# Maximum area for fixed perimeter of a triangle

I'm trying to prove that the triangle of largest area for a given perimeter is equilateral, but I'm having some difficulties.

I've done 2 different proofs for a similar problem but for rectangles - one proof uses the AM-GM mean inequality and the other uses algebra and a little calculus.

I can't manage to use a similar method for the triangle problem - with the algebra/calculus method, we only have 2 equations (one for perimeter, one for area) but 3 unknowns (each side length of triangle) so it looked like I was going to have to do it with respect to 2 different variables at the same time.

Can anyone point me in the right direction as to how to approach this?

Thanks

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Do you know Heron's formula? – 23rd Dec 8 '12 at 16:35
Ahh that's a thought, thanks ! – Taimur Dec 8 '12 at 16:45
Straight AM-GM. – André Nicolas Dec 8 '12 at 17:11
@Taimur: You are welcome! I hope you have solved the problem by yourself. – 23rd Dec 8 '12 at 17:24

## 1 Answer

Using Heron's formula, $\triangle^2=s(s-a)(s-b)(s-c)$

Using $AM\ge GM$, $$\frac{s-a+s-b+s-c}3\ge \{(s-a)(s-b)(s-c)\}^\frac13$$

or $(s-a)(s-b)(s-c)\le (\frac s3)^3$ the equality occurs when $s-a=s-b=s-c$ ie when $a=b=c$

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Thanks for the answer, how come the area is abc/4R and what is R? – Taimur Dec 8 '12 at 16:44
You cannot just fix $R$ – Phira Dec 8 '12 at 16:46