1972 USAMO Problems/Problem 2
A given tetrahedron is isosceles, that is, . Show that the faces of the tetrahedron are acute-angled triangles.
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.
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.
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.
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