1971 IMO Problems/Problem 4
Problem
All the faces of tetrahedron are acute-angled triangles. We consider all closed polygonal paths of the form defined as follows: is a point on edge distinct from and ; similarly, are interior points of edges , respectively. Prove:
(a) If , then among the polygonal paths, there is none of minimal length.
(b) If , then there are infinitely many shortest polygonal paths, their common length being , where .
Solution
Rotate the triangle around the edge until are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting and . Therefore, . Summing the four equations like this, we get exactly .
Now, draw all four faces in the plane, so that is constructed on the exterior of the edge of and so on with edges and .
The final new edge (or rather ) is parallel to the original one (because of the angle equation). Call the direction on towards "right" and towards "left". If we choose a vertex on and connect it to the corresponding vertex on A'B'. This works for a whole interval of vertices if lies to the left of and and lies to the right of . It is not hard to see that these conditions correspond to the fact that various angles are acute by assumption.
Finally, regard the sine in half the isosceles triangle which gives the result with the angles around instead of , but the role of the vertices is symmetric.
See Also
1971 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |