# Smallest possible perimeter of a triangle if area is 135

What is the smallest possible perimeter of a (flat, regular) triangle if the area is 135?

I tried various equations but I always ended up with two unknown variables.

-

Hint: You want an equilateral triangle. That should give you a single equation.

-
You can see this by first fixing the base of the triangle, and this fixes the height $h$ (since the area is given). So take a line parallel to the base and find the point on it which minimises the perimeter - the area takes care of itself - you should get an isosceles triangle. This shows that if your triangle has two sides of different lengths, you can choose the third side as a base and find a triangle of smaller perimeter on the same base with the same area by equalising the sides. If there is a minimum, the sides must all be equal. – Mark Bennet Jun 6 '12 at 20:56
Thank you very much! I got it. – libjup Jun 6 '12 at 20:57

The following proof is more complicated than the calculus approach. However, it uses some interesting ideas.

Our problem is equivalent to showing that among all triangles with a given perimeter, the equilateral triangle has maximum area.

Let the perimeter be $p$, and let $s=p/2$ be the half-perimeter. If the sides are $a$, $b$, and $c$, then $a+b+c=2s$. By Heron's Formula, the area of the triangle is $$\sqrt{s(s-a)(s-b)(s-c)}.$$

We want to maximize area, so equivalently we want to maximize $s(s-a)(s-b)(s-c)$. But since the perimeter is given, so constant, we want to maximize $(s-a)(s-b)(s-c)$ under the constraint $a+b+c=2s$.

Now we quote the case $n=3$ of the important Arithmetic Mean - Geometric Mean Inequality known as AM-GM to high school students who are serious about math contests.

Let $x$, $y$, and $z$ be positive. Then $$\frac{x+y+z}{3} \ge \sqrt[3]{xyz},$$ with equality only when $x=y=z$.

Let $x=s-a$, $y=s-b$, and $z=s-c$. Then $x+y+z=s$, and therefore $$\frac{s}{3} \ge \sqrt[3]{(s-a)(s-b)(s-c)},$$ with equality precisely when $s-a=s-b=s-c$, meaning that $a=b=c$. Thus $$\sqrt{s(s-a)(s-b)(s-c)}\le \sqrt{s(s^3/27)}=\frac{s^2}{3\sqrt{3}},$$ with equality only for the equilateral triangle with perimeter $2s$.

-