Difference between revisions of "2021 AIME II Problems/Problem 5"
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==Solution 1== | ==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 <math>\sqrt{84}</math> exclusive, and the larger bound is between <math>\sqrt{116}</math> and 14, exclusive. The area of these triangles are from 0 (straight line) to <math>2\sqrt{84}</math> on the first "small bound" and the larger bound is between 0 and 20. | + | We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given <math>4</math> and <math>10</math> as the sides, so we know that the 3rd side is between <math>6</math> and <math>14</math>, 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 <math>6</math> and <math>\sqrt{84}</math> exclusive, and the larger bound is between <math>\sqrt{116}</math> and <math>14</math>, exclusive. The area of these triangles are from <math>0</math> (straight line) to <math>2\sqrt{84}</math> on the first "small bound" and the larger bound is between <math>0</math> and <math>20</math>. |
<math>0 < s < 2\sqrt{84}</math> is our first equation, and <math>0 < s < 20</math> is our 2nd equation. Therefore, the area is between <math>\sqrt{336}</math> and <math>\sqrt{400}</math>, so our final answer is <math>\boxed{736}</math>. | <math>0 < s < 2\sqrt{84}</math> is our first equation, and <math>0 < s < 20</math> is our 2nd equation. Therefore, the area is between <math>\sqrt{336}</math> and <math>\sqrt{400}</math>, so our final answer is <math>\boxed{736}</math>. | ||
Revision as of 07:19, 15 June 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 and
as the sides, so we know that the 3rd side is between
and
, 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
and
exclusive, and the larger bound is between
and
, exclusive. The area of these triangles are from
(straight line) to
on the first "small bound" and the larger bound is between
and
.
is our first equation, and
is our 2nd equation. Therefore, the area is between
and
, so our final answer is
.
~ARCTICTURN
Solution 2 (Inequalities and Casework)
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 we fix
and
Without the loss of generality, we consider
on only one side of
As shown below, all locations for at which
is an obtuse triangle are indicated in red, excluding the endpoints.
- The region in which
is obtuse is determined by construction.
- The region in which
is obtuse is determined by the corollaries of the Inscribed Angle Theorem.
For any fixed value of the height from
is fixed. We need obtuse
to be unique, so there can only be one possible location for
As shown below, all possible locations for
are on minor arc
including
but excluding
Let the brackets denote areas:
- If
then
will be minimized (attainable). By the same base and height and the Inscribed Angle Theorem, we have
- If
then
will be maximized (unattainable). For this right triangle, we have
Finally, we get from which
~MRENTHUSIASM (credit given to Snowfan)
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.