AMC 10 · 2006 · #24
Grade 8 geometry-3dCenters of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?
Pick an answer.
AMC 10 2006 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 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.
Givens: The cube is a unit cube (every edge has length $1$); The octahedron's $6$ vertices are the centers of the cube's $6$ faces; Edges of the octahedron join centers of adjacent (edge-sharing) faces, and the result is a regular octahedron; Answer choices: (A) $\tfrac{1}{8}$, (B) $\tfrac{1}{6}$, (C) $\tfrac{1}{4}$, (D) $\tfrac{1}{3}$, (E) $\tfrac{1}{2}$
Unknowns: The volume of the octahedron
Understand
Restated: 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.
Givens: The cube is a unit cube (every edge has length $1$); The octahedron's $6$ vertices are the centers of the cube's $6$ faces; Edges of the octahedron join centers of adjacent (edge-sharing) faces, and the result is a regular octahedron; Answer choices: (A) $\tfrac{1}{8}$, (B) $\tfrac{1}{6}$, (C) $\tfrac{1}{4}$, (D) $\tfrac{1}{3}$, (E) $\tfrac{1}{2}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #17 Visualize Spatial Relationships, #1 Draw a Diagram, #8 Analyze the Units
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.
Execute — Answer: B
8.G.B.8 Step 1 Put the cube on coordinates
- Center the unit cube at the origin, so it runs from $-\tfrac12$ to $\tfrac12$ along each axis.
- Then each face center lies on an axis: the two faces $x=\pm\tfrac12$ give centers $(\pm\tfrac12,0,0)$, the faces $y=\pm\tfrac12$ give $(0,\pm\tfrac12,0)$, and the faces $z=\pm\tfrac12$ give $(0,0,\pm\tfrac12)$.
- These six points are the octahedron's vertices.
- Notice four of them, $(\pm\tfrac12,0,0)$ and $(0,\pm\tfrac12,0)$, all sit in the flat plane $z=0$; the remaining two, $(0,0,\pm\tfrac12)$, stick out above and below that plane.
💡 Centering the cube on the origin makes each face center land neatly on an axis, so the shape is easy to read off by eye.
7.G.B.6 Step 2 Split into two square pyramids
- The four vertices in the plane $z=0$ form a square — call it the octahedron's equator.
- The two remaining vertices, $(0,0,\tfrac12)$ and $(0,0,-\tfrac12)$, are the apexes above and below that square.
- So the whole octahedron is exactly two square pyramids glued base-to-base on that equator square: an upper pyramid with apex $(0,0,\tfrac12)$ and a lower pyramid with apex $(0,0,-\tfrac12)$.
- They are mirror images, hence equal in volume, so I only need one pyramid's volume and then double it.
💡 A shape you can't measure directly often breaks into two copies of a shape you can.
6.G.A.1 Step 3 Area of the equator square
- The equator square has vertices $(\tfrac12,0,0)$, $(0,\tfrac12,0)$, $(-\tfrac12,0,0)$, $(0,-\tfrac12,0)$.
- Its two diagonals run along the axes: from $(\tfrac12,0,0)$ to $(-\tfrac12,0,0)$ is length $1$, and from $(0,\tfrac12,0)$ to $(0,-\tfrac12,0)$ is also length $1$.
- (These diagonals connect centers of opposite cube faces, which are a full edge apart, so each diagonal is $1$.) A square with perpendicular diagonals of length $d$ splits into four right triangles and has area $\tfrac{d^2}{2}$, so the base area is $\tfrac{1^2}{2}=\tfrac12$.
💡 When a square is tilted so its diagonals lie flat, half the product of the diagonals is the tidiest way to get its area.
8.G.B.8 Step 4 Height of one pyramid
- The equator square lies in the plane $z=0$.
- The upper apex is at $(0,0,\tfrac12)$, so its height above that plane is just its $z$-coordinate, $\tfrac12$.
