Difference between revisions of "2022 AIME I Problems/Problem 15"
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Now, all the cosines in here are fairly standard: <math>\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}</math>, <math>\;</math> <math>\cos \frac{\pi}{12} = \frac{\sqrt{6} + \sqrt{2}}{4}</math>,<math>\;</math> and <math>\cos \frac{5\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}</math>. With some final calculations: | Now, all the cosines in here are fairly standard: <math>\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}</math>, <math>\;</math> <math>\cos \frac{\pi}{12} = \frac{\sqrt{6} + \sqrt{2}}{4}</math>,<math>\;</math> and <math>\cos \frac{5\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}</math>. With some final calculations: | ||
<cmath>(-1)^2\left(\frac{\sqrt{2}}{2}\right)^2\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 = \left(\frac{1}{2}\right)\left(\frac{2 + \sqrt{3}}{4}\right)\left(\frac{2 - \sqrt{3}}{4}\right) = \frac{\left(2 - \sqrt{3}\right)\left(2 + \sqrt{3}\right)}{2\cdot4\cdot4} = \frac{1}{32}.</cmath> | <cmath>(-1)^2\left(\frac{\sqrt{2}}{2}\right)^2\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 = \left(\frac{1}{2}\right)\left(\frac{2 + \sqrt{3}}{4}\right)\left(\frac{2 - \sqrt{3}}{4}\right) = \frac{\left(2 - \sqrt{3}\right)\left(2 + \sqrt{3}\right)}{2\cdot4\cdot4} = \frac{1}{32}.</cmath> | ||
− | This is our answer in simplest form <math>\frac{m}{n}</math>, so <math>m + n = 1 + 32 = \boxed{ | + | This is our answer in simplest form <math>\frac{m}{n}</math>, so <math>m + n = 1 + 32 = \boxed{033}.</math> |
- Oxymoronic15 | - Oxymoronic15 |
Revision as of 20:49, 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, 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