Difference between revisions of "2022 AIME I Problems/Problem 15"
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Revision as of 00:21, 18 February 2022
Contents
Problem
Let
and
be positive real numbers satisfying the system of equations:
Then
can be written as
where
and
are relatively prime positive integers. Find
Solution 1 (geometric interpretation)
First, we note that we can let a triangle exist with side lengths ,
, and opposite altitude
. This shows that the third side, which is the nasty square-rooted sum, is going to have the length equal to the sum on the right - let this be
for symmetry purposes. So, we note that if the angle opposite the side with length
has a value of
, then the altitude has length
and thus
so
and the triangle side with length
is equal to
.
We can symmetrically apply this to the two other triangles, and since by law of sines, we have is the circumradius of that triangle. Hence. we calculate that with
, and
, the angles from the third side with respect to the circumcenter are
, and
. This means that by half angle arcs, we see that we have in some order,
,
, and
(not necessarily this order, but here it does not matter due to symmetry), satisfying that
,
, and
. Solving, we get
,
, and
.
We notice that
- kevinmathz
Solution 2 (pure algebraic trig, easy to follow)
(This eventually whittles down to the same concept as Solution 1)
Note that in each equation in this system, it is possible to factor ,
, or
from each term (on the left sides), since each of
,
, and
are positive real numbers. After factoring out accordingly from each terms one of
,
, or
, the system should look like this:
This should give off tons of trigonometry vibes. To make the connection clear,
,
, and
is a helpful substitution:
From each equation
can be factored out, and when every equation is divided by 2, we get:
which simplifies to (using the Pythagorean identity
):
which further simplifies to (using sine addition formula
):
Without loss of generality, taking the inverse sine of each equation yields a simple system:
giving solutions
,
,
. Since these unknowns are directly related to our original unknowns, there are consequent solutions for those:
,
, and
. When plugging into the expression
, noting that
helps to simplify this expression into:
Now, all the cosines in here are fairly standard: ,
,
and
. With some final calculations:
This is our answer in simplest form
, so
-Oxymoronic15
solution 3
Let , rewrite those equations
;
square both sides, get three equations:
Getting that
Subtract first and third equation, getting ,
Put it in first equation, getting ,
Since , the final answer is
the final answer is
~bluesoul
Solution 4
Denote ,
,
.
Hence, the system of equations given in the problem can be written as
Each equation above takes the following form:
Now, we simplify this equation by removing radicals.
Denote and
.
Hence, the equation above implies
Hence, .
Hence,
.
Because and
, we get
.
Plugging this into the equation
and simplifying it, we get
Therefore, the system of equations above can be simplified as
Denote .
The system of equations above can be equivalently written as
Taking , we get
Thus, we have either or
.
:
.
Equation (2') implies .
Plugging and
into Equation (2), we get contradiction. Therefore, this case is infeasible.
:
.
Plugging this condition into (1') to substitute , we get
Taking , we get
Taking (4) + (5), we get
Hence, .
Therefore,
Therefore, the answer is .
\end{solution}
~Steven Chen (www.professorchenedu.com)
Video Solution
https://www.youtube.com/watch?v=ihKUZ5itcdA
~Steven Chen (www.professorchenedu.com)
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by 2021 AIME I, II |
Followed by 2022 AIME II | |
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.