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What would be the most elementary proof of the following:

A triangle has been drawn inside the circle with radius $r$ and its area is as large as possible. Prove that the triangle is equilateral.

I mean, can we avoid topology and compactness to prove that the maximal triangle can be found? And can we avoid derivatives and trigonometry to find the maximal area?

amWhy
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mathprogrammer
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3 Answers3

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How about this for a simple proof? Let $A, B, C$ be the vertices of the triangle on a circle.

Fix a side, say $AB$, and then we see that the area is maximised by having $AC = BC$ (the base is fixed, height would be maximised if $C$ is on the perpendicular bisector and at the farthest point).

As we can choose any side for the base and the other two sides must equal for maximum area, unless the triangle is equilateral we can always find a better triangle...

Macavity
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  • This is a good argument but is this an entirely rigorous proof that the equilateral triangle has maximum area? We can consider a sequence of areas of triangles undergoing these symmetrizing transformations, $S_k$, and we can say that this sequence is monotonically growing and it is bounded (less than the circle area). So it has a limit (the Weierstrass theorem of calculus). But how do we prove that this limit is equal to the area of equilateral triangle? – Maxim Umansky Feb 05 '15 at 18:39
  • @MaximUmansky the answer given implies if there is a maximum, it has to be reached when the triangle is equilateral. That there is such a maximum is stated in the problem statement, so this suffices. $$ $$ Of course, if we had to also show that there exists such a maximum, there would be more work to do. – Macavity Feb 06 '15 at 10:23
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Suppose a triangle with vertices $A,B,C$ lie inside a circle whose area is maximaum among all the triangles. Without loss of generality assume $A,B,C$ lie on the circle. Suppose $ABC$ is not equilateral. Then there exist a vertex (say $A$) such that the straight line passing through $A$ and parallel to $BC$ cuts the circle at point $E$ different from $A$. Choose a point $P$ on the line $AE$ which is different from $A$ and $E$. Then the triangle $BPC$ has same area as $ABC$. Suppose now extended $BP$ meets the circle at $F$. Then we have got a triangle, namely $BCF$, whose area is strictly larger than $ABC$ and this is a contradiction.

pritam
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Suppose we an angle $\;\theta\;$ of the triangle $\;\Delta ABC\;$ with sides $\;a,b,c\;$ and respective opposite angles $\;\alpha,\beta,\gamma\;$ and inscribed in a circle of radius $\;r\;$ . The sinus law tells us that

$$\frac a{\sin\alpha}=\frac b{\sin\beta}=\frac c{\sin\gamma}=2r$$

If we now use the area formula

$$S(r,\alpha,\beta):=S_{\Delta ABC}=\frac12ab\sin\gamma=\frac12(2r\sin\alpha)(2r\sin\beta)\sin\gamma=$$

$$=2r^2\sin\alpha\sin\beta\sin(\alpha+\beta)$$

$$\begin{align*}\text{I}\;\;S'_r&=4r\sin\alpha\sin\beta\sin(\alpha+\beta)\stackrel ?=0\\ \text{II}\;\;S'_\alpha&=2r^2\sin\beta\underbrace{\left[\cos\alpha\sin(\alpha+\beta)+\sin\alpha\cos(\alpha+\beta)\right]}_{\sin(2\alpha+\beta)}\stackrel ?=0\\ \text{III}\;\;S'_\beta&=2r^2\sin\alpha\underbrace{\left[\cos\beta\sin(\alpha+\beta)+\sin\beta\cos(\alpha+\beta)\right]}_{\sin(\alpha+2\beta)}\stackrel ?=0\end{align*}$$

Since we can assume $\;r,\sin\alpha,\sin\beta\neq 0\;,\;\;0<\alpha\,,\,\beta <\pi\;$ (why?) , we get

$$\begin{align*}\text{II}\implies&\sin(2\alpha+\beta)=0\iff 2\alpha+\beta=\pi\\ \text{III}\implies&\sin(\alpha+2\beta)=0\iff\alpha+2\beta=\pi\end{align*}$$

Solving the above easy linear system we get

$$3\beta=\pi\implies \beta=\frac\pi3$$

and etc....

DonAntonio
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