1966 IMO Problems/Problem 3
Prove that the sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space.
Solution
We will need the following lemma to solve this problem:
Suppose there is a point in a regular tetrahedron such that the distances from this point to the faces , , , and are, respectively, , , , and . Then, the value is constant.
We will compute the volume of in terms of the areas of the faces and the distances from the point to the faces:
This value is constant, so the proof of the lemma is complete.
Let our tetrahedron be , and the center of its circumscribed sphere be . Construct a new regular tetrahedron, , such that the centers of the faces of this tetrahedron are at , , , and .
For any point in ,
with equality only occurring when , , , and are perpendicular to the faces of , meaning that . This completes the proof.
~mathboy100
See Also
1966 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |