AMC 8 · 2023 · #12
Grade 7 geometry-2dProblem
The figure below shows a large white circle with a number of smaller white and shaded circles in its
interior. What fraction of the interior of the large white circle is shaded?
Pick an answer.
AMC 8 2023 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: A large white circle is sitting on a unit grid. Inside it there is one big shaded circle, two white circles cut out of the shaded one, and three small shaded circles floating elsewhere. Using the grid to read off each radius, what fraction of the area of the large white circle is actually shaded?
Givens: Large white circle has radius $3$ (diameter $6$ on the grid); Large shaded circle has radius $2$ and sits inside the large white circle; Two inner white circles each have radius $1$ and lie inside the large shaded circle; Three small shaded circles each have radius $\tfrac{1}{2}$ (diameter $1$) and lie outside the large shaded circle; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{11}{36}$, (C) $\tfrac{1}{3}$, (D) $\tfrac{19}{36}$, (E) $\tfrac{5}{9}$
Unknowns: The fraction $\dfrac{\text{shaded area}}{\text{area of the large white circle}}$
Understand
Restated: A large white circle is sitting on a unit grid. Inside it there is one big shaded circle, two white circles cut out of the shaded one, and three small shaded circles floating elsewhere. Using the grid to read off each radius, what fraction of the area of the large white circle is actually shaded?
Givens: Large white circle has radius $3$ (diameter $6$ on the grid); Large shaded circle has radius $2$ and sits inside the large white circle; Two inner white circles each have radius $1$ and lie inside the large shaded circle; Three small shaded circles each have radius $\tfrac{1}{2}$ (diameter $1$) and lie outside the large shaded circle; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{11}{36}$, (C) $\tfrac{1}{3}$, (D) $\tfrac{19}{36}$, (E) $\tfrac{5}{9}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The figure is the problem. Tool #1 (Draw a Diagram) tells us to read radii straight off the grid — every circle's diameter is a whole or half number of grid squares, so no algebra is needed. Tool #7 (Identify Subproblems) splits the shaded region into three easy pieces: (a) the big shaded disk, (b) MINUS the two inner white disks that punch holes in it, (c) PLUS the three little shaded disks elsewhere. Compute the three pieces, combine, then divide by the area of the large white circle.
Execute — Answer: B
5.G.A.2 Step 1 - Read radii off the grid.
- Tool #1: use the dashed unit grid as a ruler.
- The large white circle spans $6$ units across, so $r_{\text{big}}=3$.
- The large shaded circle spans $4$ units, so $r_{\text{shaded}}=2$.
- The two inner white circles each span $2$ units, so each has radius $1$.
- The three little shaded circles each span $1$ unit, so each has radius $\tfrac{1}{2}$.
💡 Reading a length straight from a coordinate grid is the Grade 5 "graph points on the plane" skill.
7.G.B.4 Step 2 - Area of the big white circle (this is the denominator).
- Apply the circle-area formula $A=\pi r^2$ once: with $r=3$ we get $9\pi$.
- This is the WHOLE we are comparing the shaded area to.
💡 Knowing $A=\pi r^2$ is the Grade 7 circle-area formula — this is the only "older" skill the problem needs.
7.G.B.4 Step 3 - Subproblem A — area of the big shaded disk.
- Same formula with $r=2$: $\pi(2)^2=4\pi$.
- This is what we would have if the two inner white circles weren't there yet.
💡 Same Grade 7 circle-area formula, just plugged in with a different radius.
7.G.B.4 Step 4 - Subproblem B — the two inner white circles.
- Each has $r=1$, so each has area $\pi(1)^2=\pi$.
- Two of them total $2\pi$.
- These are CUT OUT of the big shaded disk (they appear white), so we will subtract $2\pi$ from $4\pi$.
💡 Repeating the same circle-area formula and adding equal pieces is still a Grade 7 area calculation.
5.NF.B.4 Step 5 - Subproblem C — the three little shaded circles.
- Each has $r=\tfrac{1}{2}$, so each has area $\pi\left(\tfrac{1}{2}\right)^2=\tfrac{\pi}{4}$.
- Three of them total $\tfrac{3\pi}{4}$.
