Competition · AMC preparation · step 4 of 4

AMC 10 · 2012B · #17

Grade 8 geometry-3d

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

volume-conecircular-sectorsimilar-figures identify-subproblems ↑ Prerequisites: volume-cone

📏 Medium solution 💡 3 insights

Problem

A flat paper disk of radius 12 is cut along two of its radii, splitting it into two sectors: one with central angle 120° and the other with the leftover angle. Each sector is then rolled up so its two straight edges meet, forming the curved side (lateral surface) of a cone. Compare the two cones by volume and report the ratio (smaller volume) : (larger volume).

Pick an answer.

(A)

$\frac{1}{8}$

(B)

$\frac{1}{4}$

(C)

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

(D)

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

(E)

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

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Rolling never stretches paper, so each straight edge keeps length 12 and becomes the cone's slant edge; the arc closes into the base circle.

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

2STEP 2

Split the circumference by angle

The rim 24π splits by angle: the 120° sector takes 8π, the 240° sector takes 16π, and together they use it all up.

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

3STEP 3

Turn each arc into a base radius

Each arc is a base circumference 2π r, so π cancels twice: 8π gives r₁ = 4 and 16π gives r₂ = 8.

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

4STEP 4

Check each cone exists, then find its height

A cut through the tip leaves legs r, h and hypotenuse 12, and both radii stay under 12, so h₁ = 8√(2) and h₂ = 4√(5).

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

5STEP 5

Compute both volumes

Plug into V = 1/3π r² h: V₁ = 128√(2)/3π and V₂ = 256√(5)/3π, and a quick compare shows V₁ is the smaller.

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 1/3 and π cancel, leaving √(2)/2√(5); rationalizing the denominator turns it into √(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

A parent dashboard for the family lives at sensimlab.com.