2022 AIME I Problems/Problem 15
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, let define a triangle with side lengths , , and , with altitude from 's equal to . , the left side of one equation in the problem.
Let be angle opposite the side with length . Then the altitude has length and thus , so and the side length is equal to .
We can symmetrically apply this to the two other equations/triangles.
By law of sines, we have , with as the circumradius, same for all 3 triangles. The circumcircle's central angle to a side is , so the 3 triangles' , have angles , respectively.
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 ):
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:
When plugging into the expression , noting that helps to simplify this expression into:
Now, all the cosines in here are fairly standard:
With some final calculations:
This is our answer in simplest form , so .
~Oxymoronic15
solution 3
Let , rewrite those equations
;
and solve for
Square both sides and simplify, to get three equations:
Square both sides again, and simplify to get three equations:
Subtract first and third equation, getting ,
Put it in first equation, getting ,
Since , and so 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)
Solution 5
Let , , and . Then,
Notice that , , and . Let , , and where , , and are real. Substituting into , , and yields Thus, so . Hence,
so , for a final answer of .
Remark
The motivation for the trig substitution is that if , then , and when making the substitution in each equation of the initial set of equations, we obtain a new equation in the form of the sine addition formula.
~ Leo.Euler
Solution 6 (Geometric)
In given equations, so we define some points: Notice, that and each points lies in the first quadrant.
We use given equations and get some scalar products: So
Points and are simmetric with respect to
Case 1 Case 2
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
~Math Gold Medalist
Video Solution
https://www.youtube.com/watch?v=ihKUZ5itcdA
~Steven Chen (www.professorchenedu.com)
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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