AMC 10 · 2007 · #19

Grade 11 geometry-3d
Archetype Net Folding / Face Adjacency └ Sub-archetype Net-Edge Matching for Length, Area, or Volume
volume-cylindercircle-circumferencetrigonometric-ratios spatial-visualizationconvert-to-algebra ↑ Prerequisites: volume-cylindertrigonometric-ratios
📏 Long solution 💡 3 insights
Problem
A rhombus is bent without stretching into a cylinder by taping one side onto the opposite side. The cylinder's volume is given. Find the sine of the rhombus's angle.

Pick an answer.

(A)
$\frac{\pi}{9}$
(B)
$\frac{1}{2}$
(C)
$\frac{\pi}{6}$
(D)
$\frac{\pi}{4}$
(E)
$\frac{\sqrt{3}}{2}$
How to solve
Strategy Visualize Spatial Relationships

Once you know the can's radius and height in terms of the rhombus, this is one line of algebra. So the entire problem is the sentence before that line: which length of the rhombus goes around the can, and which one goes up it. Both candidates are 6-ish quantities living in the same figure, and guessing wrong changes the answer, so the temptation is to assert the answer to that question from a picture. Tool #17 (Visualize Spatial Relationships) and Tool #10 (Create a Physical Representation) are used instead to derive it: first track what the tape actually identifies — B becomes C, so the untaped side BC closes into a loop — and then unroll the finished can, which is bending run backwards and therefore hands the rhombus back. In that unrolled picture the two rims are straight lines and the answer to 'which direction goes around' is forced, not chosen. Tool #1 (Draw a Diagram) supplies the right triangle that turns the rhombus's height into 6sinθ, Tool #4 (Introduce a Variable) keeps the unknown lap count in the algebra instead of quietly setting it to 1, and Tool #13 (Convert to Algebra) writes the single volume equation. Tool #14 (Extreme Principle) then does two jobs at the end that are easy to skip: the bound sinθ ≤ 1 is what kills every lap count past the first, and running the construction forwards shows that a rhombus with the answer's angle really does roll into a can of volume 6 — necessity alone would leave the problem's 'is rolled to form a cylinder of volume 6' unearned. Tool #3 (Eliminate Possibilities) settles the sharp-versus-blunt ambiguity and checks the four rejected choices by computing the volume each one would give.

1STEP 1

Track what the tape identifies

The taping turns two sides into the two rims.

A ≡ D, B ≡ C ⟹ BC and AD close into loops = the two rims
2STEP 2

Unroll the can to fix the directions

Unrolling fixes which direction goes around.

around-direction = direction of BC, h = dist(line BC, line AD)
3STEP 3

A closed trip around the rim gives the radius

A closed trip around the rim gives the radius.

6 = k · 2π r ⟹ r = 3/kπ, π r² = π·9/k²π² = 9/k²π
4STEP 4

The height is six times the sine

The height is the side times the sine.

h = AF = 6sinθ
5STEP 5

One equation for the volume

The volume becomes one equation in that sine.

V = π r²h = 9/k²π · 6sinθ = 54sinθ/k²π = 6 ⟹ sinθ = k²π/9
6STEP 6

One lap only, and it really works

More than one wrap would push the sine past one, so only one lap works.

k ≥ 2 → sinθ ≥ 4π/9 ≈ 1.396 ＞ 1 (impossible); k=1: r=3/π, h=2π/3, V = 9/π·2π/3 = 6
7STEP 7

Both rhombi, one sine

Either rhombus gives the same sine, π/9, choice (B).

sin(180°-θ) = sinθ; V = 54sinθ/π: π/9↦ 6, 1/2↦27/π≈ 8.59, π/6↦ 9, π/4↦27/2, √(3)/2↦27√(3)/π≈ 14.89
Answer
π/9
Rebuild the can from the answer and look at it. Radius 3/π ≈ 0.955, so it is about 1.9 across and 2π/3 ≈ 2.09 tall — a squat can, a little taller than it is wide, holding 9/π·2π/3 = 6. Next, bracket the problem. Since V = 54sinθ/π with sinθ ≤ 1, the largest can any side-6 rhombus can make is 54/π ≈ 17.19, reached when the rhombus is a square, and the volume shrinks to 0 as the rhombus flattens. So 6 is attainable, and it should sit at the fraction 6/17.19 ≈ 0.349 of the way up the sine scale — which is π/9, the answer, arrived at without solving anything. A third check tests the no-stretching claim: the rhombus's area is 6 · 6sinθ = 4π ≈ 12.57, and the can's curved surface has area 2π r h = 2π·3/π·2π/3 = 4π. Equal, as they must be if paper was only bent. Now the traps. The rhombus this problem describes has a 20.4° corner — a thin sliver, not the comfortable wide rhombus one instinctively draws — and choices (B) 1/2 and (E) √(3)/2 are exactly the sines of the two comfortable-looking rhombi, 30° and 60°, which give volumes 8.59 and 14.89. Both can also be dismissed in one line without any geometry: V = 54sinθ/π is irrational whenever sinθ is rational, and the given volume 6 is not irrational, so the answer has to carry a π. Choices (C) π/6 and (D) π/4 survive that filter but overshoot, giving exactly 9 and 13.5. The most instructive near-miss is not on the list at all: bending the sheet the other way, so that the fold lines run parallel to AB, makes the circumference 6sinθ and gives 54sin²θ/π = 6, hence sinθ = √(π)/3 ≈ 0.591. That number appears nowhere among the choices, which is a quiet confirmation that 'cylinder' was meant literally — that other tube has spiralling rims and is not one.
💡Key takeaway

The whole problem is deciding which side of the rhombus becomes the circle: the sides you tape turn into a seam, so the sides you do not tape are the rims, and after that it is one volume formula.

  • Track what the tape identifies
  • Unroll the can to fix the directions
  • A closed trip around the rim gives the radius
  • The height is six times the sine
  • One equation for the volume
  • One lap only, and it really works
  • Both rhombi, one sine

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