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2011 AMC 10A Problems/Problem 24

Revision as of 12:10, 9 February 2014 by Saphira 7 (talk | contribs) (→‎Solution 2)

Problem 24

Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?

$\text{(A)}\,\frac{1}{12} \qquad\text{(B)}\,\frac{\sqrt2}{12} \qquad\text{(C)}\,\frac{\sqrt3}{12} \qquad\text{(D)}\,\frac{1}{6} \qquad\text{(E)}\,\frac{\sqrt2}{6}$

Solution

The two tetrahedra look somewhat like this.CubeAndStel.gif

Solution 1

A regular unit tetrahedron can be split into eight tetrahedra that have lengths of $\frac{1}{2}$. The volume of a regular tetrahedron can be found using base area and height:

For a tetrahedron of side length 1, its base area is $\frac{\sqrt{3}}{4}$, and its height can be found using Pythagoras' Theorem. Its height is $\sqrt{1^2-\left(\frac{\sqrt3}{3}\right)^2}=\frac{\sqrt2}{\sqrt3}$. Its volume is $\frac13\times\frac{\sqrt{3}}{4}\times\frac{\sqrt2}{\sqrt3}=\frac{\sqrt{2}}{12}$.

The tetrahedron actually has side length $\sqrt2$, so the actual volume is $\frac{\sqrt{2}}{12}\times\sqrt2^3=\frac13$.

On the eight small tetrahedra, the four tetrahedra on the corners of the large tetrahedra are not inside the other large tetrahedra. Thus, $\frac{4}{8}=\frac{1}{2}$ of the large tetrahedra will not be inside the other large tetrahedra.

The intersection of the two tetrahedra is thus $\frac12\times\frac13=\frac{1}{6}=\boxed{\text{(D)}}$.

Solution 2

The intersection of the two tetrahedra is an octahedron that has points touching the center of each face of the original cube. We can split the octahedron into two square pyramids, and from there we can find the volume of each pyramid. The sides of the square face of the pyramid will have lengths of $\frac{\sqrt2}{2}$, so the volume of both pyramids (the whole octahedra) will be $(\frac{\sqrt2}{2})^2\times2\times\frac{1}{2}\times\frac{1}{3}=1/6=\boxed{\text{(D)}}$.

See Also

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AMC 10 Problems and Solutions

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