AMC 10 · 2014 · #23
Grade 8 geometry-3dA sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?
Pick an answer.
AMC 10 2014 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 sphere sits inside a truncated right circular cone (a frustum), touching the top disk, the bottom disk, and the slanted side all at once. The frustum's volume is exactly twice the sphere's volume. Find the ratio of the bottom base radius to the top base radius.
Givens: A sphere is inscribed in the frustum: it is tangent to the top base, the bottom base, and the lateral (slanted) surface.; Volume of the frustum $= 2 \times$ volume of the sphere.; The solid is a right circular frustum, so the axial cross-section is symmetric.; Answer choices: (A) $\dfrac{3}{2}$, (B) $\dfrac{1+\sqrt{5}}{2}$, (C) $\sqrt{3}$, (D) $2$, (E) $\dfrac{3+\sqrt{5}}{2}$
Unknowns: The ratio $\dfrac{R}{r}$ of the bottom base radius $R$ to the top base radius $r$.
Understand
Restated: A sphere sits inside a truncated right circular cone (a frustum), touching the top disk, the bottom disk, and the slanted side all at once. The frustum's volume is exactly twice the sphere's volume. Find the ratio of the bottom base radius to the top base radius.
Givens: A sphere is inscribed in the frustum: it is tangent to the top base, the bottom base, and the lateral (slanted) surface.; Volume of the frustum $= 2 \times$ volume of the sphere.; The solid is a right circular frustum, so the axial cross-section is symmetric.; Answer choices: (A) $\dfrac{3}{2}$, (B) $\dfrac{1+\sqrt{5}}{2}$, (C) $\sqrt{3}$, (D) $2$, (E) $\dfrac{3+\sqrt{5}}{2}$
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #1 Draw a Diagram, #4 Introduce a Variable, #13 Convert to Algebra
A sphere inside a cone is hard to reason about in 3D, so the load-bearing move is Tool #17 (Visualize Spatial Relationships): slice the solid with the vertical plane through its axis. The sphere becomes a circle inscribed in an isosceles trapezoid — a flat picture (Tool #1) where tangency turns into two facts we can measure: the height equals the sphere's diameter, and the slant leg equals the sum of the two base radii. Name the three radii (Tool #4), then Tool #13 (Convert to Algebra) turns 'frustum volume = twice sphere volume' into a single equation in the ratio $R/r$, which factors to a quadratic.
Execute — Answer: E
7.G.A.3 Step 1 Slice through the axis
- Cut the frustum with the vertical plane through its central axis.
- The cross-section is an isosceles trapezoid: bottom side $2R$, top side $2r$.
- The sphere cuts to a circle inscribed in that trapezoid, touching the top, the bottom, and both slanted sides.
- Since the circle touches both horizontal sides, its diameter spans the full height, so $h = 2\rho$ where $\rho$ is the sphere's radius.
💡 A round ball wedged between two parallel plates just fits — the gap between the plates is exactly one ball-diameter.
8.G.B.7 Step 2 Slant leg from tangent lengths
- Look at the right half of the trapezoid.
- From the bottom corner, the two tangent segments to the circle are equal, and the horizontal one has length $R$; from the top corner they are equal too, with horizontal length $r$.
- So the slanted leg is $R + r$ long.
- The leg also spans a horizontal change $R - r$ and a vertical change $h$, so by the Pythagorean theorem $(R+r)^2 = (R-r)^2 + h^2$.
- Expanding gives $h^2 = 4Rr$, and with $h = 2\rho$ this says $\rho = \sqrt{Rr}$.
💡 Two tangent lines drawn from the same point to a circle always have equal reach, so the sloped edge is just the two base radii laid end to end.
8.G.C.9 Step 3 Write both volumes
- Use the standard formulas.
- The sphere: $V_{\text{sph}} = \tfrac{4}{3}\pi \rho^3 = \tfrac{4}{3}\pi (Rr)^{3/2}$.
- The frustum: $V_{\text{fr}} = \tfrac{\pi h}{3}\left(R^2 + Rr + r^2\right)$, and substituting $h = 2\sqrt{Rr}$ gives $V_{\text{fr}} = \tfrac{2\pi\sqrt{Rr}}{3}\left(R^2 + Rr + r^2\right)$.
💡 The frustum volume is the average-ish cross-section rule $R^2+Rr+r^2$, and the sphere volume is the familiar four-thirds-pi-r-cubed.
8.EE.C.7 Step 4 Apply the volume condition
- Set frustum volume equal to twice the sphere volume: $\tfrac{2\pi\sqrt{Rr}}{3}\left(R^2 + Rr + r^2\right) = 2 \cdot \tfrac{4}{3}\pi (Rr)^{3/2}$.
- Every term carries $\tfrac{2\pi}{3}\sqrt{Rr}$, so divide it out from both sides (note $(Rr)^{3/2} = Rr\sqrt{Rr}$).
