Competition · AMC preparation · step 4 of 4

AMC 10 · 2023B · #20

Grade 8 geometry-3d
great-circle-arcspatial-visualizationpythagorean-theoremsymmetry-argument identify-subproblemssymmetry-argument ↑ Prerequisites: pythagorean-theoremspatial-visualization
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
Four congruent semicircles, joined end-to-end on a sphere of radius 2, form a closed curve that splits the sphere's surface into two congruent pieces. The total length of the curve is π√(n). Find n.

Pick an answer.

(A)
32
(B)
12
(C)
48
(D)
36
(E)
27

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The total length is 4 times one semicircle's arc length, which is π times its radius, which is half its diameter — and the diameter is a chord of the sphere joining adjacent junction points. So the problem chains: (a) pin down where the four junction points sit, (b) compute the chord between adjacent junctions, (c) convert to semicircle radius and arc length, (d) total length and solve for n. Tool #7 (Identify Subproblems) names each rung; Tool #1 (Diagram) of the great-circle cross-section makes the chord visible; Tool #17 (Visualize Spatial Relationships) confirms the four points sit at the corners of a square on a great circle.

1STEP 1

Locate the four junction points

Two congruent halves force maximal symmetry, so the four junctions sit at the corners of a square inscribed in a great circle of radius 2.

A, B, C, D on a great circle of radius R = 2, ∠ AOB = 90°
2STEP 2

Find the chord between neighbors

In the great-circle plane OA = OB = 2 with ∠ AOB = 90°, so the Pythagorean theorem gives the chord AB = 2√(2).

AB² = OA² + OB² = 2² + 2² = 8 → AB = 2√(2)
3STEP 3

Turn the chord into a radius

The chord AB is each semicircle's diameter, so the radius is half of it: r = √(2).

r = AB/2 = 2√(2)/2 = √(2)
4STEP 4

Find one semicircle's length

One semicircle is half a circle of radius √(2), so its arc length is π r = π√(2).

L_semi = π r = π√(2)
5STEP 5

Add the four semicircles

Four congruent arcs total 4π√(2) = π√(32), and matching π√(n) reads off n.

L_total = 4 · π√(2) = π√(32), so n = 32 → (A) 32
Answer
32
Cross-check the magnitudes: the sphere's full great-circle circumference is 2π R = 4π, so our curve length 4π√(2) ≈ 5.66π is bigger than one great circle — sensible, since the four semicircles bulge around the sphere rather than tracing the great circle exactly. The n = 32 choice matches perfectly: (4√(2))² = 32. Also 32 is the only answer choice whose square root produces a rational multiple of √(2), which the geometry forces.
💡Key takeaway

Symmetry forces the four junction points to sit at the corners of a square inscribed in a great circle of the sphere. The Pythagorean theorem in the great-circle plane gives the chord between adjacent corners as 2√(2), so each semicircle has radius √(2) and arc length π√(2). Four of them total 4π√(2) = π√(32), so n = (A) 32.

  • Locate the four junction points
  • Find the chord between neighbors
  • Turn the chord into a radius
  • Find one semicircle's length
  • Add the four semicircles

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