Circumradius and Inradius

We also get to know from the solution that M and N are the mid-points of sides AB and AC respectively because MN is half of BC.

Let $\omega$ , $O$ and $I$ be the circumcircle, the circumcenter and the incenter of $\triangle ABC$ respectively. Let $D$ be the midpoint of intersection of the line $BI$ and the circle $\omega$ such that $D$ $\not =$ $B$ . Then $D$ is the midpoint of the arc $AC$ . Hence $OD$ is $perpendicular$ to $CN$ and $OD$ $=$ $R$ .

We first show the $\triangle MNC$ and $\triangle IOD$ are similar. Because $BC$ $=$ $BM$ , the line $BI$ ( the bisector of $\angle MBC$ ) is perpendicular to the line $CM$ . Because $OD$ is $perpendicular$ to $CN$ and $ID$ is $perpendicular$ to $MC$ .

So, $\angle ODI$ $=$ $\angle NCM$ ............... $(1)$

Let $\angle ABC$ $=$ $2\beta$ . In the $\triangle BCM$ .

So, $\frac{CM}{NC}$ $=$ $\frac{CM}{BC}$ $=$ $2\beta$ ...................... $(2)$

Since $\angle DIC$ $=$ $\angle DCI$ , we have $ID$ $=$ $CD$ $=$ $AD$ . LetE be the point of intersection of line $DO$ and the circle $\omega$ such that $E$ $\not=$ $D$ . Then $DE$ is a diameter of $\omega$ and also $\angle DEC$ $=$ $\angle DBC$ $=$ $\beta$ .

So, $\frac{DI}{OD}$ $=$ $\frac{CD}{OD}$ $=$ $\frac{2Rsin\beta}{R}$ $=$ $2 sin \beta$ ................. $(3)$

So, from above equations it is clear that $\triangle MNC$ and $\triangle IOD$ are similar.

It follows that, $\frac{MN}{BC}$ $=$ $\frac{MN}{NC}$ $=$ $\frac{IO}{OD}$ $=$ $\frac{IO}{R}$

Now from $Euler's$ formula sates that, $OI^{2}$ $=$ $R^{2}$ $-$ $2Rr$

therefore , $\frac{MN}{BC}$ $=$ $\large \sqrt(1-\frac{2r}{R})$

So, $\large \sqrt(1-\frac{2*6}{16})$

$Answer$ $is$ $\sqrt{\frac{4}{16}}$ $=$ $\frac{1}{2}$ $=$ $0.50$