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, we have ..., from which ....
By the Pythagorean Inequality Theorem, we get ..., so that ....
Taking the intersection produces ... for this case.
Case (2): The longest side has length
By the Triangle Inequality Theorem, we have ..., from which ....
By the Pythagorean Inequality Theorem, we get ..., so that ....
Taking the intersection produces ... for this case.
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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