Surrounded Perimeter

The way the problem was stated is maybe open for misinterpretation, because one might look for the lowest value for $\frac{p}{r}$ . Therefore I rephrase the question: If A is the smallest value such that $\frac{p}{r}\le A$ will hold for any triangle, what is $A^2$ ?

So we need to identify what shape the triangle with maximum $\frac{p}{r}$ has. If you start with a fixed circle and ask which inscribed triangle has maximum perimeter, it becomes clearer that that would be an equilateral triangle.

An equilateral triangle with side 1 has perimeter $p=3$ . The distance from its centre to a vertex is the radius of its circumscribed circle: $r=\frac{1}{3}\sqrt{3}$ . The ratio between these is $A=\frac{p}{r}=3\sqrt{3}$ , so that $A^2=\boxed{27}$ .

For other triangles this ratio will be lower.

Now, $\begin{aligned} \frac{\rm Perimeter}{\rm Circumradius} &= \frac{a+b+c}{R} \\ &= \frac{a}{R} + \frac{b}{R} + \frac{c}{R} ----> eq1 \\ \end{aligned}$

From Sine rule, $\begin{aligned} \frac{eq1}{2} &= \frac{a}{2R} + \frac{b}{2R} + \frac{c}{2R} \\ &= SinA + Sin B + Sin C \end{aligned}$

Sorry guys, I am not able to attach the graph but maybe you can draw and understand on your own.

Now, in a concave graph such as sin(x), any three points on the graph when connected will always form a Triangle on the inside or on the graph.

Coordinates of the centroid will be ( $\frac{A + B + C }{3}$ , $\frac{SinA + SinB + SinC }{3}$ ). One can easily observe that the Centroid of any Triangle will always lie on the line x = $\frac{\pi}{3}$ . And y coordinates on the line x = $\frac{\pi}{3}$ has a minimum value of 0 and maximum value of Sin((\frac{\pi}{3})).

Therefore, $\begin{aligned} Max--> \frac{SinA + Sin B + Sin C}{3} &= Sin(\frac{\pi}{3}) \\ SinA + SinB + SinC &= \frac{3\sqrt3}{2} \\ eq1 &= 3\sqrt3 \\ A &= 3\sqrt3 \\ A^2 &= 27 \end{aligned}$

Intuitive hint: think equilateral triangles!

Relation of circum-radius and lenght of triangle is: r * cos30 = length/2 A=3*√3