at $x = \frac{20}{7}$, which is $\frac{20}{7} - 2 = \frac67$ along a width of $2$, so $FP = \frac37 EF$. Point $Q$ is at $x = \frac{40}{13}$, which is $\frac{40}{13} - 2 = \frac{14}{13}$ along, so $FQ = \frac{7}{13} EF$. Then
$$\frac{PQ}{EF} = \frac{7}{13} - \frac37 = \frac{49 - 39}{91} = \frac{10}{91}.$$
The same fractions also come out of similar triangles without coordinates.

Place $D = (0,0)$, $C = (5,0)$, $B = (5,4)$, $A = (0,4)$. Then $E = (4,4)$ (one unit left of $B$), $G = (5,1)$ (one unit above $C$), and $F = (2,0)$ (two units right of $D$).

Line $EF$ through $(2,0)$ and $(4,4)$ has slope $2$: $y = 2x - 4$.

Line $AC$ through $(0,4)$ and $(5,0)$: $y = 4 - \tfrac45 x$. Intersect with $EF$:
$$2x - 4 = 4 - \tfrac45 x \;\Longrightarrow\; \tfrac{14}{5}x = 8 \;\Longrightarrow\; x_P = \tfrac{20}{7}.$$

Line $AG$ through $(0,4)$ and $(5,1)$: $y = 4 - \tfrac35 x$. Intersect with $EF$:
$$2x - 4 = 4 - \tfrac35 x \;\Longrightarrow\; \tfrac{13}{5}x = 8 \;\Longrightarrow\; x_Q = \tfrac{40}{13}.$$

Since $P$, $Q$, $E$, $F$ all lie on the same line, the ratio of lengths equals the ratio of horizontal changes:
$$\frac{PQ}{EF} = \frac{x_Q - x_P}{x_E - x_F} = \frac{\frac{40}{13} - \frac{20}{7}}{4 - 2} = \frac{\frac{280 - 260}{91}}{2} = \frac{10}{91}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{10}{91}}$.