Product of the Radii

We use the two well known formulae for the area of a triangle:

$A=\frac{abc}{4R},A=\frac{a+b+c}{2} r$

$\frac{abc}{4R}=\frac{a+b+c}{2} r \Rightarrow R r=\frac{abc}{2(a+b+c)}$

The question gives us:

$a+b+c=\frac{29}{6}, abc=4$

$R r=\frac{4}{2 \frac{29}{6}}= \frac{12}{29} \Rightarrow m=12,n=29$

$m+n=12+29=\boxed{41}$

Note: An example traingle where the statement is true is $\frac{4}{3},\frac{3}{2},2$ . (Just to show the configuration is possible)

Let $(a,b,c)$ , $\triangle$ , $S$ , $R$ , $r$ be the sides,area of triangle,semi-perimeter,circumradius and inradius respectively.

We know that , $R=\frac{a b c}{4\triangle} \cdots (i)$

$and$

$r=\frac{\triangle}{S} \cdots (ii)$

Now, $(i)\times (ii)$

$R\times r=\left (\frac{a b c}{4\triangle}\right ) \times \left( \frac{\triangle}{S}\right )$

Substituting the values of given parameters, we get

$\boxed{R\times r=\frac{12}{29}}$

$\implies \boxed {m+n=41}$

Let A be the area of the triangle and s be the semi-perimeter( $\frac{a+b+c}{2}$ ) We know that, r = $\frac{A}{s}$ * * R = $\frac{abc}{4A}$ From here, R * r = $\frac{abc}{4A}$ * $\frac{A}{s}$ = $\frac{4}{4}$ * $\frac{12}{29}$ = $\frac{12}{29}$ From here, m = 12, n = 29 Thus, m + n = 12 + 29 = 41 Note : - Since abc = 4 and ( $\frac{a+b+c}{2}$ ) = $\frac{29}{12}$

We have

$\triangle = sr$

$\triangle= abc/4R$ .

From both equations $Rr= abc/4s$ .

Hence $Rr=1/s= 12/29$ .

Therefore answer= $\boxed{41}$

$Please-Note$ :: All are in usual notations