Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively. let $r_1,r_2$ and$r_3$ be the ex-radii of the triangle.

The value of $\frac{r_1}{bc}+\frac{r_2}{ca}+\frac{r_3}{ab}$ is equal to?

• A. $\frac{1}{2R}-\frac{1}{r}$
• B. $2R-r$
• C. $r-2R$
• D. $\frac{1}{r}-\frac{1}{2R}$

Answer 1

Answer: (D)

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

$\frac{r_1}{bc}+\frac{r_2}{ca}+\frac{r_3}{ab}=a=2R\sin A,r_1=s\tan \left(\frac{A}{2}\right),c=2R\sin C,r_2=s\tan \left(\frac{C}{2}\right)b=2R\sin B,r_2=s\tan \left(\frac{B}{2}\right)\frac{a{r}_1+b{r}_{2}c{r}_3}{\left(abc\right)}=\frac{\left(2Rs\right)}{\left(abc\right)}[\sin A\tan \left(\frac{A}{2}\right)+\sin B\tan \left(\frac{B}{2}\right)+\sin C\tan \left(\frac{C}{2}\right)]=\frac{2Rs}{\left(abc\right)}2[{\sin }^2\left(\frac{A}{2}\right)+{\sin }^2\frac{B}{2}+{\sin }^2\left(\frac{C}{2}\right)]=\frac{2Rs}{\left(abc\right)}\left[2{\sin }^2\left(\frac{A}{2}\right){\sin }^2\left(\frac{B}{2}\right){\sin }^2\left(\frac{C}{2}\right)\right]=\frac{2Rs}{\left(abc\right)}(1-\cos A)+(1-\cos B)+(1-\cos C)=\frac{2Rs}{\left(abc\right)}[3-(\cos A+\cos B+\cos C\left)\right]=\frac{2s}{abc}[3R-R(\cos A+\cos B+\cos C)]=\frac{2s}{abc}[3R-R(1+4\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right))]=\frac{2s}{abc}[2R-4R\sin \left(\frac{A}{2}\right)+\sin \left(\frac{B}{2}\right)\sin \frac{C}{2}]=\frac{2s}{abc}(2R-r)=s[\frac{4R}{abc}-\frac{2r}{abc}]=s[\frac{1}{△}-\frac{2r}{4r△}]=\frac{s}{△}[1-\frac{r}{2R}]=\frac{1}{r}[1-\frac{r}{2R}]=\frac{1}{r}-\frac{1}{2R}$