AMC 8 · 2013 · #24

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-rectanglesarea-trianglescoordinate-geometry area-differencecoordinate-geometryidentify-subproblems ↑ Prerequisites: area-rectanglesarea-trianglescoordinate-geometry

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Three squares of equal area are arranged in an L-shape: square EFGH sits on the left, square GHIJ on the right (sharing side GH), and a third square ABCD sits on top of the line EHI with its base DC landing exactly on the midpoints of HE and IH. Find the ratio of the shaded pentagon AJICB to the total area of all three squares.

Pick an answer.

(A)

$hspace{.05in}\frac{1}{4}$

(B)

$hspace{.05in}\frac{7}{24}$

(C)

$hspace{.05in}\frac{1}{3}$

(D)

$hspace{.05in}\frac{3}{8}$

(E)

$hspace{.05in}\frac{5}{12}$

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The shaded pentagon is awkward — its diagonal side AJ cuts across two squares. Tool #7 (Identify Subproblems) is the geometry workhorse: slice the pentagon along the horizontal line through C and I so it becomes a clean trapezoid on top plus a right triangle on the bottom. Tool #1 (Diagram) gives us coordinates so we can read lengths off the page instead of doing algebra; Tool #9 (Easier Problem) tells us to pick a friendly side length — set s = 2 — so every distance is a whole or half number and the ratio at the end still comes out the same.

1STEP 1

Set s = 2 with G at the origin; the midpoint conditions then fix every vertex — see the coordinates below.

A(-1,4), B(1,4), C(1,2), D(-1,2), E(-2,2), H(0,2), I(2,2), J(2,0)

2STEP 2

Shrink the problem (Tool #9): each square is 2 × 2 = 4, so the three together have area 12 — now just find the pentagon.

3 × (2 × 2) = 12

3STEP 3

Cut the pentagon along y = 2 (through D, H, C, I). The slant AJ crosses that line at K( $\frac{1}{2}$ , 2), splitting it into top AKCB and bottom KIJ.

K = (

\frac{1}{2}

, 2)

4STEP 4

Top piece AKCB is a trapezoid: parallel sides AB = 2 and KC = $\frac{1}{2}$ , height 2, so its area is $\frac{5}{2}$ .

Area(AKCB) =

\frac{1}{2}

(2 +

\frac{1}{2}

) × 2 =

\frac{5}{2}

5STEP 5

Bottom piece KIJ is a right triangle with legs IJ = 2 and KI = $\frac{3}{2}$ , so its area is $\frac{3}{2}$ .

Area(KIJ) =

\frac{1}{2}

\frac{3}{2}

× 2 =

\frac{3}{2}

6STEP 6

Add the pieces: pentagon = $\frac{5}{2}$ + $\frac{3}{2}$ = 4 — exactly one square — so the ratio is one third, choice (C).

Pentagon/(Three squares) =

\frac{4}{12}

\frac{1}{3}

→ (C)

Answer

hspace{.05in}

\frac{1}{3}

The pentagon's area came out to 4, which is exactly the area of one square. That feels right when you look at the picture: the pentagon overlaps the right square fully and steals a triangular slice from the right side of the top square, while leaving an equally sized triangular slice of the top square un-shaded on the left. The two missing/added triangles are congruent (both have legs 1 and

\frac{3}{2}

at s=2), so the trade is even and the pentagon ends up with exactly one square's worth of area. One square out of three squares is

\frac{1}{3}

, matching (C).

💡Key takeaway

This AMC 8 problem only needs Grade 6 polygon-area decomposition — split, add, compare — that you already know!

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