AMC 10 · 2012 · #19

Grade 8 geometry-3d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

spatial-visualizationcoordinate-geometrysymmetry-argumentpythagorean-theorem convert-to-algebraspatial-visualizationcoordinate-geometry ↑ Prerequisites: coordinate-geometrypythagorean-theorem

📏 Long solution 💡 4 insights 📊 Diagram

Problem

A regular eight-faced solid sits inside a cube with one vertex on each of six segments. Find its edge length.

Pick an answer.

(A)

$\frac{3\sqrt{2}}{4}$

(B)

$\frac{7\sqrt{6}}{16}$

(C)

$\frac{\sqrt{5}}{2}$

(D)

$\frac{2\sqrt{3}}{3}$

(E)

$\frac{\sqrt{6}}{2}$

How to solve

Strategy Introduce a Variable

Put the cube on axes with P₁ at the origin (tool #1, tool #4): the six octahedron vertices then need only six unknown distances along the six edges. Spatial reasoning (tool #17) does the heavy lifting twice. First it settles which vertices are opposite in the octahedron — the three near P₁ must be mutually adjacent, so each pairs off with one of the three near P₁'. That immediately forces the three near P₁ to form an equilateral triangle, which collapses all six unknowns into a single a (tool #13). Then one cross-edge distance gives one equation; the squared terms cancel and it is linear (tool #4 again). Finally tool #17 returns to verify the result: two symmetries of the configuration show that one equation really does force all twelve edges equal, and the three-diagonal test confirms the solid is a genuine regular octahedron. Matching the exact value against the list (tool #3) picks the choice — necessary here, because two of the options agree to within 1%.

1STEP 1

Put the cube on axes

Coordinates place all six vertices.

A=(a,0,0), B=(0,b,0), C=(0,0,c); A'=(1-p,1,1), B'=(1,1-q,1), C'=(1,1,1-r)

2STEP 2

Sort out which vertices are opposite

The two triples sit on opposite corners.

A,B,C: x+y+z ≤ 1 A',B',C': x+y+z ≥ 2

3STEP 3

Equal triangles collapse six unknowns to one

Equal triangles collapse six unknowns to one.

a=b=c=p=q=r, s = a√(2)

4STEP 4

Measure one cross edge

One cross edge is measured directly.

AB'² = (1-a)² + (1-a)² + 1² = 2(1-a)² + 1

5STEP 5

The squares cancel — solve for a

The squared terms cancel in the equation.

2(1-a)² + 1 = 2a² → 2a² - 4a + 3 = 2a² → 4a = 3 → a = 3/4

6STEP 6

One equation forces all twelve edges

One equation fixes all twelve edges.

(x,y,z)↦(z,x,y) and (x,y,z)↦(1,1,1)-(x,y,z) preserve the six points

7STEP 7

Confirm it is regular, then read off the answer

The edge is 3√2/4, choice (A).

s = a√(2) = 3/4√(2) = 3√(2)/4 → (A)

Answer

3√(2)/4

Size checks pass: the edge 3√2/4 ≈ 1.061 is less than the cube's face diagonal √2 ≈ 1.414, and the octahedron's longest diagonal s√2 = 3/2 = 1.5 is less than the cube's space diagonal √3 ≈ 1.732, so the solid fits. The octahedron's centre landed on the cube's centre, the symmetry one would expect. Estimation alone cannot finish this problem: (A) ≈ 1.061 and (B) 7√6/16 ≈ 1.072 differ by about 1%, so the exact value is what separates them; (C) ≈ 1.118, (D) ≈ 1.155, (E) ≈ 1.225 are further off. Worth noting where a fast argument can go wrong: forcing just one distance to match is only a necessary condition, and it is the symmetry check in step 6 plus the perpendicular-diagonal test in step 7 that make the answer sufficient — they are what prove such an octahedron exists at all, rather than assuming it.

💡Key takeaway

When a shape is squeezed into a symmetric box, let the symmetry name the unknowns first — here it forced all six vertices to sit the same distance 3/4 from their corner, leaving one easy equation instead of six hard ones.

Put the cube on axes
Sort out which vertices are opposite
Equal triangles collapse six unknowns to one
Measure one cross edge
The squares cancel — solve for a
One equation forces all twelve edges
Confirm it is regular, then read off the answer