Competition · AMC preparation · step 4 of 4

AMC 10 · 2019B · #20

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
area-circlescoordinate-geometrypythagorean-theoremthirty-sixty-ninety-triangle area-differenceidentify-subproblems ↑ Prerequisites: area-circlescoordinate-geometrypythagorean-theorem
📏 Long solution 💡 5 insights 📊 Diagram
Problem
Three congruent semicircles of radius 1 sit side by side on segment AD (with AB = BC = CD = 2), their flat edges on AD and their tops at E, F, G. A large circle of radius 2 is centered at F. The shaded area equals (area inside the big circle) - (area inside the three semicircles). Express it as ab\frac{a}{b}π - √(c) + d with gcd(a, b) = 1, then find a + b + c + d.

Pick an answer.

(A)
13
(B)
14
(C)
15
(D)
16
(E)
17

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

How to solve
Strategy Draw a Diagram

Tool #1 (Diagram): set up coordinates from the asy figure — F = (0, 0), big circle of radius 2 at origin, small semicircles centered (-2, -1), (0, -1), (2, -1) each radius 1. Tool #7 (Subproblems): split into (a) middle semicircle area (entirely inside big circle), (b) area of each side semicircle that lies inside the big circle. Tool #16 (Complement): shaded = (big disk) - (semicircle parts inside big disk).

1STEP 1

Set up the circle equations

Put F = (0, 0): big circle x² + y² = 4, small semicircles at (0, -1), (±2, -1). By symmetry the two side ones contribute equally.

big: x² + y² = 4; small: (x ± 2)² + (y+1)² = 1 or x² + (y+1)² = 1
2STEP 2

Handle the middle semicircle

The middle semicircle's farthest point is only √(2) from F, under radius 2, so it sits fully inside — its whole area π2\frac{\pi}{2} counts.

middle semicircle area inside big = π/2
3STEP 3

Find the intersection points

Intersecting the big circle with the right small circle gives 5x² - 16x + 12 = 0, so they meet at G = (2, 0) and P = (65\frac{6}{5}, -85\frac{8}{5}).

(2, 0) = G, (6/5, -8/5) = P
4STEP 4

Locate the point below the diameter

Since P sits below the diameter line y = -1, ignore it; the real cut is where the big circle meets y = -1, at x = √(3), inside [1, 3].

√(3) ≈ 1.732 ∈ [1, 3]
5STEP 5

Break the region into pieces

The right semicircle ∩ big disk splits into a triangle plus two circular segments: one big-circle bulge, one small-circle bulge.

triangle T = C(1,-1), V(√(3),-1), G(2,0)
6STEP 6

Find the triangle's area

Shoelace on (1, -1), (√(3), -1), (2, 0) gives triangle area 3−12\frac{\sqrt{3} - 1}{2}.

T = (√(3) - 1)/2
7STEP 7

Find the big-circle segment

The big-circle chord V–G subtends π6\frac{\pi}{6} at F, so its sector is π3\frac{\pi}{3} and the segment (sector minus triangle) is π3\frac{\pi}{3} - 1.

big segment = π/3 - 1
8STEP 8

Find the small-circle segment

The small-circle arc from G to C is a quarter circle: sector π4\frac{\pi}{4} minus the spanning right triangle 12\frac{1}{2} gives segment π4\frac{\pi}{4} - 12\frac{1}{2}.

small segment = π/4 - 1/2
9STEP 9

Add the three pieces

Adding triangle + two segments: right semicircle ∩ big disk = 32\frac{\sqrt{3}}{2} + 7π12\frac{7\pi}{12} - 2; the left side is the same by symmetry.

right semicircle ∩ big = √(3)/2 + 7π/12 - 2
10STEP 10

Total the semicircle overlap

Total semicircle area inside the big disk = π2\frac{\pi}{2} + 2(32\frac{\sqrt{3}}{2} + 7π12\frac{7\pi}{12} - 2) = 5π3\frac{5\pi}{3} + √(3) - 4.

semicircles ∩ big = 5π/3 + √(3) - 4
11STEP 11

Subtract to get the shaded area

Shaded = 4π - (5π3\frac{5\pi}{3} + √(3) - 4) = 7π3\frac{7\pi}{3} - √(3) + 4, so a = 7, b = 3, c = 3, d = 4 and a + b + c + d = 17 — choice (E).

shaded = 7π/3 - √(3) + 4 → a + b + c + d = 17
Answer
17
Numerically: shaded ≈ 7⋅3.141593\frac{7 \cdot 3.14159}{3} - 1.732 + 4 ≈ 7.330 - 1.732 + 4 ≈ 9.598. Big disk area = 4π ≈ 12.566. Difference ≈ 2.97, which should equal the total semicircle area inside the big disk. Middle gives π2\frac{\pi}{2} ≈ 1.571, two side semicircles give 2 · (0.866 + 1.833 - 2) = 2 · 0.699 ≈ 1.398. Total ≈ 2.969. ✓ matches. All four constants are positive integers and gcd(7, 3) = 1. Final sum is 17.
💡Key takeaway

This AMC 10 problem only needs Grade 8 coordinate-geometry you already know — set F at the origin, find where the two circles meet (G and where the big circle hits y = -1 at x = √(3)), then add triangle + two circular segments per side. Shaded area = 7π3\frac{7\pi}{3} - √(3) + 4, so a + b + c + d = 7 + 3 + 3 + 4 = 17. The answer is (E).

  • Set up the circle equations
  • Handle the middle semicircle
  • Find the intersection points
  • Locate the point below the diameter
  • Break the region into pieces
  • Find the triangle's area
  • Find the big-circle segment
  • Find the small-circle segment
  • Add the three pieces
  • Total the semicircle overlap
  • Subtract to get the shaded area

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