AMC 10 · 2012 · #23

Grade 8 geometry-3d

Archetype Solid Volume by Decomposition └ Sub-archetype Fraction or Difference of a Bounding Solid

spatial-visualizationpythagorean-theoremequilateral-triangle identify-subproblems ↑ Prerequisites: spatial-visualization

📏 Long solution 💡 4 insights

Problem

A wooden unit cube has a corner sliced off by one flat cut through three of its vertices: two diagonally opposite corners of one face and the far corner of the opposite face that touches neither of those two. The sliced-off tetrahedron is thrown away, and the leftover solid is set on a table resting on the flat triangular cut. How tall does the leftover solid stand?

Pick an answer.

(A)

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

(B)

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

(C)

(D)

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

(E)

$\sqrt{2}$

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Visualize Spatial Relationships

The hard part is picturing which corner gets shaved and how the flipped solid sits. Once you see that the slice removes a single cube corner and that the cube's long diagonal runs straight through that corner perpendicular to the cut face, the height is just that diagonal minus the sliver the cut removes. Setting coordinates and breaking the sliver's height out of a two-way volume turns the spatial picture into a short calculation.

1STEP 1

Locate the sliced-off corner

Set the cube as $[0,1]^3$ ; the cut hits $P=(1,0,0)$ , $Q=(0,1,0)$ , $R=(1,1,1)$ — the neighbors of $O=(1,1,0)$ , so tetrahedron $OPQR$ goes.

O=(1,1,0), P=(1,0,0), Q=(0,1,0), R=(1,1,1)

2STEP 2

Measure the cut triangle

$OP$ , $OQ$ , $OR$ are perpendicular unit edges, so every side of $PQR$ is $\sqrt{2}$ — the cut face is equilateral.

PQ=QR=RP=√(1²+1²)=√(2)

3STEP 3

Find the tall diagonal through the corner

Across the cube from $O$ sits $S=(0,0,1)$ ; the space diagonal $OS$ has length $\sqrt{3}$ and meets the cut face at a right angle.

OS=√(1²+1²+1²)=√(3)

4STEP 4

Area of the equilateral cut face

An equilateral triangle of side $s$ has area $\frac{\sqrt{3}}{4}s^2$ ; with $s^2=2$ the cut face has area $\frac{\sqrt{3}}{2}$ .

[PQR]=√(3)/4 (√(2))²=√(3)/2

5STEP 5

Height of the discarded tetrahedron

Volume of $OPQR$ two ways: $\frac{1}{6}=\frac{1}{3}\cdot\frac{\sqrt{3}}{2}\cdot h$ , so the drop from $O$ to the cut face is $\frac{\sqrt{3}}{3}$ .

1/6=1/3·√(3)/2 · h → h=1/√(3)=√(3)/3

6STEP 6

Subtract along the diagonal

Subtracting along the diagonal, $\sqrt{3}$ minus $\frac{\sqrt{3}}{3}$ leaves $\frac{2\sqrt{3}}{3}$ — choice (D).

√(3)-√(3)/3=(3√(3)-√(3))/3=2√(3)/3

Answer

(2 √(3))/3

The height 2√3/3 is about 1.155 — a little taller than the cube's own edge of 1, yet well below the full space diagonal √3 ≈ 1.732. That fits: only a small corner sliver was removed, so the flipped object should stand almost as tall as the whole diagonal, but not quite. It also equals 2/√3, matching a direct check that the farthest leftover vertex S=(0,0,1) sits a perpendicular distance (|0+0-1-1|)/√3 = 2/√3 from the cutting plane x+y-z=1.

💡Key takeaway

Slice a corner off a cube, and the leftover block stands as tall as the cube's long diagonal minus the little bit the slice removed.

Locate the sliced-off corner
Measure the cut triangle
Find the tall diagonal through the corner
Area of the equilateral cut face
Height of the discarded tetrahedron
Subtract along the diagonal