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)
Every obtuse triangle must satisfy both of the following:
- Triangle Inequality Theorem: If and are the side-lengths of a triangle with then
- Pythagorean Inequality Theorem: If and are the side-lengths of an obtuse triangle with then
For one such obtuse triangle, let and be the side-lengths and be the area. We will use casework on the longest side:
Case (1): The longest side has length
By the Triangle Inequality Theorem, ...
By the Pythagorean Inequality Theorem, ...
Case (2): The longest side has length \boldsymbol{x.}
By the Triangle Inequality Theorem, ...
By the Pythagorean Inequality Theorem, ...
Solution in progress. No edit please
~MRENTHUSIASM
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 | ||
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