1968 IMO Problems/Problem 4
Problem
Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which are the sides of a triangle.
Solution
Let the edges of one of the faces of the tetrahedron have lengths , , and . Let , , and be the lengths of the sides that are not adjacent to the sides with lengths , , and , respectively.
Without loss of generality, assume that . I shall now prove that either or , by proving that if , then .
Assume that . The triangle inequality gives us that , so must be greater than . We also have from the triangle inequality that . Therefore . Therefore either or .
If , then the vertex where the sides of length , , and meet satisfies the given condition. If , then the vertex where the sides of length , , and meet satisfies the given condition. This proves the statement.
See Also
1968 IMO (Problems) • Resources) | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |