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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 $R=2r$.

enter image description here

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.

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  • $\begingroup$ You can use Herons Formula to get rid of the area in the term R/r. $\endgroup$ – Leonhard Jun 23 '14 at 9:59
  • $\begingroup$ @Leonhard .. ok let me try $\endgroup$ – Debashish Jun 23 '14 at 10:02
  • $\begingroup$ So that gives me $8(s-a)(s-b)(s-c)=abc$ $\endgroup$ – Debashish Jun 23 '14 at 10:04
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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.

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  • $\begingroup$ Thanks for the response. But is there any result that for any triangle $R\geq 2r$ ? $\endgroup$ – Debashish Jun 24 '14 at 10:48
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    $\begingroup$ 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. $\endgroup$ – Peter Woolfitt Jun 24 '14 at 16:44
  • $\begingroup$ Thanks a lot :) I will check these tomorrow in computer. Mobile browser doesn't display the contents properly :( $\endgroup$ – Debashish Jun 24 '14 at 19:20
  • $\begingroup$ @ 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. $\endgroup$ – Debashish Jun 25 '14 at 10:39
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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 $R$:

$\hspace{3.6cm}$enter image description here

Since each side of the triangle is the same, the triangle is equilateral.

So the answer is true.

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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, $R=2r$.

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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

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