AMC 10 · 2011 · #24

Grade 8 geometry-3d
Archetype Solid Volume by Decomposition └ Sub-archetype Fraction or Difference of a Bounding Solid
spatial-visualizationvolume-rectangular-prismpythagorean-theorem physical-representationidentify-subproblems ↑ Prerequisites: spatial-visualization
📏 Long solution 💡 3 insights
Problem
A unit cube has 8 corners. Two different regular tetrahedra are built using only these corners as their vertices. Find the volume of the solid region that lies inside both tetrahedra at once.

Pick an answer.

(A)
$\frac{1}{12}$
(B)
$\frac{\sqrt2}{12}$
(C)
$\frac{\sqrt3}{12}$
(D)
$\frac{1}{6}$
(E)
$\frac{\sqrt2}{6}$

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Visualize Spatial Relationships

This is a pure 3D picture problem, so the main move is to see the two tetrahedra sitting in the cube and picture where they overlap. Placing the cube on coordinates pins down every vertex. Seeing that the shared region is an octahedron whose corners are the cube's face centers is the key insight, and then the volume becomes an easier subproblem: cut the octahedron into two simple square pyramids.

1STEP 1

Pick the two tetrahedra by alternating corners

Two-color the cube's 8 corners by even or odd coordinate sum: the 4 even corners form one tetrahedron, the 4 odd ones the other.

Tetra 1:(0,0,0),(1,1,0),(1,0,1),(0,1,1) Tetra 2:(1,0,0),(0,1,0),(0,0,1),(1,1,1)
2STEP 2

Each edge is a face diagonal of length root two

Any two of its corners differ in exactly two coordinates, so all six edges are face diagonals of the same length root two.

√(1²+1²+0²)=√(2) → all 6 edges=√(2)
3STEP 3

See the overlap as an octahedron at the face centers

On every face the two tetrahedra's edges cross at its center, so the shared region is an octahedron with the 6 face centers as corners.

corners=6 face centers: (1/2,1/2,0),(1/2,1/2,1),(1/2,0,1/2),(1/2,1,1/2),(0,1/2,1/2),(1,1/2,1/2)
4STEP 4

Find the square through the middle of the octahedron

Slice halfway up: the 4 side-face centers form a square whose two diagonals are both 1, so its area is one half.

d₁=d₂=1 → square area=(d₁ · d₂)/2=1/2
5STEP 5

Add the two square pyramids

That square splits it into two pyramids of height one half, each one twelfth, so the overlap is 1/6 — choice (D).

V=2·1/3·1/2·1/2=1/6
Answer
1/6
One tetrahedron has volume one third (the unit cube with four corner pieces of volume one sixth each cut off). The overlap must be smaller than a single tetrahedron, and one sixth is indeed less than one third and less than the cube's volume of 1, so the size is sensible. The clean value 1/6 matches choice (D).
💡Key takeaway

Two tetrahedra sharing a cube overlap in an octahedron whose corners are the cube's face centers; split it into two square pyramids and the volume is one sixth.

  • Pick the two tetrahedra by alternating corners
  • Each edge is a face diagonal of length root two
  • See the overlap as an octahedron at the face centers
  • Find the square through the middle of the octahedron
  • Add the two square pyramids

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