AMC 10 · 2019 · #20
Grade 8 geometry-2dProblem
As shown in the figure, line segment AD is trisected by points B and C so that AB=BC=CD=2. Three semicircles of radius 1, \overarcAEB,\overarcBFC, and \overarcCGD, have their diameters on AD, lie in the same halfplane determined by line AD, and are tangent to line EG at E,F, and G, respectively. A circle of radius 2 has its center on F. The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form
ba⋅π−c+d,
where a,b,c, and d are positive integers and a and b are relatively prime. What is a+b+c+d?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Three congruent semicircles of radius $1$ sit side by side on segment $\overline{AD}$ (with $AB = BC = CD = 2$), their flat edges on $\overline{AD}$ 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 $\frac{a}{b}\pi - \sqrt{c} + d$ with $\gcd(a, b) = 1$, then find $a + b + c + d$.
Givens: Three semicircles of radius $1$ centered at $(-2, -1)$, $(0, -1)$, $(2, -1)$, top half (using the asy coordinates: $F = (0, 0)$ is the top of the middle semicircle); Big circle of radius $2$ centered at $F = (0, 0)$; Shaded region = (big disk) $\setminus$ (union of three semicircles); Choices: (A) $13$, (B) $14$, (C) $15$, (D) $16$, (E) $17$
Unknowns: $a + b + c + d$, where the shaded area is $\tfrac{a}{b}\pi - \sqrt{c} + d$
Understand
Restated: Three congruent semicircles of radius $1$ sit side by side on segment $\overline{AD}$ (with $AB = BC = CD = 2$), their flat edges on $\overline{AD}$ 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 $\frac{a}{b}\pi - \sqrt{c} + d$ with $\gcd(a, b) = 1$, then find $a + b + c + d$.
Givens: Three semicircles of radius $1$ centered at $(-2, -1)$, $(0, -1)$, $(2, -1)$, top half (using the asy coordinates: $F = (0, 0)$ is the top of the middle semicircle); Big circle of radius $2$ centered at $F = (0, 0)$; Shaded region = (big disk) $\setminus$ (union of three semicircles); Choices: (A) $13$, (B) $14$, (C) $15$, (D) $16$, (E) $17$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #16 Change Focus / Count the Complement
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).
Execute — Answer: E
8.G.B.8 Step 1 - Setup.
- Place $F = (0, 0)$.
- The big circle is $x^2 + y^2 = 4$.
- The middle small semicircle has center $(0, -1)$, radius $1$, upper half ($y \ge -1$).
- The right small semicircle (containing arc $\overarc{CGD}$) has center $(2, -1)$, radius $1$, upper half.
- By the left-right symmetry of the picture, the left semicircle contributes the same as the right.
💡 Pin down coordinates from the asy figure so distances and intersections become arithmetic.
7.G.B.4 Step 2 - The middle semicircle is entirely inside the big circle.
- Check: a point on the upper semicircle is $(\cos t, -1 + \sin t)$ for $t \in [0, \pi]$.
- Its distance from origin is $\sqrt{\cos^2 t + (\sin t - 1)^2} = \sqrt{2 - 2\sin t} \le \sqrt{2} < 2$.
- So the middle semicircle lies wholly inside the big disk.
- Its area is $\pi/2$.
💡 Middle bump fits inside the big circle — no clipping.
8.G.B.8 Step 3 - Find where the big circle meets the right small circle.
- Solve $x^2 + y^2 = 4$ and $(x-2)^2 + (y+1)^2 = 1$ simultaneously.
- Subtracting: $-4x + 4 + 2y + 1 = -3$, so $y = 2x - 4$.
- Substitute: $x^2 + (2x - 4)^2 = 4 \Rightarrow 5x^2 - 16x + 12 = 0 \Rightarrow x = 2$ or $x = 6/5$.
- The corresponding points are $G = (2, 0)$ and $P = (6/5, -8/5)$.
💡 Two circles meet in two points — algebra gives both.
8.G.B.7 Step 4 - Note: $P = (6/5, -8/5)$ has $y = -8/5 < -1$, so $P$ lies BELOW the diameter line $y = -1$.
