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Difference between revisions of "2003 AIME II Problems/Problem 4"
(Added alternative solution)
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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 <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
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 <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
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__TOC__
 
== Solution ==
 
== Solution ==
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=== Solution 1 ===
 
Embed the tetrahedron in 4-space (It makes the calculations easier)
 
Embed the tetrahedron in 4-space (It makes the calculations easier)
 
It's vertices are
 
It's vertices are
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<math>m+n = 1 + 27 = \boxed{028}</math>
 
<math>m+n = 1 + 27 = \boxed{028}</math>
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=== Solution 2 ===
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 +
Let the large tetrahedron be <math>ABCD</math>, and the small tetrahedron be <math>WXYZ</math>, with <math>W</math> on <math>ABC</math>, <math>X</math> on <math>BCD</math>, <math>Y</math> on <math>ACD</math>, and <math>Z</math> on <math>ABD</math>. 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 <math>AB=1</math>, for our convenience. Dropping an altitude from <math>W</math> to <math>BC</math>, and calling the foot <math>M</math>, we have <math>WM=XM=\frac{\sqrt3}{6}</math>. Since <math>\cos\angle{WMX}=\cos\angle{AMX}=MX/AM=1/3</math>. By Law of Cosines, we have <math>WX=\sqrt{1/12+1/12-2(1/12)(1/3)}=1/3</math>. Hence, the ratio of the volumes is <math>(\frac{1}{3})^3=1/27</math>. <math>m+n=1+27=\boxed{028}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2003|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2003|n=II|num-b=3|num-a=5}}

Revision as of 18:48, 31 December 2008

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 $(\frac{1}{3})^3=1/27$. $m+n=1+27=\boxed{028}$

See also

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