Difference between revisions of "2003 AIME II Problems/Problem 4"

(Solution)
m (Solution 1)
Line 7: Line 7:
 
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
<math>(1,0,0,0)</math>, <math>(0,1,0,0)</math>, <math>0,0,1,0)</math>, <math>(0,0,0,1)</math>
+
<math>(1,0,0,0)</math>, <math>(0,1,0,0)</math>, <math>(0,0,1,0)</math>, <math>(0,0,0,1)</math>
  
 
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:
 
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:

Revision as of 20:21, 25 June 2012

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

Solution 3

Consider the large tetrahedron $ABCD$ and the smaller tetrahedron $WXYZ$. Label the points as you wish, but dropping an altitude from the top vertex of $ABCD$, we see it hits the center of the base face of $ABCD$. This center is also one vertex of $WXYZ$. Consider a "side" face of $ABCD$, and the center of that face, which is another vertex of $WXYZ$. Draw the altitude of this side face (which is an equilateral triangle). These two altitudes form a right triangle. Since the center of the Side face splits the altitude of the side face into segments in the ratio of $2:1$ (centroid), and since the bases of $WXYZ$ and $ABCD$ are parallel, we can say that the altitudes of tetrahedron $ABCD$ and $WXYZ$ are in the ratio $3:1$. Thus we compute $(\frac{1}{3})^3$, and find $\frac{1}{27}$. The sum of the numerator and denominator is thus $28$.

See also

2003 AIME II (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS