In-ABC-with-usual-notation-r-1-bc-r-2-ca-r-3-ab-is-1-1-r-1-R-2-1-r-1-2R-3-1-r-1-2R-4-1-r-1-R-

Question Number 16430 by Tinkutara last updated on 22/Jun/17

In Δ A B C with usual notation

\frac{r_{1}}{b c} + \frac{r_{2}}{c a} + \frac{r_{3}}{a b} is

(1) \frac{1}{r} - \frac{1}{R}

(2) \frac{1}{r} - \frac{1}{2 R}

(3) \frac{1}{r} + \frac{1}{2 R}

(4) \frac{1}{r} + \frac{1}{R}

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17

a n s w e r (2)

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 23/Jun/17

b e c a u s e t h i s f o r m u l a i s i n a n y t r i a n v l e s

y o u c a n c o m p u t e i t f o r a e q u i l a t e r a l

t r i a n g l e a n d f i n d t h e r i g h t a n s w e r .

i n e q u i l a t e r a l t r i a n g l e :

1) R = 2 r

r_{a} = r_{b} = r_{c} = \frac{S}{p - a} = \frac{a^{2} \frac{\sqrt{3}}{4}}{\frac{3 a}{2} - a} = a \frac{\sqrt{3}}{2} = 2 R s i n A . \frac{\sqrt{3}}{2} =

= 2 R \times \frac{\sqrt{3}}{2} . \frac{\sqrt{3}}{2} = \frac{3 R}{2}

\Rightarrow Σ \frac{r_{a}}{b c} = 3 \times \frac{a \frac{\sqrt{3}}{2}}{a^{2}} = 3 \times \frac{\frac{3 R}{2}}{3 R^{2}} = \frac{3}{2 R}

a n s w e r (1) : \frac{1}{r} - \frac{1}{R} = \frac{R - r}{r . R} = \frac{R - . 5 R}{. 5 R . R} = \frac{1}{R}

a n s w e r (2) : \frac{1}{r} - \frac{1}{2 R} = \frac{2 R - r}{2 R . r} = \frac{2 R - . 5 R}{2 R \times . 5 R} = \frac{3}{2 R}

a n s w e r (3) : \frac{1}{r} + \frac{1}{2 R} = \frac{2 R + r}{2 R . r} = \frac{2 R + . 5 R}{2 R \times . 5 R} = \frac{5}{2 R}

a n s w e r (4) : \frac{1}{r} + \frac{1}{R} = \frac{R + r}{R . r} = \frac{R + . 5 R}{R \times . 5 R} = \frac{3}{R} .

Answered by Tinkutara last updated on 06/Jul/17

Using r_{1} = 4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} and

r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}, we get

\frac{r_{1}}{b c} = \frac{4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{16 R^{2} \sin \frac{B}{2} \sin \frac{C}{2} \cos \frac{B}{2} \cos \frac{C}{2}}

= \frac{\sin \frac{A}{2}}{4 R \sin \frac{B}{2} \sin \frac{C}{2}}

Similarly we write all the terms .

\frac{r_{1}}{b c} + \frac{r_{2}}{c a} + \frac{r_{3}}{a b} = \frac{1}{4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} (\sin^{2} \frac{A}{2} + \sin^{2} \frac{B}{2} + \sin^{2} \frac{C}{2})

= \frac{1}{2 r} (1 - \cos A + 1 - \cos B + 1 - \cos C)

= \frac{1}{2 r} (3 - 1 - \frac{r}{R}) = \frac{1}{r} - \frac{1}{2 R}