Prove-that-r-1-r-2-r-3-4R-r-

Question Number 19122 by Tinkutara last updated on 05/Aug/17

Prove that r_{1} + r_{2} + r_{3} = 4 R + r

Answered by ajfour last updated on 05/Aug/17

2 Rsin A = a

r_{1} = \frac{a}{\tan \frac{B}{2} + \tan \frac{C}{2}}

r = \frac{a}{\cot \frac{B}{2} + \cot \frac{C}{2}} = \frac{a (\tan \frac{B}{2} \tan \frac{C}{2})}{\tan \frac{B}{2} + \tan \frac{C}{2}}

so r_{1} = \frac{r}{\tan \frac{B}{2} \tan \frac{C}{2}}

\Rightarrow r_{1} + r_{2} + r_{3} = r (\frac{Σ \tan \frac{A}{2}}{Π \tan \frac{A}{2}}) \dots \dots (i)

4 Rsin \frac{A}{2} \cos \frac{A}{2} = a = r (\frac{\tan \frac{B}{2} + \tan \frac{C}{2}}{\tan \frac{B}{2} \tan \frac{C}{2}})

\Rightarrow \frac{4 R}{r} = \frac{(\tan \frac{B}{2} + \tan \frac{C}{2}) (1 + \tan^{2} \frac{A}{2})}{Π \tan \frac{A}{2}}

\frac{4 R}{r} = \frac{Σ \tan \frac{A}{2}}{Π \tan \frac{A}{2}} + \frac{\tan^{2} \frac{A}{2} (\tan \frac{B}{2} + \tan \frac{C}{2}) - \tan \frac{A}{2}}{Π \tan \frac{A}{2}}

\frac{4 R}{r} = \frac{r_{1} + r_{2} + r_{3}}{r} + \frac{\tan \frac{A}{2} (Σ \tan \frac{B}{2} \tan \frac{C}{2} - 1 - \tan \frac{B}{2} \tan \frac{C}{2})}{Π \tan \frac{A}{2}}

but Σ \tan \frac{B}{2} \tan \frac{C}{2} - 1 = 0, so

\frac{4 R}{r} = \frac{r_{1} + r_{2} + r_{3}}{r} - 1

\Rightarrow r_{1} + r_{2} + r_{3} = 4 R + r .

Commented by Tinkutara last updated on 06/Aug/17

Thank you very much Sir! I really

appreciate .

Commented by ajfour last updated on 05/Aug/17

Commented by ajfour last updated on 05/Aug/17

from figure :

Rsin A = \frac{a}{2}

r (\cot \frac{B}{2} + \cot \frac{C}{2}) = a

\Rightarrow r = a (\frac{\tan \frac{B}{2} \tan \frac{C}{2}}{\tan \frac{B}{2} + \tan \frac{C}{2}})

and r_{1} (\tan \frac{B}{2} + \tan \frac{C}{2}) = a

or r_{1} = \frac{r}{\tan \frac{B}{2} \tan \frac{C}{2}} .

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