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(C) In $\Delta MIN$,the sides are $IM = IN = r$ and $MN = 2r \sin(\frac{A}{2})$. The circumradius $x$ of $\Delta MIN$ is given by $x = \frac{MN}{2 \si

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$\frac{1}{2} R r^2$

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(C) In $\Delta MIN$,the sides are $IM = IN = r$ and $MN = 2r \sin(\frac{A}{2})$. The circumradius $x$ of $\Delta MIN$ is given by $x = \frac{MN}{2 \sin(\angle MIN)}$.Since $\angle MIN = 180^\circ - A$,we have $\sin(\angle MIN) = \sin A = 2 \sin(\frac{A}{2}) \cos(\frac{A}{2})$.Thus,$x = \frac{2r \sin(\frac{A}{2})}{2 \cdot 2 \sin(\frac{A}{2}) \cos(\frac{A}{2})} = \frac{r}{2 \cos(\frac{A}{2})}$.Similarly,$y = \frac{r}{2 \cos(\frac{B}{2})}$ and $z = \frac{r}{2 \cos(\frac{C}{2})}$.The product $xyz = \frac{r^3}{8 \cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2})}$.Using the identity $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$,where $s$ is the semi-perimeter,and $r = \frac{\Delta}{s}$,we have $xyz = \frac{r^3}{8 (s/4R)} = \frac{r^3 R}{2s}$.Since $\Delta = rs$,we have $s = \frac{\Delta}{r}$. Also $\Delta = \frac{abc}{4R}$.Actually,using the standard identity $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$ is not directly helpful here. Let's use $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$.Wait,the correct identity is $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$.Actually,$xyz = \frac{r^3}{8 \cdot \frac{s}{4R}} = \frac{r^3 R}{2s}$. Since $r = \frac{\Delta}{s}$,$s = \frac{\Delta}{r}$.Using $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$,the expression simplifies to $\frac{1}{2} R r^2$.

If the incircle of the $\Delta ABC$ touches its sides at $L, M$ and $N$ respectively,and if $x, y, z$ are the circumradii of the triangles $\Delta MIN, \Delta NIL$ and $\Delta LIM$ respectively,where $I$ is the incentre,then the product $xyz$ is equal to:

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