Problem:
Let be a point in the plane of triangle such that the segments , and are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to . Prove that is acute.
Solution:
First Solution: Let be the origin. For a point . denote by the vector . and denote by the length of . The given condictions may be written as
or
Adding on both sides of the last inequality gives
Since the left-hand side of the last inequality is nomnegative, the right-luand side is positive. Hence
that is. is acute.
Second Solution: For the sake of contradiction. let's assume to the contrary that is not acute. Let , and . Then . We claim that the quadrilateral is convex. Now applying the generalized Ptolemys Theorem to the convex quadrilateral yields
where the second inequality is by Cauchy-Schwarz. This implies . in contradiction with the facts that , and are the sides of on obtuse triangle and .
We present two arguments to prove our claim.
First argument Without loss of generality, we may assume that . and are in counterclockwise order. Let line and be the perpendicular bisectors of segments and , respectively. Then and meet at , the circumcenter of triangle . Lines and cut the plane into four rewions and is in the interior of one of these regions. Since and must be in the interior of the region that opposes . Since is not acute, ray does not meet and ray does not meet . Hence and must lie in the interiors of the regions adjacent to . Let denote the region containing . Then , and are the four regions in counterclockwise order. Since . either is on side or and are on opposite sides of line . In either case and are on opposite sides of line . Also, since ray does not meet and ray does not meet , it follows that is entirely in the interior of . Hence and are on opposite sides of . Therefore is convex.
Second argument: Since and . A cannot be inside or on the sides of triangle . Since , we have and hence . Hence cannot be inside or on the sides of triangle . Symmetrically, cannot be inside or on the sides of triangle . Finally since and , we have
Therefore cannot be in inside or on the sides of triangle . Since this covers all four cases, is convex.
The problems on this page are the property of the MAA's American Mathematics Competitions