2003 AIME II Problems/Problem 4
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 , where and are relatively prime positive integers. Find .
Embed the tetrahedron in 4-space to make calculations easier. Its vertices 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: ,,,.
The side length of the large tetrahedron is by the distance formula. The side length of the smaller tetrahedron is by the distance formula.
Their ratio is , so the ratio of their volumes is .
Let the large tetrahedron be , and the small tetrahedron be , with on , on , on , and on . 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 , for our convenience. Dropping an altitude from to , and calling the foot , we have . Since . By Law of Cosines, we have . Hence, the ratio of the volumes is .
Consider the large tetrahedron and the smaller tetrahedron . Label the points as you wish, but dropping an altitude from the top vertex of , we see it hits the center of the base face of . This center is also one vertex of . Consider a "side" face of , and the center of that face, which is another vertex of . 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 (centroid), and since the bases of and are parallel, we can say that the altitudes of tetrahedron and are in the ratio . Thus we compute , and find . The sum of the numerator and denominator is thus .
Video Solution by Sal Khan
|2003 AIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|