AMC 10 · 2007 · #21

Grade 8 geometry-3d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
space-diagonal-formulapythagorean-theoremsurface-areasimilar-figures identify-subproblems ↑ Prerequisites: space-diagonal-formula
📏 Medium solution 💡 3 insights
Problem
A cube has surface area 24 square meters. A sphere fits snugly inside it, just touching all six faces. Then a second, smaller cube fits snugly inside that sphere, with all eight of its corners on the sphere. Find the surface area of this inner cube.

Pick an answer.

(A)
3
(B)
6
(C)
8
(D)
9
(E)
12

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Visualize Spatial Relationships

Nothing here needs clever algebra; the whole problem is a chain of nested shapes, so Tool #17 (Visualize Spatial Relationships) is what unlocks it. The single load-bearing picture is that one length passes through all three shapes: the outer cube's edge equals the sphere's diameter (the sphere touches the faces), and that same diameter equals the inner cube's space diagonal (the corners sit on the sphere). Once you see that, Tool #7 (Identify Subproblems) breaks the work into four small, ordinary steps: outer edge from its surface area, diameter from that edge, inner edge from the diagonal, and inner surface area from the inner edge. Tool #1 (Draw a Diagram) keeps track of which length is which, and Tool #4 (Introduce a Variable) names the inner cube's edge so the space-diagonal relationship becomes an equation.

1STEP 1

Find the outer cube's edge

Six identical faces share the 24, so each face has area 4 and the outer edge is √(4) = 2 meters.

6 a² = 24 → a² = 4 → a = 2
2STEP 2

Link the three shapes with one length

The sphere spans face to face, so its diameter is that edge; the inner cube's corners ride on it, so its space diagonal is also 2.

outer edge = sphere diameter = inner space diagonal = 2
3STEP 3

Write the inner cube's space diagonal

Let s be the inner edge. Pythagoras across one face gives 2s², then up to the far corner gives 3s², so the diagonal is s√(3).

face diagonal² = s² + s² = 2s², space diagonal² = 2s² + s² = 3s²
4STEP 4

Solve for the inner edge squared

Set s√(3) = 2 and square it: 3s² = 4, so s² = 4/3 — the area needs only s², never s itself.

s√(3) = 2 → 3s² = 4 → s² = 4/3
5STEP 5

Compute the inner surface area

Six faces of area s² give 6s² = 6 · 4/3 = 8 square meters, which is choice (C).

6 s² = 6 · 4/3 = 8 → (C)
Answer
8
The inner cube should be a good deal smaller than the outer one, so its surface area should be well under 24; getting 8 fits that expectation. A sharper check compares the two cubes directly: the outer edge is 2 and the inner space diagonal is also 2, but a cube's space diagonal is √(3) times its edge, so the inner edge is 2/√(3) and the edge ratio is 1/√(3). Areas scale as the square of that ratio, 1/3, and indeed 1/3 · 24 = 8. Both routes agree, so 8 is correct.
💡Key takeaway

Track one length as it passes through nested shapes: the big cube's edge is the sphere's width is the small cube's corner-to-corner diagonal, and everything else follows from that single line.

  • Find the outer cube's edge
  • Link the three shapes with one length
  • Write the inner cube's space diagonal
  • Solve for the inner edge squared
  • Compute the inner surface area

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