AMC 8 · 2024 · #18
Grade 7 geometry-2dProblem
Three concentric circles centered at O have radii of 1, 2, and 3. Points B and C lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angles BOC, as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of ∠BOC in degrees?
Pick an answer.
AMC 8 2024 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 circles share the same center $O$, with radii $1$, $2$, and $3$. The whole inner ring (between the radius-$1$ and radius-$2$ circles) is shaded. Inside the outer ring (between the radius-$2$ and radius-$3$ circles), only a pie-slice with central angle $\angle BOC = x$ degrees is shaded. We are told the shaded area equals the unshaded area, and we must find $x$.
Givens: Three concentric circles with radii $r_1 = 1$, $r_2 = 2$, $r_3 = 3$; The entire inner ring (between radii $1$ and $2$) is shaded; Inside the outer ring (between radii $2$ and $3$), a sector with central angle $\angle BOC = x^{\circ}$ is shaded; Shaded area equals unshaded area (each is half of the whole big circle); Answer choices: (A) 108, (B) 120, (C) 135, (D) 144, (E) 150
Unknowns: The measure of $\angle BOC$ in degrees
Understand
Restated: Three circles share the same center $O$, with radii $1$, $2$, and $3$. The whole inner ring (between the radius-$1$ and radius-$2$ circles) is shaded. Inside the outer ring (between the radius-$2$ and radius-$3$ circles), only a pie-slice with central angle $\angle BOC = x$ degrees is shaded. We are told the shaded area equals the unshaded area, and we must find $x$.
Givens: Three concentric circles with radii $r_1 = 1$, $r_2 = 2$, $r_3 = 3$; The entire inner ring (between radii $1$ and $2$) is shaded; Inside the outer ring (between radii $2$ and $3$), a sector with central angle $\angle BOC = x^{\circ}$ is shaded; Shaded area equals unshaded area (each is half of the whole big circle); Answer choices: (A) 108, (B) 120, (C) 135, (D) 144, (E) 150
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities
The shaded region is a **compound shape** (the inner ring plus a sector of the outer ring), so the right move is Tool #7 — split it into two pieces we already know how to compute, then add. We use Tool #1 to keep the picture straight (the inner ring is fully shaded, the outer ring only partially). After computing $x$, Tool #3 confirms it equals one of (A)-(E). We deliberately avoid Tool #13 (Algebra) — once we see that the outer ring needs to supply only $1.5\pi$ out of its $5\pi$, a simple **fraction-of-a-circle** argument ($\frac{1.5}{5} = \frac{3}{10}$ of $360^{\circ}$) gives the angle without ever writing an equation in $x$.
Execute — Answer: A
7.G.B.4 Step 1 - Find the area of each circle using $A = \pi r^{2}$.
- The smallest has area $\pi(1)^2 = \pi$, the middle one has area $\pi(2)^2 = 4\pi$, and the largest has area $\pi(3)^2 = 9\pi$.
- The whole figure (the entire largest disk) therefore has total area $9\pi$.
💡 The formula $A = \pi r^{2}$ for the area of a circle is exactly the Grade 7 circle-area standard.
7.G.B.4 Step 2 - Split the figure into its two natural rings (subproblem #1).
- The **inner ring** (between radii $1$ and $2$) is what's left of the middle disk after cutting out the small one: area $= 4\pi - \pi = 3\pi$.
- The **outer ring** (between radii $2$ and $3$) is what's left of the big disk after cutting out the middle one: area $= 9\pi - 4\pi = 5\pi$.
- The whole inner ring ($3\pi$) is shaded; only part of the outer ring ($5\pi$) is shaded.
💡 Finding a ring's area by subtracting the inner circle from the outer circle is a direct use of the Grade 7 circle-area formula.
3.G.A.2 Step 3 - Translate the condition "shaded = unshaded" into a number.
- If two parts are equal and together make $9\pi$, each part must be half of $9\pi$.
- So the total shaded area must be $\frac{1}{2} \cdot 9\pi = \frac{9\pi}{2} = 4.5\pi$.
💡 "Two equal pieces make the whole" is the Grade 3 partition-into-equal-parts idea, applied to the area instead of to a shape outline.
5.NF.B.3 Step 4 - The inner ring already supplies $3\pi$ of shaded area for free.
- So the shaded sector of the outer ring must supply the rest: $4.5\pi - 3\pi = 1.5\pi$.
