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

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

I have gotten to the reduced equation

$n \sqrt(3) - a = \sqrt(a^2+n^2)$

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 $a$.

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Could you try do square both sides? – Nivan Jul 18 '12 at 22:03
It is a bit cleaner if you divide by $n$ and let $b=\frac an$. Then you have $\sqrt 3 - b = \sqrt{b^2+1}$ – Ross Millikan Jul 18 '12 at 22:06

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

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In fact it is linear. The $a^2$'s cancel – Ross Millikan Jul 18 '12 at 22:38
I had done that, but I made a mistake in my FOIL, writing $n^2*3-2a n \sqrt(3) + 2a$ instead of the problem-solving $n^2*3-2a n \sqrt(3) + a^2$. – SG1 Jul 18 '12 at 23:15
The final solution being from 1) my formula to 2) the $a^2+n^2$ = FOIL given above, 3) $2n^2=2a n \sqrt(3)$, 4) $a=n/\sqrt(3)$ (which is equivalent to the a priori answer given in-problem. Thanks! – SG1 Jul 18 '12 at 23:17

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.

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Your first solution won't work because I'm not actually using the 2:1 ratio, it is known to be true but not within the scope of the problem, and I can use it to check the result I get. – SG1 Jul 19 '12 at 17:38
The second solution is intriguing. It did work, and much more directly than the equation solving. However, I think your proportion is incorrect: I get $h/n=2n/H$, from which I derive the correct answer as you show. – SG1 Jul 19 '12 at 17:40