2001 USAMO Problems/Problem 4
Contents
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
Solution 1
We know that and we wish to prove that . It would be sufficient to prove that Set , , , . Then, we wish to show
which is true by the trivial inequality.
Solution 2
Let be the origin. For a point , denote by the vector , and denote by the length of . The given conditions may be written as or Adding on both sides of the last inequality gives Since the left-hand side of the last inequality is nonnegative, the right-hand side is positive. Hence that is, is acute.
Solution 3
For the sake of contradiction, let's assume to the contrary that . Let , , and . Then . We claim that the quadrilateral is convex. Now applying the generalized Ptolemy's 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 an 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 lines and be the perpendicular bisectors of segments and , respectively. Then and meet at , the circumcenter of triangle . Lines and cut the plane into four regions 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 , 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 inside or on the sides of triangle . Since this covers all four cases, is convex.
Solution 4
Let be the origin in vector space, and let denote the position vectors of respectively. Then the obtuse triangle condition, , becomes using the fact that the square of a vector (the dot product of itself and itself) is the square of its magnitude. Now, notice that to prove is acute, it suffices to show that , or . But this follows from the observation that which leads to and therefore our desired conclusion.
Solution 5
Let be midpoints of and , respectively. For the points ; let's apply Euler's quadrilateral formula, Given that . Thus, and we get is acute.
(Lokman GÖKÇE)
Solution 6
Without loss of generality, assume that in a Cartesian coordinate system, is at the point and is at the point . Let be at the point and be at the point . Without loss of generality, also assume that .
Now, assume for contradiction that is not acute. Since , , and are the sides of an obtuse triangle, with the longest side, it follows that , implying that . This inequality simplifies to . Note that since and are both perfect squares, all terms of this inequality except for are already guaranteed to be nonnegative.
If , then would be closer to than to , but since , this is not possible. Therefore, . Since not being acute implies that , it follows that . But now since all terms of are guaranteed to be nonnegative, this entire expression cannot be negative, leading to a contradiction. Therefore, is acute.
See also
2001 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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