- So $P$ is not on the boundary of the upper semicircle.
- The upper semicircle is bounded above by its arc and below by the segment $y = -1$ from $C = (1, -1)$ to $D = (3, -1)$.
- The big circle crosses $y = -1$ where $x^2 + 1 = 4 \Rightarrow x = \sqrt{3}$ (taking the positive root, which is in $[1, 3]$).
💡 The big circle slices the diameter line of the right semicircle inside the chunk $[1, 3]$.
7.G.B.6 Step 5 - The region (right semicircle $\cap$ big disk) is bounded by three curves: (i) the segment on $y = -1$ from $C = (1, -1)$ to $(\sqrt{3}, -1)$, (ii) the big-circle arc from $(\sqrt{3}, -1)$ up to $G = (2, 0)$, (iii) the small-circle arc from $G$ back around to $C$ (the upper-left quarter of the small circle).
- Decompose into a triangle plus two circular segments.
💡 Pick the chord chord chord polygon, then add the bulges back where the actual boundary is a circular arc.
8.G.B.8 Step 6 - Triangle area: vertices $(1,-1), (\sqrt{3},-1), (2, 0)$.
- Use the shoelace formula.
- $\text{Area} = \tfrac{1}{2}\,\big|\,1 \cdot (-1 - 0) + \sqrt{3} \cdot (0 - (-1)) + 2 \cdot (-1 - (-1))\,\big| = \tfrac{1}{2}\,|-1 + \sqrt{3} + 0| = \tfrac{\sqrt{3} - 1}{2}$.
💡 Shoelace handles any triangle from its three coordinates.
7.G.B.4 Step 7 - Big-circle segment (between chord $V$–$G$ and the arc).
- The chord endpoints make angles $-\pi/6$ (for $V = (\sqrt{3}, -1)$, since $\sqrt{3}^2 + 1 = 4$ and $\arctan(-1/\sqrt{3}) = -\pi/6$) and $0$ (for $G = (2, 0)$).
- Sector area $= \tfrac{1}{2} \cdot 2^2 \cdot \tfrac{\pi}{6} = \tfrac{\pi}{3}$.
- Triangle area from origin: $\tfrac{1}{2}|0 \cdot (-1 - 0) + \sqrt{3} \cdot (0 - 0) + 2 \cdot (0 - (-1))| = 1$.
- Segment $= \tfrac{\pi}{3} - 1$.
💡 Sector minus triangle gives the bulge outside the chord.
7.G.B.4 Step 8 - Small-circle segment (between chord $C$–$G$ and the upper-left quarter arc).
- The arc spans angles $\pi/2$ to $\pi$ around center $(2, -1)$ — a quarter circle.
- Sector area $= \tfrac{1}{4}\pi \cdot 1^2 = \tfrac{\pi}{4}$.
- Triangle from center $(2,-1)$ to $C = (1,-1)$ to $G = (2, 0)$: $\tfrac{1}{2}|2(-1 - 0) + 1(0 - (-1)) + 2(-1 - (-1))| = \tfrac{1}{2}|-2 + 1 + 0| = \tfrac{1}{2}$.
- Segment $= \tfrac{\pi}{4} - \tfrac{1}{2}$.
💡 Quarter-sector minus the spanning right triangle.
7.G.B.6 Step 9 - Sum: right semicircle $\cap$ big disk $= T + \text{big segment} + \text{small segment} = \tfrac{\sqrt{3}-1}{2} + (\tfrac{\pi}{3} - 1) + (\tfrac{\pi}{4} - \tfrac{1}{2}) = \tfrac{\sqrt{3}}{2} + \tfrac{7\pi}{12} - 2$.
- By symmetry, the left semicircle contributes the same.
💡 All three pieces are inside the region — add them up.
7.G.B.6 Step 10 Total semicircle area inside big disk $= \tfrac{\pi}{2} + 2 \cdot (\tfrac{\sqrt{3}}{2} + \tfrac{7\pi}{12} - 2) = \tfrac{\pi}{2} + \sqrt{3} + \tfrac{7\pi}{6} - 4 = \tfrac{3\pi + 7\pi}{6} + \sqrt{3} - 4 = \tfrac{5\pi}{3} + \sqrt{3} - 4$.
