2021 AIME II Problems/Problem 5
Contents
Problem
For positive real numbers , let denote the set of all obtuse triangles that have area and two sides with lengths and . The set of all for which is nonempty, but all triangles in are congruent, is an interval . Find .
Solution 1
We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given 4 and 10 as the sides, so we know that the 3rd side is between 6 and 14, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the 3rd side. So the triangles sides are between 6 and exclusive, and the larger bound is between and 14, exclusive. The area of these triangles are from 0 (straight line) to on the first "small bound" and the larger bound is between 0 and 20. is our first equation, and is our 2nd equation. Therefore, the area is between and , so our final answer is .
~ARCTICTURN
Solution 2 (Casework: Detailed Explanation of Solution 1)
If and are the side-lengths of an obtuse triangle with then both of the following must be satisfied:
- Triangle Inequality Theorem:
- Pythagorean Inequality Theorem:
For one such obtuse triangle, let and be its side-lengths and be its area. We apply casework to its longest side:
Case (1): The longest side has length so
By the Triangle Inequality Theorem, we have from which
By the Pythagorean Inequality Theorem, we have from which
Taking the intersection produces for this case.
At the obtuse triangle degenerates into a straight line with area at the obtuse triangle degenerates into a right triangle with area Together, we obtain or
Case (2): The longest side has length so
By the Triangle Inequality Theorem, we have from which
By the Pythagorean Inequality Theorem, we have from which
Taking the intersection produces for this case.
At the obtuse triangle degenerates into a straight line with area at the obtuse triangle degenerates into a right triangle with area Together, we obtain or
Answer
It is possible for non-congruent obtuse triangles to have the same area. Therefore, all such positive real numbers are in exactly one of or Taking the exclusive disjunction, the set of all such is from which
~MRENTHUSIASM
Solution 3
We have the diagram below.
We proceed by taking cases on the angles that can be obtuse, and finding the ranges for that they yield .
If angle is obtuse, then we have that . This is because is attained at , and the area of the triangle is strictly decreasing as increases beyond . This can be observed from by noting that is decreasing in .
Then, we note that if is obtuse, we have . This is because we get when , yileding . Then, is decreasing as increases by the same argument as before.
cannot be obtuse since .
Now we have the intervals and for the cases where and are obtuse, respectively. We are looking for the that are in exactly one of these intervals, and because , the desired range is giving
Solution 4
Note: Archimedes15 Solution which I added an answer here are two cases. Either the and are around an obtuse angle or the and are around an acute triangle. If they are around the obtuse angle, the area of that triangle is as we have and is at most . Note that for the other case, the side lengths around the obtuse angle must be and where we have . Using the same logic as the other case, the area is at most . Square and add and to get the right answer
Solution 5 (Diagrams)
For we fix and Without the loss of generality, we consider on only one side of
As shown below, all locations for at which is an obtuse triangle are indicated in red, excluding the endpoints.
- The region in which is obtuse is determined by construction.
- The region in which is obtuse is determined by the corollaries of the Inscribed Angle Theorem.
For any fixed value of we need obtuse to be unique, or there can only be one possible location for As shown below, all possible locations for are on minor arc including but excluding
Let the brackets denote areas.
- If then will be minimized. By the same base and height and the Inscribed Angle Theorem, we have
- If then will be maximized. For this right triangle, we have
Finally, we get from which
~MRENTHUSIASM (credit given to ...)
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See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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