So I have seen this question asked before but with variations (circle of radius 4, and an equilateral triangle) and so I am hoping for an answer on how to do this.

After looking around I saw that people assume that the maximum perimeter of such a triangle is equilateral, meaning you have all the degrees. I wanted to ask how to actually prove that or something. I was given this problem on an exam and I usually sit down and do them just because I like solving these kinds of problems but I couldn't get it to work because I got too many messy equations and I had no time to clean up. What I'd like to ask is what is the best way of solving this, if you don't assume this?

When I tried to solve it I thought that I could do it like I would do with a square:

Find an equation

 f(x) = 2*(sqrt((1 - cos x)² + sin² x) + sin x)  => perimeter

and then get

f'(x) = 0 

and find what angle would satisfy those conditions. What I got though is a mess of trigonometric stuff that I found impossible to solve (my memory is bad so I easily forget formulas for trig. functions).

Anyway, was I doing the right thing but I may have messed up with the formulas or is there something I could do instead?

Visually what I did was thinking of the triangle's height being the x-coordinate from $x = 1$, so with an angle of $2\pi/3$ I get height = 1½ for example. And then the base would be just $\sin{ (2\pi/3)} = \sqrt{3}/2$

Anyway, that was just an example to try to explain how I was thinking when I set the equation up. I bet there's a better way that I'm not seeing. And I want to know how to prove things so if you want to tell me something like "this is always true for ..." please give me a link to an/or explanation of why that is so.

  • $\begingroup$ Isosceles triangle include equilaterals right? Then, there are proofs for this, let me search for one. $\endgroup$ – Sawarnik Mar 13 '14 at 12:50
  • $\begingroup$ Work with the angles at the centre of the circle to the points of the triangle. A general inscribed triangle is given by 3 angles which must add to 360. Write down the perimeter in terms of these angles and maximise with respect to the two angles. Gives you two simple equations to solve. $\endgroup$ – user121049 Mar 13 '14 at 12:53
  • $\begingroup$ cut-the-knot.org/triangle/InscribedEquilateral.shtml $\endgroup$ – Sawarnik Mar 13 '14 at 12:53
  • $\begingroup$ Okey thanks guys I'll look it up once I'm done with the current one I'm solving :) $\endgroup$ – Simon Mar 13 '14 at 12:55

I thought an image could help: enter image description here

The perimeter is $$\overline{BC} + 2\overline{AB} = 2\sin \alpha + 2\cdot 2\sin \beta = 2\sin \alpha + 2\cdot 2\sin \left(\frac{\pi}{2} - \frac{\alpha}{2}\right) = 2\left(\sin \alpha+2\cos\left(\frac{\alpha}{2} \right) \right)$$ For further simplifications let $\alpha = 2t$, then you have to maximize $$f(t) = 2\left(\sin 2t+2\cos t) \right) = 2(2\sin t \cos t + 2\cos t ) = 4(\sin t \cos t + \cos t) $$ Finding $f'(t)$ is easy: $f'(t) = 4(\cos t \cdot \cos t + \sin t \cdot -\sin t -\sin t) = 4(\cos^2 t - \sin^2 t -\sin t )=4(-2\sin^2 t -\sin t +1)$ and it equals $0$ when $\sin t = \frac12 \Rightarrow t = \frac{\pi}{6} \Rightarrow \alpha = 2t = \frac{\pi}{3} $ so the triangle is equilater!


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