1968 IMO Problems/Problem 4
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.
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.
|1968 IMO (Problems) • Resources)|
|1 • 2 • 3 • 4 • 5 • 6||Followed by|
|All IMO Problems and Solutions|