Difference between revisions of "2021 AIME II Problems/Problem 5"
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Note: Archimedes15 Solution which I added an answer | Note: Archimedes15 Solution which I added an answer | ||
here are two cases. Either the <math>4</math> and <math>10</math> are around an obtuse angle or the <math>4</math> and <math>10</math> are around an acute triangle. If they are around the obtuse angle, the area of that triangle is <math><20</math> as we have <math>\frac{1}{2} \cdot 40 \cdot \sin{\alpha}</math> and <math>\sin</math> is at most <math>1</math>. Note that for the other case, the side lengths around the obtuse angle must be <math>4</math> and <math>x</math> where we have <math>16+x^2 < 100 \rightarrow x < 2\sqrt{21}</math>. Using the same logic as the other case, the area is at most <math>4\sqrt{21}</math>. Square and add <math>4\sqrt{21}</math> and <math>20</math> to get the right answer <cmath>a^2+b^2= \boxed{736}\Box</cmath> | here are two cases. Either the <math>4</math> and <math>10</math> are around an obtuse angle or the <math>4</math> and <math>10</math> are around an acute triangle. If they are around the obtuse angle, the area of that triangle is <math><20</math> as we have <math>\frac{1}{2} \cdot 40 \cdot \sin{\alpha}</math> and <math>\sin</math> is at most <math>1</math>. Note that for the other case, the side lengths around the obtuse angle must be <math>4</math> and <math>x</math> where we have <math>16+x^2 < 100 \rightarrow x < 2\sqrt{21}</math>. Using the same logic as the other case, the area is at most <math>4\sqrt{21}</math>. Square and add <math>4\sqrt{21}</math> and <math>20</math> to get the right answer <cmath>a^2+b^2= \boxed{736}\Box</cmath> | ||
+ | |||
+ | ==Solution 5 (Diagrams)== | ||
+ | For obtuse <math>\triangle ABC,</math> we fix <math>AB=10</math> and <math>BC=4.</math> Without the loss of generality, we consider <math>C</math> on only one side of <math>\overline{AB}.</math> | ||
+ | |||
+ | <b>WILL BE BACK SOON. NO EDIT PLEASE.</b> | ||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|num-b=4|num-a=6}} | {{AIME box|year=2021|n=II|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:45, 14 May 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)
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 obtuse we fix
and
Without the loss of generality, we consider
on only one side of
WILL BE BACK SOON. NO EDIT PLEASE.
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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