Competition · AMC preparation · step 4 of 4

AMC 10 · 2021B · #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)

(B)

(C)

(D)

(E)

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

Spot the equilateral triangles

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

Find the pentagon's angles

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 into three triangles

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

Find the left triangle's area

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

[ABC] = 1/2(2)(2)sin 120° = 2 · √(3)/2 = √(3)

5STEP 5

Copy it for the right triangle

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

[AED] = √(3)

6STEP 6

Find the two diagonals

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

Find the middle triangle's area

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

AM = √(12 - 1) = √(11); [ACD] = 1/2(2)(√(11)) = √(11)

8STEP 8

Add the three areas

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

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.

Spot the equilateral triangles
Find the pentagon's angles
Cut the pentagon into three triangles
Find the left triangle's area
Copy it for the right triangle
Find the two diagonals
Find the middle triangle's area
Add the three areas

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