- What remains is $R^2 + Rr + r^2 = 4Rr$, i.e.
- $R^2 - 3Rr + r^2 = 0$.
💡 Both sides share the same messy factor $\sqrt{Rr}$, so cancelling it strips the problem down to a clean relation between $R$ and $r$.
8.EE.A.2 Step 5 Solve for the ratio
- Divide $R^2 - 3Rr + r^2 = 0$ by $r^2$ and let $x = \tfrac{R}{r}$: $x^2 - 3x + 1 = 0$.
- By the quadratic formula $x = \tfrac{3 \pm \sqrt{9-4}}{2} = \tfrac{3 \pm \sqrt{5}}{2}$.
- Since $R > r$ the ratio must exceed $1$, so keep the larger root $x = \tfrac{3+\sqrt{5}}{2} \approx 2.62$.
- That is choice $\textbf{(E)}$.
💡 The whole equation only depends on the ratio $R/r$, so calling it $x$ collapses two unknowns into one quadratic with a clean radical answer.
7.G.A.3 Cut the frustum with the vertical plane through its central axis. The cross-sect 8.G.B.7 Look at the right half of the trapezoid. From the bottom corner, the two tangent 8.G.C.9 Use the standard formulas. The sphere: $V_{\text{sph}} = \tfrac{4}{3}\pi \rho^3 8.EE.C.7 Set frustum volume equal to twice the sphere volume: $\tfrac{2\pi\sqrt{Rr}}{3}\l 8.EE.A.2 Divide $R^2 - 3Rr + r^2 = 0$ by $r^2$ and let $x = \tfrac{R}{r}$: $x^2 - 3x + 1 Review
Reasonableness: Plug $x = \tfrac{3+\sqrt{5}}{2}$ back in: $x^2 - 3x + 1 = 0$ holds exactly, and numerically $x \approx 2.618$, comfortably bigger than $1$ as required for a bottom base wider than the top. Sanity on the volumes: with $r = 1$, $R = 2.618$, the height is $h = 2\sqrt{Rr} \approx 3.24$, giving frustum volume $\approx 12.13$ and sphere volume $\approx 6.06$ — a ratio of $2.00$, matching the condition. The smaller root $\tfrac{3-\sqrt{5}}{2} \approx 0.38$ is just the reciprocal (the same shape viewed upside down), correctly rejected because $R>r$. Answer (E) stands.
Alternative: Instead of tangent lengths, use the geometric-mean fact directly: a sphere inscribed in a frustum always satisfies $\rho^2 = Rr$ (its radius is the geometric mean of the base radii), which drops out of the right triangle formed by the slant height, the radius difference, and the height. Feeding $\rho = \sqrt{Rr}$ and $h = 2\rho$ into the volume equation gives the identical quadratic $x^2 - 3x + 1 = 0$ and the same answer (E).
CCSS standards used (min grade 8)
7.G.A.3Describe the two-dimensional figures that result from slicing three-dimensional figures (Taking the vertical cross-section through the axis so the frustum becomes an isosceles trapezoid and the sphere becomes an inscribed circle.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Relating the slant leg $R+r$, the horizontal change $R-r$, and the height $h$ to get $h^2 = 4Rr$ and hence $\rho = \sqrt{Rr}$.)8.G.C.9Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve problems (Writing the sphere volume $\tfrac{4}{3}\pi\rho^3$ and the frustum (cone-based) volume $\tfrac{\pi h}{3}(R^2+Rr+r^2)$.)8.EE.C.7Solve linear equations in one variable, including using the distributive property and collecting like terms (Cancelling the shared factor $\tfrac{2\pi}{3}\sqrt{Rr}$ from the volume equation to reduce it to $R^2 - 3Rr + r^2 = 0$.)8.EE.A.2Use square root and cube root symbols to represent solutions and evaluate roots; know that $\sqrt{2}$ is irrational (Handling the radicals $\sqrt{Rr}$ and the final irrational root $\tfrac{3+\sqrt{5}}{2}$ from the quadratic $x^2 - 3x + 1 = 0$.)
⭐ Cut the ball-in-a-cone straight down the middle: the ball becomes a circle inside a trapezoid, the height equals the ball's diameter, and the slanted side equals the two radii added together. Those facts turn 'volume is double' into $x^2 - 3x + 1 = 0$, whose big root $\tfrac{3+\sqrt{5}}{2}$ is the answer.
⭐ Cut the ball-in-a-cone straight down the middle: the ball becomes a circle inside a trapezoid, the height equals the ball's diameter, and the slanted side equals the two radii added together. Those facts turn 'volume is double' into $x^2 - 3x + 1 = 0$, whose big root $\tfrac{3+\sqrt{5}}{2}$ is the answer.
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