- The outer ring's total area is $5\pi$, and the shaded sector is the fraction $\frac{1.5\pi}{5\pi} = \frac{1.5}{5} = \frac{3}{10}$ of it.
💡 Interpreting the ratio $\frac{1.5}{5}$ as the fraction $\frac{3}{10}$ (a piece divided by the whole) is the Grade 5 "fraction as division" idea.
4.NF.B.4 Step 5 - A sector of a circle (or ring) takes the same fraction of the full $360^{\circ}$ around the center that it takes of the total area.
- So if the shaded sector is $\frac{3}{10}$ of the ring, its central angle is $\frac{3}{10}$ of $360^{\circ}$, which is $108^{\circ}$.
💡 Multiplying a fraction by a whole number $\bigl(\tfrac{3}{10} \times 360\bigr)$ is the Grade 4 fraction-times-whole-number standard.
4.NBT.A.2 Step 6 - Match $108^{\circ}$ to the answer choices.
- Of $108, 120, 135, 144, 150$, our value $108$ is exactly choice (A).
- The other choices correspond to bigger fractions of the outer ring than we need: e.g.
- $150^{\circ}$ would shade $\frac{150}{360} \cdot 5\pi \approx 2.08\pi$ of the outer ring, giving a total shaded area $\approx 5.08\pi > 4.5\pi$, too much.
💡 Comparing the number $108$ against five three-digit answer choices is a Grade 4 multi-digit comparison.
7.G.B.4 Find the area of each circle using $A = \pi r^{2}$. The smallest has area $\pi(1 7.G.B.4 Split the figure into its two natural rings (subproblem #1). The **inner ring** 3.G.A.2 Translate the condition "shaded = unshaded" into a number. If two parts are equa 5.NF.B.3 The inner ring already supplies $3\pi$ of shaded area for free. So the shaded se 4.NF.B.4 A sector of a circle (or ring) takes the same fraction of the full $360^{\circ}$ 4.NBT.A.2 Match $108^{\circ}$ to the answer choices. Of $108, 120, 135, 144, 150$, our val Review
Reasonableness: Plug $x = 108^{\circ}$ back into the shaded-area expression: $3\pi + \frac{108}{360} \cdot 5\pi = 3\pi + \frac{3}{10} \cdot 5\pi = 3\pi + 1.5\pi = 4.5\pi$, which is exactly half of $9\pi$. So shaded equals unshaded, as required. The angle is also reasonable by feel: the outer ring ($5\pi$) is bigger than the inner ring ($3\pi$), so we only need a modest slice of it — about $\tfrac{1}{3}$ of the full circle, and indeed $108^{\circ}$ is slightly less than $120^{\circ} = \tfrac{1}{3} \cdot 360^{\circ}$.
Alternative: An alternative is Tool #13 (Convert to Algebra): let $x$ be the angle and write $3\pi + \frac{x}{360} \cdot 5\pi = \frac{9\pi}{2}$, then solve for $x$. The algebra gives the same $x = 108$, but it hides the punchy idea that the outer ring only has to supply $\frac{1.5\pi}{5\pi} = \frac{3}{10}$ of itself. The subproblem-plus-fraction path is friendlier for elementary students.
CCSS standards used (min grade 7)
3.G.A.2Partition shapes into equal parts with equal areas (Turning "shaded equals unshaded" into "shaded is half of the total", i.e. $\tfrac{1}{2} \cdot 9\pi = 4.5\pi$.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Comparing $108$ against the five three-digit answer choices to pick (A).)4.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a whole number (Computing $\tfrac{3}{10} \cdot 360^{\circ} = 108^{\circ}$ to convert the area fraction into a central angle.)5.NF.B.3Interpret a fraction as division of the numerator by the denominator (Reading the ratio $\frac{1.5\pi}{5\pi}$ as the fraction $\frac{1.5}{5} = \frac{3}{10}$ of the outer ring.)7.G.B.4Know the formulas for area and circumference of a circle (Computing $A = \pi r^{2}$ for each of the three circles ($\pi$, $4\pi$, $9\pi$) and subtracting to get the two ring areas ($3\pi$ and $5\pi$).)
⭐ This AMC 8 problem only needs the Grade 7 circle-area formula $A = \pi r^{2}$ you already know!
⭐ This AMC 8 problem only needs the Grade 7 circle-area formula $A = \pi r^{2}$ you already know!
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