AMC 10 · 2016 · #19

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area

coordinate-geometryslope-interceptpythagorean-theorem convert-to-algebra ↑ Prerequisites: coordinate-geometry

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A 5-by-4 rectangle ABCD has three marked points: E on AB with EB=1, G on BC with CG=1, and F on CD with DF=2. The diagonal AC and the segment AG both cross the segment EF, at points P and Q. Find what fraction of EF the piece PQ takes up.

Pick an answer.

(A)

$~\frac{\sqrt{13}}{16}$

(B)

$~\frac{\sqrt{2}}{13}$

(C)

$~\frac{9}{82}$

(D)

$~\frac{10}{91}$

(E)

$~\frac19$

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The figure is full of named points and crossing segments, so the cleanest move is to drop the whole rectangle onto a coordinate grid. Once every point has numbers, each line becomes an equation, and finding where two segments cross becomes solving two equations together. The big question PQ/EF then splits into small subproblems: locate P, locate Q, then measure.

1STEP 1

Put the rectangle on a grid

Put D at the origin, so A=(0,4) and C=(5,0); the three marked points become E=(4,4), F=(2,0), G=(5,1).

A=(0,4), B=(5,4), C=(5,0), D=(0,0), E=(4,4), G=(5,1), F=(2,0)

2STEP 2

Write the line through E and F

From F=(2,0) to E=(4,4) the slope is 2, so line EF is y=2x-4; every point of EF, P and Q included, satisfies it.

slope_EF=(4-0)/(4-2)=2, y=2x-4

3STEP 3

Find P where AC meets EF

Line AC is y=4-( $\frac{4}{5}$ )x; set it equal to y=2x-4 to get x= $\frac{20}{7}$ , so P=( $\frac{20}{7}$ , $\frac{12}{7}$ ).

\frac{4}{5}

x=2x-4 →

\frac{14}{5}

x=8 → x=

\frac{20}{7}

, P=(

\frac{20}{7}

\frac{12}{7}

)

4STEP 4

Find Q where AG meets EF

Line AG is y=4-( $\frac{3}{5}$ )x; set it equal to y=2x-4 to get x= $\frac{40}{13}$ , so Q=( $\frac{40}{13}$ , $\frac{28}{13}$ ).

\frac{3}{5}

x=2x-4 →

\frac{13}{5}

x=8 → x=

\frac{40}{13}

, Q=(

\frac{40}{13}

\frac{28}{13}

)

5STEP 5

Measure PQ and EF, then divide

PQ=( $\frac{20}{91}$ )√5 and EF=2√5 lie on one line, so the √5 cancels: PQ/EF=( $\frac{20}{91}$ )√5 ÷ 2√5= $\frac{10}{91}$ .

PQ/EF=

\frac{20}{91}

\frac{\sqrt{5}}{2\sqrt{5}}

\frac{10}{91}

Answer

\frac{10}{91}

Numerically PQ≈0.49 and EF≈4.47, giving a ratio near 0.11, which matches

\frac{10}{91}

≈0.110. PQ is a small slice in the middle of EF, so a small fraction below

\frac{1}{8}

is sensible. The denominator 91=7·13 is exactly the product of the two denominators that appeared when locating P (7) and Q (13), a reassuring sign no arithmetic slipped.

💡Key takeaway

Put the whole picture on a grid, turn each line into an equation, and the crossing points are just where two equations agree.

Put the rectangle on a grid
Write the line through E and F
Find P where AC meets EF
Find Q where AG meets EF
Measure PQ and EF, then divide

Put the rectangle on a grid

Write the line through E and F

Find P where AC meets EF

Find Q where AG meets EF

Measure PQ and EF, then divide

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