AMC 10 · 2012 · #17
Grade 8 geometry-3dJesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?
Pick an answer.
AMC 10 2012 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 paper disk of radius $12$ is cut along two radii into two sectors. The smaller sector has a central angle of $120^\circ$; the larger has $360^\circ - 120^\circ = 240^\circ$. Each sector is rolled up so its two straight edges meet, forming the lateral (side) surface of a cone. Find the ratio of the volume of the smaller cone to the volume of the larger cone.
Givens: Disk radius is $12$ (this becomes the slant height of each cone); Smaller sector central angle: $120^\circ$; Larger sector central angle: $240^\circ$; Each sector's arc becomes the circumference of that cone's base circle; Answer choices: (A) $\frac{1}{8}$, (B) $\frac{1}{4}$, (C) $\frac{\sqrt{10}}{10}$, (D) $\frac{\sqrt{5}}{6}$, (E) $\frac{\sqrt{5}}{5}$
Unknowns: The ratio $\dfrac{V_{\text{small}}}{V_{\text{large}}}$ of the two cone volumes
Understand
Restated: A paper disk of radius $12$ is cut along two radii into two sectors. The smaller sector has a central angle of $120^\circ$; the larger has $360^\circ - 120^\circ = 240^\circ$. Each sector is rolled up so its two straight edges meet, forming the lateral (side) surface of a cone. Find the ratio of the volume of the smaller cone to the volume of the larger cone.
Givens: Disk radius is $12$ (this becomes the slant height of each cone); Smaller sector central angle: $120^\circ$; Larger sector central angle: $240^\circ$; Each sector's arc becomes the circumference of that cone's base circle; Answer choices: (A) $\frac{1}{8}$, (B) $\frac{1}{4}$, (C) $\frac{\sqrt{10}}{10}$, (D) $\frac{\sqrt{5}}{6}$, (E) $\frac{\sqrt{5}}{5}$
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
Tool #17 (Visualize Spatial Relationships) is the key: you must mentally fold each flat sector into a cone to see what each length becomes — the disk's radius $12$ turns into the cone's slant height, and the curved arc turns into the base circle. Get that mapping right and the rest is formula work. Tool #7 (Subproblems) then splits each cone into three clean pieces: base radius from the arc, height from the Pythagorean Theorem, then volume. Tool #4 (Introduce a Variable) names the base radius $r$ so we can solve $2\pi r = \text{arc length}$.
Execute — Answer: C
7.G.B.4 Step 1 Fold the sector into a cone
- Picture rolling a flat sector until its two straight radius-edges glue together.
- Every point on the arc ends up on the rim of the base circle, and the sharp point of the sector becomes the tip of the cone.
- The straight edge (length $12$, the disk's radius) slants from tip to rim — that is the slant height $\ell = 12$.
- The arc, meanwhile, becomes the whole base circle, so the arc length equals the base circumference $2\pi r$.
💡 The disk radius becomes the slant edge, and the curved arc becomes the base rim — this is the one mapping the whole problem hinges on.
7.G.B.4 Step 2 Base radius of each cone
- The full disk has circumference $2\pi(12) = 24\pi$.
- The smaller sector is $\tfrac{120}{360} = \tfrac{1}{3}$ of the disk, so its arc is $\tfrac{1}{3}(24\pi) = 8\pi$; setting $2\pi r = 8\pi$ gives $r_{\text{small}} = 4$.
- The larger sector is $\tfrac{240}{360} = \tfrac{2}{3}$ of the disk, so its arc is $16\pi$, giving $2\pi r = 16\pi$ and $r_{\text{large}} = 8$.
💡 The base circle is exactly as long as the arc that made it, so its radius comes straight from $2\pi r = \text{arc}$.
8.G.B.7 Step 3 Height from the Pythagorean Theorem
- In a cone the base radius $r$, the height $h$, and the slant height $\ell$ form a right triangle with $\ell$ as the hypotenuse: $r^2 + h^2 = \ell^2$.
- With $\ell = 12$ for both cones, solve for $h$.
- Smaller: $h = \sqrt{12^2 - 4^2} = \sqrt{144 - 16} = \sqrt{128} = 8\sqrt{2}$.
- Larger: $h = \sqrt{12^2 - 8^2} = \sqrt{144 - 64} = \sqrt{80} = 4\sqrt{5}$.
💡 Radius, height, and slant form a right triangle, so the slant $12$ locks the height once the radius is known.
8.G.C.9 Step 4 Volume of each cone
- Use $V = \tfrac{1}{3}\pi r^2 h$.
- Smaller: $V_{\text{small}} = \tfrac{1}{3}\pi(4^2)(8\sqrt{2}) = \tfrac{1}{3}\pi(16)(8\sqrt{2}) = \tfrac{128\sqrt{2}}{3}\pi$.
