# Perimeter of Triangle inside a circle

If the circle has a radius of 4, what is the perimeter of the inscribed equilateral triangle?

Answer: $12\sqrt{3}$

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As you know, the best way to get the best possible answers is to tell us a little of the context of the problem (was it given in a class? Which class, what level?) and what your thoughts on it are (even if only to say you have no idea), rather than simply posting the problem with or without the final answer and expecting people to solve it for you. –  Arturo Magidin Jul 18 '12 at 3:39
Well am practicing for a standardized test. I have posted the answer along with the question. I actually am not exactly asking for a solution .. but for suggestions on how to solve it.. –  Rajeshwar Jul 18 '12 at 3:52
The general standardized test does not cover trigonometric ratios thus I am trying to solve it without using them –  Rajeshwar Jul 18 '12 at 3:54
Your comment to Zev gives the lie to your statement; you are looking for a specific way of doing it (avoiding trig functions). You need to give the context and any notes about the kinds of tools you have available/want to use in the question itself! Otherwise, you are wasting people's time and just increasing the general level of frustration to no avail. –  Arturo Magidin Jul 18 '12 at 4:08
@Rejshwar: Yes! If you are looking for a specific kind of approach, mention it in the question. If you are looking to avoid a particular kind of approach, mention it in the question. If you have no idea what to do, mention it in the question. If you have had some ideas but don't seem to lead anywhere, mention it in the question. The more you say about what you want, what you've tried, and what you don't want, the better chances that you will get an answer that is useful! –  Arturo Magidin Jul 18 '12 at 16:07

See here for the image I am referring to:

the angle $\angle ABC$ is $60$, so the angle $\angle AOC$ is $120$ (inscribed angles to a chord in a circle have half the value the angle from the center $O$ to that same chord).

In triangle $\triangle ACO$, $OA = OC = \text{radius}$, then $$\angle IAO = \angle ICO = (180 - \angle AOC)/2 = 30$$ ($I$ is midpoint of $AC$, and $OI$ is perpendicular to $AC$)

Triangle $\triangle AIO$ is both half an equilateral triangle, and a right triangle at $I$; then $$OI = \frac{1}{2} OA = \frac{1}{2}\cdot 4 = 2$$ $$AI = \sqrt{OA^2 - OI^2} = \sqrt{4^2 - 2^2} = \sqrt{12} = 2\sqrt{3}$$ $$AC = 2 AI\text{ (as I is the midpoint of AC)} = 4\sqrt{3}$$ so the perimeter of $\triangle ABC = 3 \cdot AC = \fbox{$12\sqrt{3}\;$}$

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Please see here for how to typeset common math expressions with LaTeX, and see here for how to use Markdown formatting. If you need help with more esoteric math expressions, there are many excellent LaTeX references on the internet, including Stack Exchange's own TeX.SE. Here's a helpful trick: if you see a math expression on this site for which you want to know the LaTeX code, you can right click on it, go to "Show Math As", then choose "TeX Commands". –  Zev Chonoles Jul 18 '12 at 4:04
I've added a somewhat cleaner version of your image, hope you don't mind (feel free to change back if you want). Note that your labeling of $B$ and $C$ is reversed from the image that the question was using. –  Zev Chonoles Jul 18 '12 at 4:14
I know the image was not very good, its the only one I had on my site; I'm unable to post actual images here due to my low rep points (I joined the site last night), a few rep points from you guys would be great! I am aware that the B and C are reversed and as I said I couldn't upload any other image so I figured you guys would correct it :) Thanks for the help. **Added: I looked at my rep points and I now have enough. –  DiscreteGenius Jul 18 '12 at 19:01

Hint: Use the law of sines on an interior triangle:

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I want to solve it without using trignometric identities –  Rajeshwar Jul 18 '12 at 3:46
Don't you think that's something you ought to have mentioned in your question? –  Zev Chonoles Jul 18 '12 at 3:53
@Rajeshwar: That is precisely why you need to give context in the question! As it is, you've succeeded only in not getting the answer you sought, and making people waste their time. –  Arturo Magidin Jul 18 '12 at 4:07

Without trigonometry: since $\,\Delta ABC\,$ is equilateral, its circumcenter is the same as its incenter is the same as the intersection point of its medians. Call this point $\,O\,\Longrightarrow \,$ if $\,AM\,$ is the whole median from $\,A\,$ to BC, then $$4=AO=\frac{2}{3}AM\Longrightarrow x = 6$$

(since the intersection point of the medians cuts each of them in a$\,1:2\,$ proportion, the longest side being always on the vertex side).

Now draw the triangle $\,\Delta AMB\,$ ,with sides $\,x\,,\,x/2\,,\,6\,\,,\,\,x=\,$ the triangle's side, and use Pythagoras (in an equilateral triangle, each median is its root vertex's angle bisector and also the height to the other side): $$x^2=6^2+\left(\frac{x}{2}\right)^2\Longrightarrow \frac{3}{4}x^2=36\Longrightarrow x=\frac{12}{\sqrt 3}\Longrightarrow P_{\Delta ABC}=3x=12\sqrt 3$$

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If you observe carefully, we are given the circumradius of the equilateral triangle. If $s$ and $h$ be the side and the height respectively of an equilateral traingle then we know circumradius is given by $\frac 23 \times h$. Thus,
$$\frac 23 \times h = 4\implies h = 6$$
Again, $h=6=\frac{\sqrt{3}}2 \times s \implies s = \frac {12}{\sqrt{3}} \implies \text{ perimeter } = 3s = 12\sqrt{3}$