In $\triangle ABC$, if $a:b:c = 4:5:6$, then the ratio of the circumradius to its inradius is

$ABC$ is an isosceles triangle with an inscribed circle with center $O$. Let $P$ be the midpoint of $BC$. If $AB = AC = 15$ and $BC = 10$, then $OP$ equals

In $\triangle ABC$,$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2} =$

In a $\triangle ABC$,$a: b: c = 4: 5: 6$. The ratio of the radius of the circumcircle to that of the incircle is

In a $\triangle ABC$,let $\angle C = \frac{\pi}{2}$. If $r$ and $R$ are the inradius and circumradius of $\triangle ABC$ respectively,then $R+r=$

In a $\triangle ABC$,$D, E$,and $F$ are the points of contact of the incircle with the sides $AB, BC$,and $CA$ respectively,such that $AD=\alpha, BE=\beta$,and $CF=\gamma$. Then $\frac{\alpha \beta \gamma}{\alpha+\beta+\gamma} = $