AMC 10 · 2007 · #21
Grade 8 geometry-3dA sphere is inscribed in a cube that has a surface area of 24 square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The outer cube has surface area $24$ square meters; A sphere is inscribed in the outer cube (it touches all six faces from inside); A second cube is inscribed in the sphere (all eight corners lie on the sphere); Answer choices: (A) $3$, (B) $6$, (C) $8$, (D) $9$, (E) $12$
Unknowns: The surface area, in square meters, of the inner cube
Understand
Restated: 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.
Givens: The outer cube has surface area $24$ square meters; A sphere is inscribed in the outer cube (it touches all six faces from inside); A second cube is inscribed in the sphere (all eight corners lie on the sphere); Answer choices: (A) $3$, (B) $6$, (C) $8$, (D) $9$, (E) $12$
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #7 Identify Subproblems, #1 Draw a Diagram, #4 Introduce a Variable
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.
Execute — Answer: C
6.G.A.4 Step 1 Find the outer cube's edge
- A cube has $6$ identical square faces, so each face has area $24 \div 6 = 4$ square meters.
- A square with area $4$ has side $\sqrt{4} = 2$, so the outer cube's edge is $2$ meters.
💡 Surface area is just six equal faces, so dividing by six and taking a square root walks straight back to the edge.
7.G.B.6 Step 2 Link the three shapes with one length
- The sphere is inscribed in the outer cube, so it touches each face at its center; the distance across the sphere is exactly the gap between two opposite faces, which is the cube's edge.
- So the sphere's diameter is $2$.
- The inner cube is inscribed in the sphere, so its eight corners all sit on the sphere; the longest diagonal of the inner cube (corner to opposite corner) stretches across the sphere and equals that same diameter.
- Therefore the inner cube's space diagonal is $2$.
💡 One straight length threads through all three shapes, so measuring it once fixes it everywhere.
8.G.B.7 Step 3 Write the inner cube's space diagonal
- Let $s$ be the inner cube's edge.
- Reach an opposite corner in two Pythagorean hops.
- First, a diagonal across one face joins two corners of a square with side $s$, so its length squared is $s^2 + s^2 = 2s^2$.
- Second, the space diagonal is the hypotenuse of a right triangle whose legs are that face diagonal and one more edge $s$, so its length squared is $2s^2 + s^2 = 3s^2$.
- Thus the space diagonal is $\sqrt{3s^2} = s\sqrt{3}$.
💡 A corner-to-corner reach through a box is just the Pythagorean theorem done twice, once across a face and once up to the far corner.
8.EE.A.2 Step 4 Solve for the inner edge squared
- The space diagonal equals $2$, so set $s\sqrt{3} = 2$.
- Squaring both sides gives $3s^2 = 4$, so $s^2 = \dfrac{4}{3}$.
- Notice we only need $s^2$, never $s$ itself, because surface area is built from $s^2$.
💡 Squaring the diagonal equation trades an awkward square root for a clean value of $s^2$, which is exactly what the area needs.
6.G.A.4 Step 5 Compute the inner surface area
- The inner cube's surface area is $6$ faces, each of area $s^2$, so it is $6s^2 = 6 \cdot \dfrac{4}{3} = 8$ square meters.
- That is choice (C).
💡 Because surface area is six times $s^2$, having $s^2$ in hand finishes the problem in one multiplication.
6.G.A.4 A cube has $6$ identical square faces, so each face has area $24 \div 6 = 4$ squ 7.G.B.6 The sphere is inscribed in the outer cube, so it touches each face at its center 8.G.B.7 Let $s$ be the inner cube's edge. Reach an opposite corner in two Pythagorean ho 8.EE.A.2 The space diagonal equals $2$, so set $s\sqrt{3} = 2$. Squaring both sides gives 6.G.A.4 The inner cube's surface area is $6$ faces, each of area $s^2$, so it is $6s^2 = Review
Reasonableness: 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 $\sqrt{3}$ times its edge, so the inner edge is $\frac{2}{\sqrt{3}}$ and the edge ratio is $\frac{1}{\sqrt{3}}$. Areas scale as the square of that ratio, $\frac{1}{3}$, and indeed $\frac{1}{3} \cdot 24 = 8$. Both routes agree, so $8$ is correct.
Alternative: Skip the inner edge entirely with the similarity shortcut. The two cubes are similar, and the length that survives all the way through is $2$: it is the outer cube's edge and also the inner cube's space diagonal. For any cube, edge and space diagonal differ by the factor $\sqrt{3}$, so the inner cube is scaled down by $\frac{1}{\sqrt{3}}$ in length and therefore by $\left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$ in area. So the inner surface area is $\frac{1}{3}$ of $24$, which is $8$, giving (C) without ever solving for $s$.
CCSS standards used (min grade 8)
6.G.A.4Represent 3D figures with nets and find surface area (Turning surface area into face area for the outer cube ($24 \div 6 = 4$, so edge $2$), and rebuilding the inner cube's surface area from $s^2$ as $6s^2 = 8$.)7.G.B.6Solve problems involving surface area and volume of three-dimensional objects (Recognizing that the inscribed sphere's diameter equals the outer cube's edge and also the inner cube's space diagonal, linking all three shapes with one length.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Applying the Pythagorean theorem twice inside the inner cube to get its space diagonal $s\sqrt{3}$ from the edge $s$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Squaring $s\sqrt{3} = 2$ to solve $3s^2 = 4$ for $s^2 = \frac{4}{3}$.)
⭐ 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.
⭐ 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.
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