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 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.
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.
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.
Proof by contradiction: Assume at least one of the tetrahedron's faces are obtuse. WLOG, assume is an obtuse angle. Using SSS congruence to prove that all four faces of the tetrahedron are congruent also shows that the angles surrounding point are congruent to angles in triangle ; namely, is congruent to , and is congruent to . Since the internal angles of triangle must add to 180 degrees, so do the angles surrounding point A. Now lay triangle on a flat surface. A diagram would make it clear that from the perspective of an aerial view, the "apparent" measures of and (which are most likely distorted visually, assuming these angles stick up vertically from the flat surface on which triangle lies) can never exceed the true measures of those angles (equality happens when these angles also lie flat on top of triangle ). This means that these two angles can never join to form side AD (because is more than the sum of and - a direct consequence of the facts that is obtuse and all three angles add up to 180 degrees), so the tetrahedron with obtuse triangle faces is impossible.
Lemma: given triangle and the midpoint of , which we will call , we can say that if , then . Proof: Since is the midpoint of , . Since it is given that , we can substitute to get two inequalities: The above inequalities imply that and . Adding these inequalities and simplifying the RHS, we have that . Adding to both sides, replacing the LHS with and dividing by gets us that . This is our desired inequality, so we are done. Note that all faces of this tetrahedron are congruent, by SSS. In particular, we will use that . WLOG, assume that is the largest angle in triangle . Because , the median from to is equal length to the median from to . These points meet at , the midpoint of . By the triangle inequality, By substituting with (this is a given in the problem) and with , and then dividing by , we get that By the lemma we showed at the start, this implies that , and since we said that was the largest angle, triangle must be acute. Since all of the faces of this tetrahedron are congruent, then, all of the faces must be acute.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
|1972 USAMO (Problems • Resources)
|1 • 2 • 3 • 4 • 5
|All USAMO Problems and Solutions