AMC 10 · 2012 · #15

Grade 8 geometry-3d

Archetype Net Folding / Face Adjacency └ Sub-archetype Net-Edge Matching for Length, Area, or Volume

cone-unrollingvolume-conecircle-circumferencepythagorean-theorem spatial-visualizationidentify-subproblemsconvert-to-algebra ↑ Prerequisites: volume-conecircle-circumference

📏 Medium solution 💡 3 insights

Problem

Two sectors of one disk are each rolled into a cone. Find the ratio of their volumes.

Pick an answer.

(A)

$\frac{1}{8}$

(B)

$\frac{1}{4}$

(C)

$\frac{\sqrt{10}}{10}$

(D)

$\frac{\sqrt{5}}{6}$

(E)

$\frac{\sqrt{5}}{5}$

How to solve

Strategy Visualize Spatial Relationships

The whole problem hinges on one act of mental folding, which is Tool #17 (Visualize Spatial Relationships): a flat sector rolls into a cone, and rolling preserves lengths. That single fact pins down two quantities at once — the sector's straight radius becomes the cone's SLANT height (not its vertical height), and the sector's arc becomes the cone's base circle. Getting that pairing right is the entire problem; almost every wrong answer here comes from treating 12 as a vertical height, or from assuming the two cones are similar. Tool #7 (Identify Subproblems) then breaks the job into four clean pieces done twice: arc length → base radius → vertical height → volume. Tool #4 (Introduce a Variable) lets the two cases share one set of formulas driven by the fraction f of the disk used, so the answer can be checked a second way without redoing arithmetic. Tool #14 (Extreme Principle) supplies the existence check: a sector rolls into a genuine cone only when its base radius stays strictly below the slant height 12, and it verifies which of the two cones is actually the smaller one rather than assuming it.

1STEP 1

See what rolling preserves

The disk's radius becomes each cone's slant.

ℓ = 12 (both cones), arc length of sector = 2π r (base circumference)

2STEP 2

Split the circumference by angle

The angles split the circumference proportionally.

arc₁₂₀ = 1/3(24π) = 8π, arc₂₄₀ = 2/3(24π) = 16π, 8π + 16π = 24π ✓

3STEP 3

Turn each arc into a base radius

Each arc gives a base radius, 4 and 8.

2π r₁ = 8π → r₁ = 4; 2π r₂ = 16π → r₂ = 8

4STEP 4

Check each cone exists, then find its height

Both cones really exist, giving their heights.

h = √(12² - r²) exists ⇔ r ＜ 12. h₁ = √(144-16) = √(128) = 8√(2), h₂ = √(144-64) = √(80) = 4√(5)

5STEP 5

Compute both volumes

The volumes follow directly.

V₁ = 1/3π(4)²(8√(2)) = 128√(2)/3π, V₂ = 1/3π(8)²(4√(5)) = 256√(5)/3π, V₁ ＜ V₂

6STEP 6

Form and simplify the ratio

The ratio simplifies to √10/10, choice (C).

V₁/V₂ = 128√(2)/256√(5) = √(2)/2√(5) = (√(2)·√(5))/(2 · 5) = √(10)/10 → (C)

Answer

√(10)/10

Numerically √(10)/10 ≈ 0.316, and the volumes give 128√(2)/256√(5) ≈ 181.0/572.4 ≈ 0.316 — the same value, so the algebra and the arithmetic agree. The ratio is less than 1, as it must be for smaller-over-larger. Two wrong choices are exactly the traps this problem is built around. Choice (A) 1/8 is what you get by assuming the cones are similar: base radii 4 and 8 scale by 1/2, so volumes would scale by (1/2)³ = 1/8. But the cones are NOT similar — they share slant height 12 while their radii differ, so their shapes differ (h₁ = 8√(2) is not half of h₂ = 4√(5)). Choice (B) 1/4 is the ratio of base areas alone, 16/64, forgetting that the narrow cone is also the taller one. The true answer 0.316 sits sensibly between those two: the narrow cone loses on base area but claws some back on height, so the ratio should exceed 1/4 yet stay well under 1.

💡Key takeaway

When flat paper rolls into a cone, the straight edge becomes the slanted side and the curved edge becomes the base circle — so both cones here have slant 12, which means they are not scaled copies and you must find each height with the Pythagorean theorem before comparing volumes.

See what rolling preserves
Split the circumference by angle
Turn each arc into a base radius
Check each cone exists, then find its height
Compute both volumes
Form and simplify the ratio