# Does this proof work to prove that the greatest area of a triangle inside a circle is when the triangle is equilateral?

Does this proof work to prove that the greatest area of a triangle inside a circle is when the triangle is equilateral? I gather it doesn't because most of the proofs I've seen use derivatives etc. If so why doesn't it work?

Consider a triangle $ABC$ inscribed inside a circle $\Gamma$. Assume WLOG one of the sided is fixed then one can easily see that the other two sides being equal maximizes the area of the triangle (it has the greatest height). Now consider one of the two equal sides - using that side as a base we can see from the same argument as before that the triangle must in fact be equilateral to maximize the area (this triangle has the greatest height as before).

Thanks in advance.

• Yes, I think this can work...yet it still looks a little shaky, doesn't it? I mean, you could try something some basic analytic geometry and, perhaps, some calculus to make it more formal...? Commented Jun 19, 2013 at 12:10
• Whoops! I was making my own assumptions about the problem. As @ChristianBlatter points out, you have proven that a non-equilateral triangle's area can be improved by your process (and that an equilateral triangle's cannot), but consider your process: starting with a triangle with side lengths $(a,b,c)$ and "base" $a$, you improve the area by constructing $(a,d,d)$; then, you improve that area (relative to the "first" base $d$) to get $(e,d,e)$. It could be that $d\neq e$, so you can keep trying, ever-increasing the area, but where's the guarantee that the process ends (or even converges)?
– Blue
Commented Jun 19, 2013 at 15:48

## 2 Answers

You have proven that as long as the three sides are not all equal the triangle does not have maximal area. If you assume without proof that a maximal triangle exists you are done. But note that the following could still be true: There is no maximal triangle at all, and someone comes up with some method enlarging even an equilateral triangle.

In order to prove that there is in fact a maximal triangle one can use a compactness argument: The area of the triangle with vertices $re^{i\phi_k}$ $\>(1\leq k\leq3)$ is a continuous function on $[0,2\pi]^3$; therefore it assumes a maximum for certain choices of the $\phi_k$.

A completely elementary proof that the equilateral triangle has maximal area is not so easy. You can find one (in German) here, pp. 60/61:

http://www.math.ethz.ch/~blatter/SolutionsI.pdf

Let us some analytic geometry using your idea: suppose our circle is $\,x^2+y^2=R^2\;$ , and we're going to fix, as you did, one of our sides as being parallel to the $\,x-$axis for simplicity, so that its end points are $\,A=(a,b)\;,\;\;B=(-a,b)\;,\;\;a>0\;,\;a^2+b^2=R^2\;,\;\;b\le 0\;$ .

Let the other vertex be $\,C=(0,R)\;$ since, as you remarked, for the above two points of the triangle, the maximal height is obtained when the other two sides are equal, which means the third vertex is on the sides perpendicular bisector, which is easy to see is the $\,y-$axis.

Thus, the area of the triangle depends on the distance $\,R-b=R-\sqrt{R^2-a^2}\,$ and on the horizonal side length's $\,2a\;$ :

$$f(a):=a\left(R-\sqrt{R^2-a^2}\right)\implies f'(a)=R-\sqrt{R^2-a^2}+\frac{a^2}{\sqrt{R^2-a^2}}\stackrel ?=0\iff$$

$$R\sqrt{R^2-a^2}-R^2+a^2+a^2=0\implies(2a^2-R^2)^2=R^2(R^2-a^2)\implies$$

$$R^4-4R^2a^2+4a^4=R^4-R^2a^2\implies a^2\left(4a^2-3R^2\right)=0\implies$$

$$\implies a=\frac{\sqrt3}2R$$

so that the slope of the line

$$CA\,\;,\;\;A=\left(\frac{\sqrt3}2R\,,\,-\frac12R\right)\;,\;C=(0,R)\;\,\;\;\text{is}$$

$$-\frac{\frac32R}{\frac{\sqrt3}2R}=-\sqrt3=\tan\frac{-\pi}3$$

and we're done...