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 obtuse. 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.
Solution 4
Suppose for the sake of contradiction that is not acute. Since all three sides of triangles and are congruent, those two triangles are congruent, meaning . Construct a sphere with diameter . Since angles and are both not acute, and both lie on or inside the sphere. We seek to make to satisfy the conditions of the problem. This can only occur when is a diameter of the sphere, since both points lie on or inside the sphere. However, for to be a diameter, all four points must be coplanar, as all diameters intersect at the center of the sphere. This would make tetrahedron degenerate, creating a contradiction. Thus, all angles on a face of an isosceles tetrahedron are acute.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.