- Larger: $V_{\text{large}} = \tfrac{1}{3}\pi(8^2)(4\sqrt{5}) = \tfrac{1}{3}\pi(64)(4\sqrt{5}) = \tfrac{256\sqrt{5}}{3}\pi$.
💡 Once radius and height are in hand, the cone volume formula does the rest for each cone.
8.EE.A.2 Step 5 Form and simplify the ratio
- Divide the volumes; the $\tfrac{1}{3}\pi$ cancels: $\dfrac{V_{\text{small}}}{V_{\text{large}}} = \dfrac{128\sqrt{2}}{256\sqrt{5}} = \dfrac{\sqrt{2}}{2\sqrt{5}}$.
- Rationalize by multiplying top and bottom by $\sqrt{5}$: $\dfrac{\sqrt{2}\cdot\sqrt{5}}{2\cdot 5} = \dfrac{\sqrt{10}}{10}$.
- This matches choice (C).
💡 Cancel the shared $\tfrac{1}{3}\pi$, then clear the root from the bottom to land on a clean $\frac{\sqrt{10}}{10}$.
7.G.B.4 Picture rolling a flat sector until its two straight radius-edges glue together. 7.G.B.4 The full disk has circumference $2\pi(12) = 24\pi$. The smaller sector is $\tfra 8.G.B.7 In a cone the base radius $r$, the height $h$, and the slant height $\ell$ form 8.G.C.9 Use $V = \tfrac{1}{3}\pi r^2 h$. Smaller: $V_{\text{small}} = \tfrac{1}{3}\pi(4^ 8.EE.A.2 Divide the volumes; the $\tfrac{1}{3}\pi$ cancels: $\dfrac{V_{\text{small}}}{V_{ Review
Reasonableness: The smaller cone should hold less than the larger, so the ratio must be under $1$ — and $\frac{\sqrt{10}}{10} = \frac{1}{\sqrt{10}} \approx 0.316$ is. A quick numeric pass agrees: $V_{\text{small}} \approx \frac{128(1.414)}{3}\pi \approx 60.3\pi$ and $V_{\text{large}} \approx \frac{256(2.236)}{3}\pi \approx 190.8\pi$, whose ratio $\approx 0.316$ — exactly $\frac{\sqrt{10}}{10}$. The trap answers $\frac{1}{8}$ and $\frac{1}{4}$ come from wrongly assuming volume scales like the angle ratio $\frac{1}{2}$ cubed or squared, ignoring that the heights differ too.
Alternative: Tool #7 with a general ratio formula. If a fraction $k$ of the disk (radius $R$) forms a cone, then $r = kR$ and $h = R\sqrt{1 - k^2}$, so $V = \tfrac{1}{3}\pi (kR)^2 R\sqrt{1-k^2} = \tfrac{1}{3}\pi R^3 k^2\sqrt{1-k^2}$. Here $k_{\text{small}} = \tfrac{1}{3}$ and $k_{\text{large}} = \tfrac{2}{3}$, so the ratio is $\dfrac{(1/3)^2\sqrt{1-1/9}}{(2/3)^2\sqrt{1-4/9}} = \dfrac{\tfrac{1}{9}\cdot\tfrac{\sqrt{8}}{3}}{\tfrac{4}{9}\cdot\tfrac{\sqrt{5}}{3}} = \dfrac{\sqrt{8}}{4\sqrt{5}} = \dfrac{2\sqrt{2}}{4\sqrt{5}} = \dfrac{\sqrt{2}}{2\sqrt{5}} = \dfrac{\sqrt{10}}{10}$. Same result.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Turning each sector's arc (a fraction of the disk's $24\pi$ circumference) into the base circumference $2\pi r$ and solving for the base radius.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding each cone's height from the right triangle formed by base radius, height, and slant height: $r^2 + h^2 = 12^2$.)8.G.C.9Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve problems (Computing each cone volume with $V = \tfrac{1}{3}\pi r^2 h$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Simplifying the radical heights $\sqrt{128} = 8\sqrt{2}$, $\sqrt{80} = 4\sqrt{5}$ and rationalizing the final ratio to $\frac{\sqrt{10}}{10}$.)
⭐ When a paper sector rolls into a cone, the disk's radius becomes the slant height and the arc becomes the base circle — recover the base radius from the arc, get the height with the Pythagorean Theorem, and the volume ratio works out to $\frac{\sqrt{10}}{10}$.
⭐ When a paper sector rolls into a cone, the disk's radius becomes the slant height and the arc becomes the base circle — recover the base radius from the arc, get the height with the Pythagorean Theorem, and the volume ratio works out to $\frac{\sqrt{10}}{10}$.
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