AMC 10 · 2019 · #8

Grade 8 geometry-2d

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

equilateral-trianglethirty-sixty-ninety-trianglearea-trianglesarea-rectanglespythagorean-theorem area-differenceidentify-subproblems ↑ Prerequisites: equilateral-trianglearea-trianglespythagorean-theorem

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A square has four equilateral triangles drawn inward — one sitting on each side, all with side length 2, and all four apexes meeting at the center of the square. Find the area inside the square but outside the triangles (the shaded region).

Pick an answer.

(A)

(B)

$12 - 4\sqrt{3}$

(C)

$3\sqrt{3}$

(D)

$4\sqrt{3}$

(E)

$16 - 4\sqrt{3}$

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The shaded shape is irregular, but it equals (whole square) − (four triangles) — Tool #16 says count the complement. Split into (a) find the square's side from the triangle height (Tool #7 sub-question), (b) find the square's area, (c) find each triangle's area, (d) subtract. Tool #1 keeps the picture honest as we work.

1STEP 1

Drop an altitude on a side-2 equilateral triangle; the 30-60-90 halves give height √(3).

h = √(4 - 1) = √(3)

2STEP 2

Every apex sits at the center, so center-to-side is √(3), half the side; the square's side is 2√(3).

side = 2 · √(3) = 2√(3)

3STEP 3

Square that side: the square's area is 12.

A_square = (2√(3))² = 4 · 3 = 12

4STEP 4

Each triangle has area √(3), and four non-overlapping ones total 4√(3).

A_△ total = 4 · √(3) = 4√(3)

5STEP 5

Subtract the four triangles from the square, leaving the shaded area 12 - 4√(3).

A_shaded = 12 - 4√(3) → (B)

6STEP 6

Numerically 4√(3) ≈ 6.93, so the shaded area ≈ 5.07 — about 42% of 12, matching the corner pockets.

12 - 4√(3) ≈ 5.07 → (B)

Answer

12 - 4√(3)

The shaded area is positive (since 4√(3) ≈ 6.93 ＜ 12) and clearly smaller than the square — good. The four triangles fill about 58% of the square; what's left forms four small kite-shaped pockets at the corners. 12 - 4√(3) matches choice (B). Choices like 4 or 4√(3) describe the triangles, not the leftover — those are trap answers.

💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagorean thinking you already know: each equilateral triangle has altitude √(3), so the square has side 2√(3) and area 12. The four triangles take up 4√(3), leaving 12 - 4√(3) for the shaded pockets.

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