💡 Middle bump plus two equal side bumps.
7.G.B.6 Step 11 - Shaded area $= 4\pi - \left(\tfrac{5\pi}{3} + \sqrt{3} - 4\right) = \tfrac{12\pi - 5\pi}{3} - \sqrt{3} + 4 = \tfrac{7\pi}{3} - \sqrt{3} + 4$.
- So $a = 7, b = 3, c = 3, d = 4$, with $\gcd(7, 3) = 1$.
- The sum is $7 + 3 + 3 + 4 = 17$.
- Answer: choice (E).
💡 Big disk minus the semicircle-overlap chunk — and the form matches the target.
8.G.B.8 Setup. Place $F = (0, 0)$. The big circle is $x^2 + y^2 = 4$. The middle small s 7.G.B.4 The middle semicircle is entirely inside the big circle. Check: a point on the u 8.G.B.8 Find where the big circle meets the right small circle. Solve $x^2 + y^2 = 4$ an 8.G.B.7 Note: $P = (6/5, -8/5)$ has $y = -8/5 < -1$, so $P$ lies BELOW the diameter line 7.G.B.6 The region (right semicircle $\cap$ big disk) is bounded by three curves: (i) th 8.G.B.8 Triangle area: vertices $(1,-1), (\sqrt{3},-1), (2, 0)$. Use the shoelace formul 7.G.B.4 Big-circle segment (between chord $V$–$G$ and the arc). The chord endpoints make 7.G.B.4 Small-circle segment (between chord $C$–$G$ and the upper-left quarter arc). The 7.G.B.6 Sum: right semicircle $\cap$ big disk $= T + \text{big segment} + \text{small se 7.G.B.6 Total semicircle area inside big disk $= \tfrac{\pi}{2} + 2 \cdot (\tfrac{\sqrt{ 7.G.B.6 Shaded area $= 4\pi - \left(\tfrac{5\pi}{3} + \sqrt{3} - 4\right) = \tfrac{12\pi Review
Reasonableness: Numerically: shaded $\approx \tfrac{7 \cdot 3.14159}{3} - 1.732 + 4 \approx 7.330 - 1.732 + 4 \approx 9.598$. Big disk area $= 4\pi \approx 12.566$. Difference $\approx 2.97$, which should equal the total semicircle area inside the big disk. Middle gives $\pi/2 \approx 1.571$, two side semicircles give $2 \cdot (0.866 + 1.833 - 2) = 2 \cdot 0.699 \approx 1.398$. Total $\approx 2.969$. ✓ matches. All four constants are positive integers and $\gcd(7, 3) = 1$. Final sum is $17$.
Alternative: Tool #15 (Reorganize): cut the big disk by the line $\overline{EG}$ (which passes through $F$ and is tangent to all three semicircles). The upper half of the big circle has area $2\pi$ and contains none of the semicircles (they sit below the tangent line). The lower half has area $2\pi$ and contains the three bumps plus the symmetric mirror regions. Working with the lower half alone halves the bookkeeping; you still end with $\tfrac{7\pi}{3} - \sqrt{3} + 4$.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for area and circumference of a circle (Computing the big-circle sector ($\pi/3$) and small-circle quarter-sector ($\pi/4$) areas.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Assembling the shaded area as (big disk) minus (semicircle overlap) and reducing to the target form.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding where $y = -1$ meets the big circle ($x = \sqrt{3}$) and the chord lengths.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Setting up circle equations and solving for the intersection points of the two circles.)
⭐ 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 $x = \sqrt{3}$), then add triangle + two circular segments per side. Shaded area $= \tfrac{7\pi}{3} - \sqrt{3} + 4$, so $a + b + c + d = 7 + 3 + 3 + 4 = 17$. The answer is $\textbf{(E)}$.
⭐ 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 $x = \sqrt{3}$), then add triangle + two circular segments per side. Shaded area $= \tfrac{7\pi}{3} - \sqrt{3} + 4$, so $a + b + c + d = 7 + 3 + 3 + 4 = 17$. The answer is $\textbf{(E)}$.
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