AMC 10 · 2011 · #22

Grade 8 geometry-3d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
spatial-visualizationsimilar-trianglespythagorean-theoremvolume-rectangular-prism identify-subproblems ↑ Prerequisites: similar-triangles
📏 Medium solution 💡 3 insights
Problem
A pyramid has a square base of side 1, and every triangular side face is equilateral. A cube sits inside with its bottom face flat on the base, and the four edges of its top face each lie on a slanted face of the pyramid. Find the cube's volume.

Pick an answer.

(A)
$5\sqrt{2} - 7$
(B)
$7 - 4\sqrt{3}$
(C)
$\frac{2\sqrt{2}}{27}$
(D)
$\frac{\sqrt{2}}{9}$
(E)
$\frac{\sqrt{3}}{9}$

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

How to solve
Strategy Draw a Diagram

A 3D pyramid-and-cube picture is hard to reason about directly, so the winning move (Tool #1, backed by Tool #9) is to slice the solid with a single flat cut and turn the whole problem into a 2D triangle diagram. The right cut passes through the apex and two opposite corners of the base; in that plane the pyramid becomes a clean triangle and the cube becomes a rectangle. Spatial visualization (Tool #17) tells us which cut to take and why the cube's tall corners land on the pyramid's slanted edges. Then naming the cube's edge s (Tool #4) turns 'corner sits on the slant' into one equation we can solve.

1STEP 1

Find the pyramid's height

Every slanted edge is 11 and the center sits 22\frac{\sqrt{2}}{2} from a corner, so Pythagoras gives height H=22H=\frac{\sqrt{2}}{2}.

H=√(1-1/2)=√(2)/2
2STEP 2

Slice through the apex and two opposite corners

The cut through the apex and two opposite corners shows a 11, 11, 2\sqrt{2} triangle — right isosceles, so its base angles are 45∘45^\circ.

1²+1²=(√(2))² → right isosceles, base angles 45°
3STEP 3

Place the cube in the 2D picture

Set the diagonal on the xx-axis so the left slant is y=xy=x; the centered cube is a rectangle s2s\sqrt{2} wide and ss tall, top corner on that line.

top-left corner=(√(2)/2-s√(2)/2, s) on y=x
4STEP 4

Solve for the cube's edge

That corner has equal height and horizontal position, so 2s+s2=22s+s\sqrt{2}=\sqrt{2}, and rationalizing gives s=2−1s=\sqrt{2}-1.

s=√(2)/(2+√(2))=√(2)-1
5STEP 5

Cube the edge to get the volume

Cube it: (2−1)2=3−22(\sqrt{2}-1)^2=3-2\sqrt{2}, and multiplying once more by 2−1\sqrt{2}-1 leaves 52−75\sqrt{2}-7, choice (A).

V=(√(2)-1)³=5√(2)-7 → (A)
Answer
5√(2) - 7
Estimate the size: √(2)≈1.414, so the edge s=√(2)-1≈0.414 and the volume ≈0.414³≈0.071. That is comfortably less than 1, which it must be since the cube fits inside a pyramid whose base is only 1×1. Checking the answer choices, 5√(2)-7≈5(1.414)-7≈0.071 matches, while (C), (D), (E) are near 0.10–0.19 (too big for a cube of edge 0.414) and (B) 7-4√(3)≈0.072 is close but comes from a 30°-flavored setup, not the 45° triangle we actually get.
💡Key takeaway

Slice a hard 3D shape with one flat cut, and it often turns into a simple 2D triangle you can actually solve.

  • Find the pyramid's height
  • Slice through the apex and two opposite corners
  • Place the cube in the 2D picture
  • Solve for the cube's edge
  • Cube the edge to get the volume

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