2003 AIME II Problems/Problem 4

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Problem

In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

Embed the tetrahedron in 4-space (It makes the calculations easier) It's vertices are $(1,0,0,0)$, $(0,1,0,0)$, $0,0,1,0)$, $(0,0,0,1)$

To get the center of any face, we take the average of the three coordinates of that face. The vertices of the center of the faces are: $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, 0)$,$(\frac{1}{3}, \frac{1}{3},0, \frac{1}{3})$,$(\frac{1}{3},0, \frac{1}{3}, \frac{1}{3})$,$(0,\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$

The side length of the large tetrahedron is $\sqrt{2}$ by the distance formula The side length of the smaller tetrahedron is $\frac{\sqrt{2}}{3}$ by the distance formula

Their ratio is $1:3$, so the ratio of their volumes is $\left(\frac{1}{3}\right)^3 = \frac{1}{27}$

$m+n = 1 + 27 = \boxed{028}$

See also

2003 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AIME Problems and Solutions