AMC 10 · 2006 · #19

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
circular-sectorpythagorean-theoremthirty-sixty-ninety-triangle area-difference ↑ Prerequisites: circular-sectorpythagorean-theorem
📏 Long solution 💡 4 insights 📊 Diagram
Problem
A circle of radius 2 is centered at O. Square OABC has side length 1, so its corners run O, A, B, C in order and B is the corner opposite O. Sides AB and CB are extended past B to meet the circle at D and at E. What is the area of the region bounded by segment BD, segment BE, and the minor arc from D to E?

Pick an answer.

(A)
$\frac{\pi}{3}+1-\sqrt{3}$
(B)
$\frac{\pi}{2}(2-\sqrt{3})$
(C)
$\pi(2-\sqrt{3})$
(D)
$\frac{\pi}{6}+\frac{\sqrt{3}+1}{2}$
(E)
$\frac{\pi}{3}-1+\sqrt{3}$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The shaded region has a curved side, so no single area formula fits. Tool #1 (Draw a Diagram) pins everything to coordinates so D and E become computable. Tool #7 (Identify Subproblems) is the key move: split the awkward region into a plain triangle BDE plus the circular segment that bulges out to the arc, and get the segment as a sector minus a triangle. Tool #3 (Eliminate Possibilities) uses the picture — the region is clearly small — to sanity-check which choice can be right.

1STEP 1

Locate D and E with coordinates

With O at the origin, B=(1,1) and line AB is x=1, so 1+y²=4 gives D=(1,√3); by symmetry E=(√3,1).

1²+y²=2² → y=√(3), D=(1,√(3)), E=(√(3),1)
2STEP 2

Split the region into two pieces

Cut the region along chord DE: toward B lies the plain triangle BDE, and out toward the arc lies the circular segment on DE.

[shaded] = [△ BDE] + [segment DE]
3STEP 3

Area of triangle BDE

Triangle BDE has its right angle at B and both legs equal √3-1, so its area is 1/2(√3-1)²=2-√3.

[△ BDE]=1/2(√3-1)²=1/2(4-2√3)=2-√3
4STEP 4

Area of the circular segment DE

Radii OD and OE make 60° and 30° with the x-axis, so sector ODE=4π/12=π/3; removing triangle ODE (area 1) leaves π/3-1.

[segment] = π/3 - 1
5STEP 5

Add the pieces and pick the choice

Add them: (2-√3)+(π/3-1)=π/3+1-√3≈0.315 — a small sliver matching the picture, which is choice (A).

[shaded]=π/3+1-√3≈ 0.315 → (A)
Answer
π/3+1-√(3)
The region is a thin sliver just outside the corner B, so its area must be small and positive. Choice (A) gives π/3+1-√3≈0.315, which is tiny — perfect for the picture. Choices (B) π/2(2-√3)≈0.42 and (C) π(2-√3)≈0.84 are noticeably bigger, and (D) ≈1.89 is far too large; (E) π/3-1+√3≈1.78 is also too big. Only (A) is small enough, confirming the answer.
💡Key takeaway

When a region has one curved side, slice it along a straight chord into a plain triangle plus a pie-slice-minus-triangle bulge, then add the parts.

  • Locate D and E with coordinates
  • Split the region into two pieces
  • Area of triangle BDE
  • Area of the circular segment DE
  • Add the pieces and pick the choice

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