Difference between revisions of "2021 AIME II Problems/Problem 5"
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− | It is possible for | + | It is possible for non-congruent obtuse triangles to have the same area. Therefore, all such positive real numbers <math>s</math> are in exactly one of <math>\left(0,2\sqrt{84}\right)</math> or <math>\left(0,20\right).</math> Taking the exclusive disjunction, the set of all such <math>s</math> is <cmath>[a,b)=\left(0,2\sqrt{84}\right)\oplus\left(0,20\right)=\left[2\sqrt{84},20\right),</cmath> from which <math>a^2+b^2=\boxed{736}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 16:36, 23 March 2021
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.
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
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.
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
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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