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

Solution 1

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 $\left(\frac{1}{3}\right)^3 = \frac{1}{27}$

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

Solution 2

Let the large tetrahedron be $ABCD$ , and the small tetrahedron be $WXYZ$ , with $W$ on $ABC$ , $X$ on $BCD$ , $Y$ on $ACD$ , and $Z$ on $ABD$ . Clearly, the two regular tetrahedrons are similar, so if we can find the ratio of the sides, we can find the ratio of the volumes. Let $AB=1$ , for our convenience. Dropping an altitude from $W$ to $BC$ , and calling the foot $M$ , we have $WM=XM=\frac{\sqrt3}{6}$ . Since $\cos\angle{WMX}=\cos\angle{AMX}=MX/AM=1/3$ . By Law of Cosines, we have $WX=\sqrt{1/12+1/12-2(1/12)(1/3)}=1/3$ . Hence, the ratio of the volumes is $(\frac{1}{3})^3=1/27$ . $m+n=1+27=\boxed{028}$

2003 AIME II (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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