1972 USAMO Problems/Problem 2
Problem
A given tetrahedron is isosceles, that is, . Show that the faces of the tetrahedron are acute-angled triangles.
Solutions
Solution 1
Suppose is fixed. By the equality conditions, it follows that the maximal possible value of occurs when the four vertices are coplanar, with on the opposite side of as . In this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable.
For the sake of contradiction, suppose is non-acute. Then, . In our optimal case noted above, is a parallelogram, so However, as stated, equality cannot be attained, so we get our desired contradiction.
Solution 2
It's not hard to see that the four faces are congruent from SSS Congruence. Without loss of generality, assume that . Now assume, for the sake of contradiction, that each face is non-acute; that is, right or isosceles. Consider triangles and . They share side . Let and be the planes passing through and , respectively, that are perpendicular to side . We have that triangles and are non-acute, so and are not strictly between planes and . Therefore the length of is at least the distance between the planes, which is . However, if , then the four points , , , and are coplanar, and the volume of would be zero. Therefore . However, we were given that in the problem, which leads to a contradiction. Therefore the faces of the tetrahedron must all be acute.
Solution 3
Let , , and . The conditions given translate to We wish to show that , , and are all positive. WLOG, , so it immediately follows that and are positive. Adding all three equations, In addition, Equality could only occur if , which requires the vectors to be coplanar and the original tetrahedron to be degenerate.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1972 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |