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For an arbitrary triangle, let $r$ = incircle radius and $R$ = circumcircle radius. Consider the ratio $r/R$? What is its maximum value? For an equilateral triangle this ratio is $1/2$. For very skewed triangles, this ratio approaches $0$. I am wondering if the maximum value is $1/2$.

This ratio is used as a measure of the quality of a triangulated surface mesh.

Ottavio
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user89699
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  • Let $a,b,c$ be the sides and $r,R$ be the inradius and circumradius, respectively. One obtains the ratio from the relation $rR=\frac{abc}{2(a+b+c)},$ see [https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#:~:text=The%20incircle%20radius%20is%20no,the%20sum%20of%20the%20altitudes.&text=The%20incenter%20lies%20in%20the,the%20midpoints%20of%20the%20sides).] – Viera Čerňanová Dec 24 '20 at 21:44

2 Answers2

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Your claim is correct.

We use the well-known area formulae $\Delta=rs=\frac{abc}{4R}=\sqrt{s(s-a)(s-b)(s-c)}$ to get $$\frac{r}{R}=\frac{4\Delta^2}{abcs}=\frac{4(s-a)(s-b)(s-c)}{abc}=\frac{4xyz}{(y+z)(z+x)(x+y)},$$ where $x=s-a, y=s-b,z=s-c$ are all positive. But $y+z\geq 2\sqrt{yz}$, $z+x\geq2\sqrt{zx}$, $x+y\geq2\sqrt{xy}$, so we get the result.

This result is called Euler's inequality. It is a corollary of the fact that $IO^2=R(R-2r)$, where $I$ is the incentre, $O$ the circumcentre.

jlammy
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We can also prove that $$ \frac{r}{R} = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}. $$

Since $\sin(x)$ is concave in $(0,\pi)$, we have $$ \begin{align} \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} & \leq\biggl(\frac{1}{3}\Bigl(\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\Bigr)\biggr)^3 \quad\text{by GM-AM inequality}\\ &\leq\biggl(\sin\frac{A/2+B/2+C/2}{3}\biggr)^3 \quad\text{by Jensen's inequality} \\ & = \frac18. \end{align} $$ Hence $\frac{r}{R}\leq\frac12$.

Stephen
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