AMC 8 · 2017 · #25

Grade 8 geometry-2d

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

area-trianglesarea-circlesreflection-symmetryangle-sum-triangle area-differenceidentify-subproblems ↑ Prerequisites: area-trianglesarea-circles

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A figure is bounded by two line segments US and UT (each of length 2, meeting at U with ∠ TUS = 60°) and by two arcs SR and TR (each a 60° arc of a circle of radius 2). The two arcs bulge inward (concave to the region) and meet at point R at the bottom. Find the area of this region.

Pick an answer.

(A)

$3\sqrt{3}-\pi$

(B)

$4\sqrt{3}-\frac{4\pi}{3}$

(C)

$2\sqrt{3}$

(D)

$4\sqrt{3}-\frac{2\pi}{3}$

(E)

$4+\frac{4\pi}{3}$

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The region mixes straight edges and curved edges, so no single area formula applies. Tool #7 (Identify Subproblems) breaks it into pieces with familiar formulas: first replace each arc by its chord to form a straight-sided figure (a kite made of two equilateral triangles), then subtract the two circular segments that the chords "hide" between chord and arc. Each segment is itself a subproblem: 60° sector minus 60° equilateral triangle. Tool #1 (Draw a Diagram) is the supporting move — sketching the chords SR and TR on top of the given figure makes the kite and the two segments visible, and shows that each chord has length 2 (chord of a 60° arc on a radius-2 circle).

1STEP 1

Each chord SR, TR closes a 60° slice of a radius-2 circle into an equilateral triangle, so SR = TR = 2 and a kite USRT of side 2 appears.

US = UT = SR = TR = 2

2STEP 2

Split the kite along ST into two equilateral triangles of side 2, each of area √(3), so the kite area is 2√(3).

Area(kite) = 2 ·

\frac{2²√(3)}{4}

= 2√(3)

3STEP 3

One segment is the 60° sector of the radius-2 circle ( $\frac{2\pi}{3}$ ) minus its equilateral triangle (√(3)), so each segment is $\frac{2\pi}{3}$ - √(3).

Segment =

\frac{2\pi}{3}

- √(3)

4STEP 4

Since the arcs are concave, subtract both segments from the kite: 2√(3) - 2( $\frac{2\pi}{3}$ - √(3)) = 4√(3) - $\frac{4\pi}{3}$ , choice (B).

Area = 2√(3) - 2 (

\frac{2\pi}{3}

- √(3)) = 2√(3) -

\frac{4\pi}{3}

+ 2√(3) = 4√(3) -

\frac{4\pi}{3}

→ (B)

Answer

4√(3)-

\frac{4\pi}{3}

Numerically, 4√(3) ≈ 6.93 and

\frac{4\pi}{3}

≈ 4.19, so the area is about 6.93 - 4.19 ≈ 2.74. As a sanity check, the kite alone has area 2√(3) ≈ 3.46, and the two concave bites should shave off a little more than 0.7 together — which matches. The answer is positive (good), strictly smaller than the kite (good, since the arcs eat into the kite), and strictly bigger than zero (the arcs do not meet so deeply that the region disappears). Choices (C) 2√(3) ignores π entirely (wrong — there are circular pieces), (E) is even larger than the kite (impossible since arcs cut area away), and (A), (D) have the wrong segment-area coefficient.

💡Key takeaway

This AMC 8 problem only needs Grade 8 work with irrational numbers like √(3) and π, plus the circle-area and triangle-area formulas you already know!

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