Difference between revisions of "2022 AIME I Problems/Problem 15"
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==Solution 1 (geometric interpretation)== | ==Solution 1 (geometric interpretation)== | ||
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First, we note that we can let a triangle exist with side lengths <math>\sqrt{2x}</math>, <math>\sqrt{2z}</math>, and opposite altitude <math>\sqrt{xz}</math>. 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 <math>l</math> for symmetry purposes. So, we note that if the angle opposite the side with length <math>\sqrt{2x}</math> has a value of <math>\sin(\theta)</math>, then the altitude has length <math>\sqrt{2z} \cdot \sin(\theta) = \sqrt{xz}</math> and thus <math>\sin(\theta) = \sqrt{\frac{x}{2}}</math> so <math>x=2\sin^2(\theta)</math> and the triangle side with length <math>\sqrt{2x}</math> is equal to <math>2\sin(\theta)</math>. | First, we note that we can let a triangle exist with side lengths <math>\sqrt{2x}</math>, <math>\sqrt{2z}</math>, and opposite altitude <math>\sqrt{xz}</math>. 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 <math>l</math> for symmetry purposes. So, we note that if the angle opposite the side with length <math>\sqrt{2x}</math> has a value of <math>\sin(\theta)</math>, then the altitude has length <math>\sqrt{2z} \cdot \sin(\theta) = \sqrt{xz}</math> and thus <math>\sin(\theta) = \sqrt{\frac{x}{2}}</math> so <math>x=2\sin^2(\theta)</math> and the triangle side with length <math>\sqrt{2x}</math> is equal to <math>2\sin(\theta)</math>. |
Revision as of 21:39, 17 February 2022
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 (easy to follow)
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, we can take the inverse sine of each equation. The resulting system is simple:
giving solutions
,
,
. Since these unknowns are directly related to our original unknowns, we also have solutions for those:
,
, and
. When plugging into the expression
(Still editing) - Oxymoronic15