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So, the area formula is: $S=\sqrt{p(p-a)(p-b)(p-c)}$

The perimeter equation is: $P=2p$

Usually, in optimization problems, we have a constraint, get the formula down to one variable, differentiate and then equal it to $0$. But here I'm totally lost.

There is also this this thread, but I still cannot understand that. Any tip or help would be really appreciated!

ⁱ ᵏⁿᵒʷ ⁱᵗ ˢᵉᵉᵐˢ ˡⁱᵏᵉ ʰᵒᵐᵉʷᵒʳᵏ ᵃⁿᵈ ˡⁱᵏᵉ ⁱ ʰᵃᵛᵉⁿ'ᵗ ᵉᵛᵉⁿ ᵗʳⁱᵉᵈ ˢᵒˡᵛⁱⁿᵍ ⁱᵗ, ᵇᵘᵗ ⁱᵗ ⁱˢⁿ'ᵗ ᵃⁿᵈ ⁱ'ᵐ ʳᵉᵃˡˡʸ ᵗʳʸⁱⁿᵍ ᵗᵒ ᵘⁿᵈᵉʳˢᵗᵃⁿᵈ ⁱᵗ.

Thomas Andrews
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EL02
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  • Perhaps you could show $\sqrt{(p-a)(p-b)} \le p-\frac{a+b}{2}$ and so conclude $a=b=c$ is optimal – Henry Sep 08 '21 at 21:02
  • You also have the triangle inequalities: $a+b>c>0$ and all cyclic permutations of $a,b,c$. But these are equivalent to $P=2p=a+b+c>c+c$, i.e., each factor in Heron’s law is positive. – Semiclassical Sep 08 '21 at 21:18
  • I will also note that “an equilateral triangle has the largest area given its perimeter” has shown up on this site a number of times. So an archive search may serve you well. – Semiclassical Sep 08 '21 at 21:22
  • An informal (and perhaps invalid) hand-waving argument is that in Real Analysis, for $x_1, x_2, \cdots, x_n \in \Bbb{R^+}$, you have that the arithmetic mean of the numbers is $\geq$ to the geometric mean, with equality only occurring when all of the numbers are equal. However, for me, this opens up a can of worms, because while I instinctly believe that area of the triangle is directly related to the geometric mean of the lengths of the triangle's 3 sides, I know of no easy (i.e. elegant) direct proof of this. – user2661923 Sep 08 '21 at 23:46

2 Answers2

2

Apply Lagrange multipliers method to optimize the function $$ p(p-a)(p-b)(p-c)$$ subject to $$a+b+c=2p$$

We get $$-p(p-b)(p-c)=\lambda $$ $$-p(p-a)(p-c)=\lambda $$ $$-p(p-a)(p-b)=\lambda $$ $$ a+b+c=2p$$ The result is $$a=b=c=2p/3$$

Mohammad Riazi-Kermani
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The isoperimetric inequality for triangles

Claim 1. Among the triangles $ABC$ with area $\Delta$, the equilateral ones have the minimum perimeter.
Proof. We show that if $ABC$ is not equilateral, its perimeter can be decreased without affecting the area.
Without loss of generality we may assume that $CA,CB$ have different lengths. If we move the vertex $C$ on a line $\ell$ parallel to $AB$ the area of $ABC$ is unchanged. Let $B'$ be the symmetric of $B$ with respect to $\ell$, let $C'$ be the intersection between $BB'$ and $\ell$. By the triangle inequality $AC+CB = AC+CB' > AB' = AC'+C'B$, so by moving $C$ to $C'$ the perimeter is decreased and the area is unchanged.

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Claim 2. Among the triangles $ABC$ with perimeter $2p$, the equilateral ones have the maximum area. Proof. This is just the dual statement of Claim 1. Assume that a triangle with perimeter $2p$ and maximum area $\Delta$ is not equilateral. Without loss of generality we may assume $AC\neq CB$. If we move $C$ to $C'$ the area stays $\Delta$ and the perimeter becomes $2p-\epsilon$ for some $\epsilon > 0$. If we dilate each side by a factor $\frac{2p}{2p-\epsilon}$, the perimeter returns to be $2p$ while the area becomes $\Delta\left(\frac{2p}{2p-\epsilon}\right)^2 > \Delta$, which contradicts the maximality of $\Delta$.

Jack D'Aurizio
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