- These are extra shading that does NOT overlap anything else, so we will ADD this.
💡 Squaring $\tfrac{1}{2}$ is the Grade 5 fraction-times-fraction skill ($\tfrac{1}{2}\times\tfrac{1}{2}=\tfrac{1}{4}$).
5.NF.A.1 Step 6 - Combine the three subproblems.
- Total shaded = (big shaded disk) - (two inner whites) + (three little shaded).
- Convert to a common denominator of $4$ and add: $4\pi-2\pi+\tfrac{3\pi}{4}=2\pi+\tfrac{3\pi}{4}=\tfrac{8\pi}{4}+\tfrac{3\pi}{4}=\tfrac{11\pi}{4}$.
💡 Combining $\pi$-terms with different denominators is just Grade 5 fraction addition with the common factor $\pi$ along for the ride.
6.RP.A.3 Step 7 - Divide to get the fraction.
- The shaded area over the whole big-white-circle area gives the answer.
- The $\pi$'s cancel and we get $\tfrac{11/4}{9}=\tfrac{11}{36}$, which matches choice (B).
💡 Setting up "part over whole" as a ratio is the Grade 6 ratio-and-rate reasoning skill.
5.G.A.2 Read radii off the grid. Tool #1: use the dashed unit grid as a ruler. The large 7.G.B.4 Area of the big white circle (this is the denominator). Apply the circle-area fo 7.G.B.4 Subproblem A — area of the big shaded disk. Same formula with $r=2$: $\pi(2)^2=4 7.G.B.4 Subproblem B — the two inner white circles. Each has $r=1$, so each has area $\p 5.NF.B.4 Subproblem C — the three little shaded circles. Each has $r=\tfrac{1}{2}$, so ea 5.NF.A.1 Combine the three subproblems. Total shaded = (big shaded disk) - (two inner whi 6.RP.A.3 Divide to get the fraction. The shaded area over the whole big-white-circle area Review
Reasonableness: Does $\tfrac{11}{36}$ make sense? The big shaded disk alone covers $\tfrac{4\pi}{9\pi}=\tfrac{4}{9}\approx 0.44$ of the big white circle, but the two inner white holes punch out half of that, dropping us close to $\tfrac{2}{9}\approx 0.22$. Adding three small shaded circles ($\tfrac{3}{4}\pi$, a smallish amount) nudges us back up to around $0.30$ — and $\tfrac{11}{36}\approx 0.306$ lands exactly there. So (B) is the only choice between $\tfrac{1}{4}$ and $\tfrac{1}{3}$ that fits, which matches our visual estimate.
Alternative: Use Tool #9 (Solve an Easier Related Problem): pretend each circle is the square it is inscribed in. The big white "square" then has area $6\times 6 = 36$, the big shaded "square" has area $4\times 4 = 16$, each inner white "square" has area $2\times 2 = 4$ (total $8$), and each small shaded "square" has area $1\times 1 = 1$ (total $3$). Shaded $= 16 - 8 + 3 = 11$ out of $36$, giving $\tfrac{11}{36}$ directly — same answer (B), and no $\pi$ needed.
CCSS standards used (min grade 7)
5.G.A.2Represent real-world and mathematical problems by graphing points (Reading each circle's radius directly off the dashed unit grid in the figure.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Squaring the radius $\tfrac{1}{2}$ to get $\tfrac{1}{4}$ for the three small shaded circles.)5.NF.A.1Add and subtract fractions with unlike denominators (Combining $4\pi-2\pi+\tfrac{3\pi}{4}$ into a single fraction $\tfrac{11\pi}{4}$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Setting up the final answer as the ratio shaded-area : total-area.)7.G.B.4Know the formulas for area and circumference of a circle (Applying $A=\pi r^2$ to every circle (the big white, the big shaded, the inner whites, and the small shaded).)
⭐ This AMC 8 problem only needs the Grade 7 circle-area formula $\pi r^2$ you already know — once you have it, the rest is just adding and subtracting circle areas like puzzle pieces!
⭐ This AMC 8 problem only needs the Grade 7 circle-area formula $\pi r^2$ you already know — once you have it, the rest is just adding and subtracting circle areas like puzzle pieces!
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