Difference between revisions of "2022 AIME I Problems/Problem 15"
Oxymoronic15 (talk | contribs) (→Solution 1 (easy to follow)) |
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\alpha + \theta &= \frac{\pi}{3}. | \alpha + \theta &= \frac{\pi}{3}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | giving solutions <math>\alpha = \frac{\pi}{8}</math>, <math>\beta = \frac{\pi}{24}</math>, <math>\theta = \frac{5\pi}{24}</math>. | + | giving solutions <math>\alpha = \frac{\pi}{8}</math>, <math>\beta = \frac{\pi}{24}</math>, <math>\theta = \frac{5\pi}{24}</math>. Since these unknowns are directly related to our original unknowns, we also have solutions for those: <math>x = 2\cos^2\left(\frac{\pi}{8}\right)</math>, <math>y = 2\cos^2\left(\frac{\pi}{24}\right)</math>, and <math>z = 2\cos^2\left(\frac{5\pi}{24}\right)</math>. When plugging into the expression |
(Still editing) - Oxymoronic15 | (Still editing) - Oxymoronic15 |
Revision as of 20:19, 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 (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