AMC 10 · 2019 · #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

\frac{a}{b}

π - √(c) + d with gcd(a, b) = 1, then find a + b + c + d.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

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

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

middle semicircle area inside big =

\frac{\pi}{2}

3STEP 3

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

(2, 0) = G, (

\frac{6}{5}

, -

\frac{8}{5}

) = P

4STEP 4

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

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

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

T =

\frac{√(3) - 1}{2}

7STEP 7

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

big segment =

\frac{\pi}{3}

- 1

8STEP 8

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

small segment =

\frac{\pi}{4}

\frac{1}{2}

9STEP 9

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

right semicircle ∩ big =

\frac{√(3)}{2}

\frac{7\pi}{12}

- 2

10STEP 10

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

semicircles ∩ big =

\frac{5\pi}{3}

+ √(3) - 4

11STEP 11

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

shaded =

\frac{7\pi}{3}

- √(3) + 4 → a + b + c + d = 17

Answer

Numerically: shaded ≈

\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

\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 = $\frac{7\pi}{3}$ - √(3) + 4, so a + b + c + d = 7 + 3 + 3 + 4 = 17. The answer is (E).

More like this