In which condition，the triangle have the maximum triangle area？

Question

These trangles have the same perimeter 2q，in which condition，can we have the trangle with the maximum area? I have tried to use the Heron's formula. $S=\sqrt{q(q-a)(q-b)(q-c)}$ I have sought the partial derivative and tried to find the stationary point but the calculation is complicated. Is there any strategy can we use to simply the calculation?

Answer 1

By AM-GM $$S\leq\sqrt{q\left(\frac{q-a+q-b+q-c}{3}\right)^3}=\frac{q^2}{3\sqrt3}.$$ The equality occurs for $q-a=q-b=q-c,$ id est, for the equilateral triangle, which says that we got a maximal value.

Answer 2

For a triangle with semiperimeter $\rho$ and the radii of the inscribed and circumscribed circles $r$ and $R$, respectively, the area \begin{align} S&=\rho\,r , \end{align}

so given $\rho$ fixed, the maximal area corresponds to the triangular shape with the largest incircle.

Since we are looking for the triangular shape with the largest $r$ with respect to the given $\rho$, we can consider $R=1$ without loss of generality.

It is well-known that in this case $\max r=\tfrac12$, which corresponds to equilateral triangle.

So, the triangle of the maximal area with given $\rho$ is equilateral with the side length $\tfrac23\,\rho$, and therefore \begin{align} \max S(\rho)&= \tfrac{\sqrt3}9\,\rho^2 . \end{align}

Answer 3

Assuming that a triangle of maximal area exists you can argue as follows: A triangle with vertices $A$, $B$ and a given perimeter has the third vertex on a certain ellipse with foci $A$, $B$. Choosing $C$ at a co-vertex of this ellipse makes the area largest and leads to an isosceles triangle. It follows that the triangle with largest area is isosceles with respect to all pairs of vertices, hence equilateral.