AMC 10 · 2021 · #20

Grade 8 geometry-2d

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

equilateral-trianglearea-trianglesthirty-sixty-ninety-trianglearea-difference identify-subproblemsarea-difference ↑ Prerequisites: area-trianglesequilateral-triangle

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A pentagon ABCDE is drawn together with two interior points F and G, using 11 line segments — every segment has length 2. The area of the pentagon can be written as √(m) + √(n) for positive integers m, n. Find m + n.

Pick an answer.

(A)

~20

(B)

~21

(C)

~22

(D)

~23

(E)

~24

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram) — sketch the pentagon and mark all 11 length-2 segments. The drawing makes the four equilateral sub-triangles (△ ABF, △ BCF, △ AGE, △ GDE) jump out, and from there the 120° angles at B and E are visible (two adjacent 60° angles). Tool #7 (Identify Subproblems) — split the pentagon into the three triangles △ ABC, △ AED, and △ ACD along the diagonals AC and AD. Each sub-area is computable on its own with elementary triangle facts, and the three add to the answer. Tool #17 (Visualize) supports the symmetry observation that lets us only compute △ ABC once (since △ AED is its mirror image).

1STEP 1

Every segment is 2, so △ ABF, △ BCF, △ AEG, △ EDG are all equilateral — every interior angle is 60°.

AB = BF = AF = 2 → △ ABF equilateral, ∠ ABF = 60°; similarly for the other three.

2STEP 2

At B, segment BF splits ∠ ABC into two 60° angles, so ∠ ABC = 120°; by symmetry ∠ AED = 120° too.

∠ ABC = 60° + 60° = 120°; ∠ AED = 120°

3STEP 3

Cut the pentagon with diagonals AC and AD into three triangles — △ ABC, △ AED, and middle △ ACD — and add their areas.

[ABCDE] = [ABC] + [ACD] + [AED]

4STEP 4

△ ABC has sides 2, 2 and included angle 120°, so [ABC] = ½·2·2·sin 120° = √(3).

[ABC] =

\frac{1}{2}

(2)(2)sin 120° = 2 ·

\frac{√(3)}{2}

= √(3)

5STEP 5

By left-right symmetry △ AED matches △ ABC, so [AED] = √(3) as well.

[AED] = √(3)

6STEP 6

Law of cosines on △ ABC gives AC² = 4 + 4 - 8cos 120° = 12, so AC = AD = 2√(3) (with CD = 2 given).

AC² = 4 + 4 + 4 = 12, AC = AD = 2√(3)

7STEP 7

△ ACD is isosceles (equal sides 2√(3), base 2); its altitude is √(12 - 1) = √(11), so [ACD] = √(11).

AM = √(12 - 1) = √(11); [ACD] =

\frac{1}{2}

(2)(√(11)) = √(11)

8STEP 8

Sum: √(3) + √(3) + √(11) = 2√(3) + √(11) = √(12) + √(11), and 2√(3) = √(12), so m + n = 23.

[ABCDE] = √(12) + √(11) → m + n = 12 + 11 = 23 → (D)

Answer

~23

Sanity-check magnitudes. √(12) ≈ 3.46 and √(11) ≈ 3.32, so the pentagon area is about 6.78. The pentagon has "width" roughly |BE| ≈ 3 (centers of two equilateral triangles separated horizontally) and "height" roughly 2 — so an area near 7 is in the right ballpark. Also each ear triangle (area √(3) ≈ 1.73) is plausible for a triangle with two sides of 2 and a 120° apex. The form √(m) + √(n) with m, n both small integers near 12 matches the answer-choice spread (20 to 24), and 23 is precisely (D).

💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagorean reasoning you already know! Cut the pentagon with diagonals AC and AD — the two outer triangles have 120° apexes and area √(3) each, and the middle isoceles triangle has altitude √(11) and area √(11). Total = 2√(3) + √(11) = √(12) + √(11), so m + n = 23.

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