2012 AMC 10B Problems/Problem 23

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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?

$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2}$


This tetrahedron has the 4 vertices in these positions: on a corner (lets call this $A$) of the cube, and the other three corners ($B$, $C$, and $D$) adjacent to this corner. We can find the height of the remaining portion of the cube by finding the height of the tetrahedron. We can find the height of this tetrahedron in perspective to the equilateral triangle base (we call this height $x$) by finding the area of the tetrahedron in two ways. $\frac{1 \times 1}{2}$ is the area of the isosceles base of the tetrahedron. Multiply by the height, $1$, and divide by $3$, we have the volume of the tetrahedron as $\frac{1}{6}$. We set this area equal to one-third the product of our desired height and the area of the equilateral triangle base. First, find the area of the equilateral triangle: $[BCD]=\frac{\sqrt{2}^2 \times \sqrt{3}}{4}=\frac{\sqrt{3}}{2}$. So we have: $\frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot x=\frac{1}{6}$, and so $x=\frac{\sqrt{3}}{3}$. Now that we have the height, our answer is (A), right? NO! we're asked to find the height of the structure. The height of the structure if the tetrahedron was still on would simply be the space diagonal of the cube, $\sqrt{3}$, so we just subtract $\frac{\sqrt{3}}{3}$ from $\sqrt{3}$ to get $\frac{2\sqrt{3}}{3}$, or $\boxed{\textbf{(D)}}$

2012 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 10 Problems and Solutions

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