True or False: The circumradius of a triangle is twice its inradius if and only if the triangle is equilateral. Let $R$ be the circumradius and $r$ be the inradius. The if part is clear to me. For an equilateral triangle, the circumcentre, the incentre and the centroid are the same point. So, by property of cebntroid $AG:GD=2:1\Rightarrow AG=2GD$. Thus $R=2r$.

But is the converse true? Whether $R=2r$ implies that the triangle must be equilateral ? We know some relations involving circumradius and inradius, like $R=\dfrac{abc}{4\Delta}, r=\dfrac{\Delta}{s}$, where $\Delta$ is the area of the triangle and $s$ is its semi-perimeter i.e. $s=\dfrac{a+b+c}{2}$. But then how to show that the triangle is equilateral if $R=2r$.
I would be thankful if anyone can help me.
 A: For a given perimeter, an equilateral triangle uniquely maximizes the inradius and minimizes the circumradius. Hence it is the only triangle in which $R=2r$, and all other triangles have $R>2r$.
The statement for the inradius is not hard to see: $r=\dfrac{\Delta}{s}$, and for a fixed perimeter the equilateral triangle uniquely maximizes the area $\Delta$, so it also uniquely maximizes $r$ since $s$ is fixed. However, things are a bit more tricky when it comes to proving that the circumradius is minimized, so I'll reference you to the Math.SE post here as a guide to that proof.
A: In this answer, it is shown that
$$
d^2=R(R-2r)
$$
where $d$ is the distance between the circumcenter and the incenter, $R$ is the circumradius and $r$ is the inradius. This is known as Euler's Circle Theorem.
If the triangle is equilateral, its circumcenter and incenter coincide; thus, $d=0$, and therefore, $R=2r$.
Suppose that $R=2r$, then $d=0$, and the incenter and the circumcenter coincide. Then each side of the triangle is
$$
2\sqrt{R^2-r^2}
$$
since the distance from the incenter to each side is $r$ and the distance from the circumcenter to each vertex is $R$:
$\hspace{3.6cm}$
Since each side of the triangle is the same, the triangle is equilateral.
So the answer is true.
A: Let $a,b,c$ are the sides of a triangle, $A=$ area of the triangle, $s=$ semi-perimeter.
$R=\dfrac{abc}{4A}, r=\dfrac{A}{s}$
We have to show $R\geq 2r$.
The relation $\dfrac{abc}{4A}\geq \dfrac{2A}{s}$ holds
if $abc\geq \dfrac{8A^2}{s}$
if $abc\geq 8(s-a)(s-b)(s-c)$
if $abc\geq (b+c-a)(c+a-b)(a+b-c)$
This is true for all triangles.
When $a=b=c$, the equality holds.This is the case of an equilateral triangle.
So, $R=2r$.
A: R is the circum-radius and r is thew in-radius. In an equilateral triangle, the cenroid also divides the median in 2:1 ratio.
so, R=(2/3)*(1.732/2)*a, where a is the side of the equilateral triangle
so, r=(1/3)*(1.732/2)*a
so, we see R=2r
