AMC 10 · 2006 · #24

Grade 8 geometry-3d
Archetype Solid Volume by Decomposition └ Sub-archetype Sum of Primitive Solids
spatial-visualizationpythagorean-theoremformula-substitution identify-subproblems ↑ Prerequisites: spatial-visualizationpythagorean-theorem
📏 Long solution 💡 3 insights
Problem
A cube with edge length 1 has a marked point at the center of each of its 6 faces. Joining the centers of faces that share an edge produces a regular octahedron sitting inside the cube. We want the volume of that octahedron.

Pick an answer.

(A)
$\frac{1}{8}$
(B)
$\frac{1}{6}$
(C)
$\frac{1}{4}$
(D)
$\frac{1}{3}$
(E)
$\frac{1}{2}$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

A whole octahedron has no ready-made volume formula in a young solver's toolkit, so the winning move is Tool #7 (Identify Subproblems): slice the octahedron across its middle into two identical square pyramids, each of which does have a simple volume rule. To slice cleanly you first have to see where the six face centers land, which is Tool #17 (Visualize Spatial Relationships) — put the cube in a coordinate box and the centers snap onto the axes. Tool #1 (Draw a Diagram) isolates the flat 'equator' square so its area is easy to read off, and Tool #8 (Analyze the Units) assembles base area and height into the pyramid-volume formula and doubles it. The problem is really one act of seeing plus one act of splitting; no heavy computation is needed.

1STEP 1

Put the cube on coordinates

Center the cube at the origin: the six face centers land on the axes at (±1/2,0,0), (0,±1/2,0), (0,0,±1/2).

vertices: (±1/2,0,0), (0,±1/2,0), (0,0,±1/2)
2STEP 2

Split into two square pyramids

Four of those points form a square in the plane z=0 and the other two are apexes, so the octahedron is two identical square pyramids.

V_oct = 2 × V_pyramid
3STEP 3

Area of the equator square

Both diagonals of that square join opposite face centers, so each is 1, and area d²/2 gives base B = 1/2.

B = d²/2 = 1²/2 = 1/2
4STEP 4

Height of one pyramid

The base lies at z=0 and the apex sits at (0,0,1/2), so each pyramid's height is h = 1/2, half the cube.

h = 1/2
5STEP 5

Volume of the octahedron

One pyramid is 1/3 · 1/2 · 1/2 = 1/12, and the octahedron is two of them: 1/6, choice (B).

V = 2(1/3 · 1/2 · 1/2) = 2 · 1/12 = 1/6
Answer
1/6
The octahedron lives entirely inside the unit cube, whose volume is 1, and it clearly fills only a modest core of that cube, so a volume of 1/6 ≈ 0.17 is the right order of size — small, but not tiny. A quick regularity check confirms the setup: every edge, whether it joins the apex to an equator vertex like (0,0,1/2) to (1/2,0,0), or two equator vertices like (1/2,0,0) to (0,1/2,0), has length √((1/2)²+(1/2)²)=1/√2, so all 12 edges match and the solid really is a regular octahedron. The trap answers come from dropping the 1/3 pyramid factor: base × height for both pyramids gives 2·1/2·1/2=1/2, which is (E), and doing it for just one gives 1/4, which is (C).
💡Key takeaway

Drop the cube onto coordinates so the face centers sit on the axes, cut the octahedron into two square pyramids with base area 1/2 and height 1/2, and add them: 2·1/3·1/2·1/2=1/6.

  • Put the cube on coordinates
  • Split into two square pyramids
  • Area of the equator square
  • Height of one pyramid
  • Volume of the octahedron

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