AMC 10 · 2007 · #18

Grade 8 geometry-2d

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

similar-trianglescoordinate-geometryarea-rectanglesarea-triangles convert-to-algebra ↑ Prerequisites: similar-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

The 12-sided polygon ABCDEFGHIJKL in the figure is a staircase shape whose vertices are labelled A, B, C, D, E, F, G, H, I, J, K, L in order around it. Each of its sides has length 4, and each two consecutive sides form a right angle. The segments AG and CH meet at a point M. What is the area of quadrilateral ABCM?

Pick an answer.

(A)

$\frac {44}{3}$

(B)

(C)

$\frac {88}{5}$

(D)

(E)

$\frac {62}{3}$

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Because every corner is a right angle, the whole figure fits perfectly on a grid, so Tool #1 (Draw a Diagram) turns each vertex into a pair of coordinates. Once the corners are numbers, Tool #4 (Introduce a Variable) lets me write each drawn segment as a line and solve for where they cross to pin down M. The region ABCM has a slanted side, so Tool #7 (Identify Subproblems) rebuilds it as one big right triangle with a small right triangle cut off — two areas I can each find with base times height.

1STEP 1

Put the corners on a grid

Every angle is right, so each side runs by 4 across or up. Put H=(0,0): then A=(0,12), B=(4,12), C=(4,8), G=(4,0).

A=(0,12), B=(4,12), C=(4,8), G=(4,0), H=(0,0)

2STEP 2

Turn the two segments into lines

Rise over run: AG drops 12 over a run of 4, so y = 12 - 3x. CH rises 8 over a run of 4, so y = 2x.

AG: y = 12 - 3x CH: y = 2x

3STEP 3

Find where they cross

M sits on both lines, so 12 - 3x = 2x, giving 5x = 12: M = (12/5, 24/5).

12 - 3x = 2x → 5x = 12 → x = 12/5, y = 24/5

4STEP 4

Rebuild ABCM as a big triangle minus a small one

CH cuts triangle ABG along CM, shaving off the small triangle MCG at corner G and leaving ABCM: [ABCM] = [ABG] - [MCG].

[ABCM] = [ABG] - [MCG]

5STEP 5

Measure the two triangles

Base BG = 12 with A a distance 4 across: [ABG] = 24. Base CG = 8 with M a distance 4 - 12/5 = 8/5 across: [MCG] = 32/5.

[ABG] = 1/2 · 12 · 4 = 24, [MCG] = 1/2 · 8 · 8/5 = 32/5

6STEP 6

Subtract to get the area

Subtract, writing 24 as 120/5: [ABCM] = 120/5 - 32/5 = 88/5, which is choice (C).

[ABCM] = 24 - 32/5 = 120/5 - 32/5 = 88/5 → (C)

Answer

88/5

The value 88/5 = 17.6 is one of the listed choices and it passes a size check: region ABCM contains the full right triangle ABC (legs AB=4 and BC=4, area 1/2·4·4 = 8) plus the extra triangle ACM below it, so the area must be comfortably more than 8 but well under the 48 of the whole rectangle ABGH — and 17.6 sits right in that band. A direct coordinate area check with the corners A(0,12), B(4,12), C(4,8), M(12/5,24/5) using the shoelace formula also gives 88/5.

💡Key takeaway

Drop a right-angled figure onto a grid, turn the crossing lines into slopes to find where they meet, then build the odd four-sided piece as one triangle with a smaller triangle cut off.

Put the corners on a grid
Turn the two segments into lines
Find where they cross
Rebuild ABCM as a big triangle minus a small one
Measure the two triangles
Subtract to get the area