2016 AMC 10B Problem 19

P16-20~6 minPrint

Rectangle $ABCD$ has $AB=5$ and $BC=4$ . Point $E$ lies on $\overline{AB}$ so that $EB=1$ , point $G$ lies on $\overline{BC}$ so that $CG=1$ . and point $F$ lies on $\overline{CD}$ so that $DF=2$ . Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$ , respectively. What is the value of $\frac{PQ}{EF}$ ?

Show solution without answering

Key insight

Put D at the origin. P and Q both lie on line EF, so PQ/EF is just the ratio of the horizontal distances: (40/13 - 20/7) divided by 2.

Solution

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

Why this works

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.

Alternative approach

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 3: Congruent Triangles
Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 5: Similar Triangles
Introduction to Algebra, Chapter 8: Graphing Lines