Deriving the formula for the radius of the circle inscribed in an equilateral triangle

Question

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 $a$.

Answer 1

Square both sides, and you end up with a quadratic equation in $a$.

Answer 2

You wrote that you knew that $H$ is divided in the ratio $2:1$. That means that $h=2a$. So by the Pythagorean Theorem $$(2a)^2=a^2+n^2.$$ It follows that $3a^2=n^2$ and therefore $a=\frac{n}{\sqrt{3}}$. If you like, then multiply top and bottom by $\sqrt{3}$ to get your preferred form.

Another way: As you did, use the Pythagorean Theorem to find that $H=\sqrt{3}\, n$. The little triangle you are focused on is similar (same angles) to the triangle which is half of the big triangle. It follows that $$\frac{h}{n}=\frac{n}{H}=\frac{n}{\sqrt{3}\,n}=\frac{1}{\sqrt{3}},$$ and now the desired result follows. The advantage is that we do not have to prove the $2:1$ property, though in fact it follows easily from the same pair of similar triangles.