# 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.

• You can use Herons Formula to get rid of the area in the term R/r.
– user70612
Jun 23, 2014 at 9:59
• @Leonhard .. ok let me try Jun 23, 2014 at 10:02
• So that gives me $8(s-a)(s-b)(s-c)=abc$ Jun 23, 2014 at 10:04

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.

• Thanks for the response. But is there any result that for any triangle $R\geq 2r$ ? Jun 24, 2014 at 10:48
• When I first did the question, I didn't look for many resources, but some Wikipedia searching turned up these two articles here to answer the question in your comment, and here to answer the original question. Jun 24, 2014 at 16:44
• Thanks a lot :) I will check these tomorrow in computer. Mobile browser doesn't display the contents properly :( Jun 24, 2014 at 19:20
• @ Peter Woolfit .. Thanks a lot for those links and your answer. So distance $d$ between incentre and circumcentre is given by $d^2=R(R-2r)$. Thus $R=2r\Rightarrow d=0$ i.e the incentre and circumcentre coincide. Thus the triangle has to be equilateral. Jun 25, 2014 at 10:39

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.

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, $R=2r$.

• Can you please tell how to get that the equality only holds when a=b=c Jun 1, 2022 at 7:56

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