Difference between revisions of "2021 AIME II Problems/Problem 5"
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Taking the intersection produces <math>6<x<\sqrt{84}</math> for this case. | Taking the intersection produces <math>6<x<\sqrt{84}</math> for this case. | ||
| − | At <math>x=6,</math> the obtuse triangle degenerates into a straight line with area <math>k=0;</math> at <math>x=\sqrt{84},</math> the obtuse triangle degenerates into a right triangle with area <math>k=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.</math> Together, we obtain <math>0<k<2\sqrt{84},</math> or <math>k\in\left(0,\sqrt{84}\right).</math> | + | At <math>x=6,</math> the obtuse triangle degenerates into a straight line with area <math>k=0;</math> at <math>x=\sqrt{84},</math> the obtuse triangle degenerates into a right triangle with area <math>k=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.</math> Together, we obtain <math>0<k<2\sqrt{84},</math> or <math>k\in\left(0,2\sqrt{84}\right).</math> |
<b>Case (2): The longest side has length <math>\boldsymbol{x.}</math></b> | <b>Case (2): The longest side has length <math>\boldsymbol{x.}</math></b> | ||
Revision as of 15:21, 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 
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 
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
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 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
