AMC 10 · 2019 · #15
Grade 8 geometry-2d
Pick an answer.
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).
Write every circle in coordinates
Write all four as equations.
Pin down coordinates from the asy figure so distances and intersections become arithmetic.
8.G.B.8Draw A DiagramHandle the middle semicircle
The middle one is fully inside.
Middle bump fits inside the big circle — no clipping.
7.G.B.4Identify SubproblemsFind the crossings
Find where an outer semicircle crosses the big circle.
Two circles meet in two points — algebra gives both.
8.G.B.8Identify SubproblemsCheck the crossing is on the arc
Check it really lies on the arc.
The big circle slices the diameter line of the right semicircle inside the chunk [1, 3].
8.G.B.7Draw A DiagramSet up the triangle
Split the overlap into a triangle and segments.
Pick the chord chord chord polygon, then add the bulges back where the actual boundary is a circular arc.
7.G.B.6Identify SubproblemsArea of the triangle
Compute that triangle's area.
Shoelace handles any triangle from its three coordinates.
8.G.B.8Draw A DiagramThe big circle's segment
Compute the big circle's segment.
Sector minus triangle gives the bulge outside the chord.
A sector with its triangle cut away leaves exactly the bulge outside the chord.
▸ Why?
A sector is the share of the circle its angle takes, so its area is fixed by the angle alone.
▸ Why?
The sector is exactly the triangle plus the bulge, so subtracting one leaves the other.
The small circle's segment
Compute the small circle's segment too.
Quarter-sector minus the spanning right triangle.
7.G.B.4Identify SubproblemsOverlap of one outer semicircle
Combine the pieces for the partial area.
All three pieces are inside the region — add them up.
7.G.B.6Draw A DiagramAdd the three semicircles
By symmetry the other side matches.
Middle bump plus two equal side bumps.
7.G.B.6Identify SubproblemsShaded area and the sum
Subtracting and tidying gives 17.
Big disk minus the semicircle-overlap chunk — and the form matches the target.
7.G.B.6Change Focus Count The ComplementThis AMC 12 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 = 7π/3 - √(3) + 4, so a + b + c + d = 7 + 3 + 3 + 4 = 17. The answer is (E).
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