2016 AMC 10B Problem 19

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

A rectangle with lots of given lengths is an invitation to coordinates: every point has integer coordinates and every line is one equation. The finishing move is the observation that segments on a common line have lengths proportional to their $x$-projections, so no distance formula is needed and no square roots appear, which also explains why the answer is a plain fraction.

Synthetically, extend $EF$ to meet line $DA$ or $CB$ and use similar triangles to locate $P$ and $Q$ as fractions of $EF$: $EP/EF = 3/7$ and $EQ/EF = 6/13$, giving $PQ/EF = \frac{6}{13} - \frac{3}{7} = \frac{10}{91}$. The coordinate route is faster under time pressure.

• Computing $PQ$ and $EF$ as separate square-root lengths, or placing $E$ at distance $1$ from $A$ rather than from $B$.
• Mixing up which intersection is $P$ (on $AC$) and which is $Q$ (on $AG$); the ratio is unaffected, but a sign slip in $x_Q - x_P$ can cause a wrong subtraction.