I am trying to derive the formula for the radius of the circle inscribed in an equilateral triangle from scratch.
Given
$2*n$ = length of a side
$H$ = the altitude of the triangle = $h + a$
$h$ = the long subdivision (from the center of the triangle to a vertex)
$a$ = the short subdivision (from the center of the triangle to a side. Also the radius of the inscribed circle)
By first deriving the altitude of the triangle
$\displaystyle \begin{align} 2 n&=\sqrt{H^2+n^2} \\ H&=\sqrt{(2 n)^2-n^2} \\ &=\sqrt{3}\;n \\ \end{align}$
I have gotten to the reduced equation
$n \sqrt(3) - a = \sqrt(a^2+n^2)$
$\displaystyle \begin{align} a+h=\sqrt{3}\;n \\ h=\sqrt{3}\;n-a \\\\ a^2+n^2=h^2 \\ h=\sqrt{a^2+n^2} \\ \end{align}$
Trying to solve for $a$, I know in advance that $a$ is $1/3$ and $h$ is $2/3$ of $H$, with
$a = n\sqrt(3)/3$
This is of course the answer I wish to derive.
In fact, plugging the equation given above into a system such as Mathematica will provide the correct answer. But I can't find out what the steps are, primarily because I know of now way to extract the $a$ term from within the square root term.
Please, no trigonometry. I know there is a fast derivation involving tangents, etc, but this is more properly an algebra problem - how to solve the equation for $a$.