- That vertical distance is the pyramid's height: $h=\tfrac12$.
- (This is exactly half the cube's height, which makes sense — the apex is a top-face center and the equator passes through the cube's middle.)
💡 The apex is one face center and the base sits at the cube's midline, so the climb between them is half a cube.
8.G.C.9 Step 5 Volume of the octahedron
- A pyramid's volume is $\tfrac13$ (base area) $\times$ (height), so one pyramid is $\tfrac13 \cdot \tfrac12 \cdot \tfrac12 = \tfrac{1}{12}$.
- The octahedron is two of these, so its volume is $2 \cdot \tfrac{1}{12} = \tfrac16$.
- That is choice $\textbf{(B)}$.
💡 One-third base times height gives a pyramid, and two pyramids make the whole diamond.
8.G.B.8 Center the unit cube at the origin, so it runs from $-\tfrac12$ to $\tfrac12$ al 7.G.B.6 The four vertices in the plane $z=0$ form a square — call it the octahedron's eq 6.G.A.1 The equator square has vertices $(\tfrac12,0,0)$, $(0,\tfrac12,0)$, $(-\tfrac12, 8.G.B.8 The equator square lies in the plane $z=0$. The upper apex is at $(0,0,\tfrac12) 8.G.C.9 A pyramid's volume is $\tfrac13$ (base area) $\times$ (height), so one pyramid i Review
Reasonableness: 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 $\tfrac16 \approx 0.17$ 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 $(0,0,\tfrac12)$ to $(\tfrac12,0,0)$, or two equator vertices like $(\tfrac12,0,0)$ to $(0,\tfrac12,0)$, has length $\sqrt{(\tfrac12)^2+(\tfrac12)^2}=\tfrac{1}{\sqrt2}$, so all $12$ edges match and the solid really is a regular octahedron. The trap answers come from dropping the $\tfrac13$ pyramid factor: base $\times$ height for both pyramids gives $2\cdot\tfrac12\cdot\tfrac12=\tfrac12$, which is $\textbf{(E)}$, and doing it for just one gives $\tfrac14$, which is $\textbf{(C)}$.
Alternative: Skip the splitting and use the regular-octahedron formula $V=\tfrac{\sqrt2}{3}\,a^3$, where $a$ is the edge length. Here $a=\tfrac{1}{\sqrt2}$, so $a^3=\tfrac{1}{2\sqrt2}$ and $V=\tfrac{\sqrt2}{3}\cdot\tfrac{1}{2\sqrt2}=\tfrac{1}{6}$. Same answer, which is a strong cross-check. A third route: the octahedron is the set of points with $|x|+|y|+|z|\le\tfrac12$, and that solid has volume $\tfrac43(\tfrac12)^3=\tfrac16$ as well.
CCSS standards used (min grade 8)
8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Placing the cube on coordinates so the face centers land on the axes, and reading the pyramid's height as the apex's distance from the base plane.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Decomposing the octahedron into two congruent square pyramids so its volume is twice one pyramid's volume.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Getting the base square's area as half the product of its diagonals, $\tfrac{d^2}{2}=\tfrac12$.)8.G.C.9Know the formulas for volumes of cones, cylinders, and spheres (Applying the one-third-base-times-height volume rule (the pyramid analog of the cone formula) and doubling for the two pyramids.)
⭐ Drop the cube onto coordinates so the face centers sit on the axes, cut the octahedron into two square pyramids with base area $\tfrac12$ and height $\tfrac12$, and add them: $2\cdot\tfrac13\cdot\tfrac12\cdot\tfrac12=\tfrac16$.
⭐ Drop the cube onto coordinates so the face centers sit on the axes, cut the octahedron into two square pyramids with base area $\tfrac12$ and height $\tfrac12$, and add them: $2\cdot\tfrac13\cdot\tfrac12\cdot\tfrac12=\tfrac16$.
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