Question

A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?

$\dfrac{\sqrt{3}}{3}$

$\dfrac{2 \sqrt{2}}{3}$

$1$

$\dfrac{2 \sqrt{3}}{3}$

$\sqrt{2}$

Solution:

The discarded tetrahedron has a right isosceles triangle of leg $1$ as one base and height $1$ , so its volume is $\dfrac13\cdot\dfrac12\cdot1=\dfrac16$ .

The cut face is an equilateral triangle of side $\sqrt2$ , so its area is $\dfrac{\sqrt3}{4}(\sqrt2)^2=\dfrac{\sqrt3}{2}$ . If $h$ is the height from the opposite vertex to this cut face, then $\dfrac13\cdot\dfrac{\sqrt3}{2}\cdot h=\dfrac16$ , so $h=\dfrac{\sqrt3}{3}$ .

The full cube diagonal has length $\sqrt3$ . After the tetrahedron is removed and the cut face is placed down, the height is $\sqrt3-\dfrac{\sqrt3}{3}=\dfrac{2\sqrt3}{3}$ .

Thus, D is the correct answer.

2012 AMC 10B Problem 23

Solution:

Problem 23 in Other Years