Question-52881

Question Number 52881 by ajfour last updated on 14/Jan/19

Commented by ajfour last updated on 14/Jan/19

T o f i n d r i n t e r m s o f a, b, c .

Commented by mr W last updated on 14/Jan/19

c a n i t b e c o n f i r m e d t h a t

r = \frac{2 b c}{a + b + c} \sqrt{\frac{(a + b - c) (a + c - b)}{(a + b + c) (b + c - a)}}

Commented by ajfour last updated on 14/Jan/19

n i c e r e s u l t S i r, h o w d^{'} y a g e t t h i s ?

Commented by behi83417@gmail.com last updated on 14/Jan/19

r = \frac{b c}{p} t g \frac{A}{2} . i t i s t r u e s i r m r W .

Commented by MJS last updated on 14/Jan/19

\dots and please re - check 52061, I get a different

result

Commented by MJS last updated on 14/Jan/19

see my answer to qu . 52806

Commented by mr W last updated on 15/Jan/19

M J S s i r : y o u r r e s u l t f o r r i s t h e s a m e .

i t c a n a l s o b e f o r m e d t o

r = \frac{2 b c}{a + b + c} \sqrt{\frac{(a + b - c) (a + c - b)}{(a + b + c) (b + c - a)}}

Commented by mr W last updated on 15/Jan/19

M J S s i r : y o u r r e s u l t f o r Q 52061 s e e m s

n o t t o b e c o r r e c t .

Commented by MJS last updated on 15/Jan/19

I will look into 52061 once more

I saw that the other result is the same as yours,

I thought you might have achieved it throuhh

a different path

Answered by ajfour last updated on 15/Jan/19

F r o m p o w e r o f p o i n t Q w . r . t .

c i r c u m c i r c l e :

(2 R \cos ϕ - \frac{r}{\sin θ}) \frac{r}{\sin θ} = r (2 R - r)

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\Rightarrow 2 R \cos ϕ = \frac{r}{\sin θ} + (2 R - r) \sin θ

\dots \dots (i)

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f u r t h e r R \cos (θ + ϕ) = \frac{b}{2} &

R \cos (θ - ϕ) = \frac{c}{2}

A d d i n g : 2 R \cos ϕ \cos θ = \frac{b + c}{2}

S o u s i n g (i)

\frac{r}{\tan θ} + R \sin 2 θ - \frac{r \sin^{2} θ}{\tan θ} = \frac{b + c}{2}

b u t \frac{\sin 2 θ}{a} = \frac{1}{2 R} \Rightarrow

\frac{{r c o s}^{2} θ}{t a n θ} = \frac{b + c - a}{2}

r = \frac{(b + c - a)}{1 + c o s A} \sqrt{\frac{1 - c o s A}{1 + c o s A}}

= \frac{(b + c - a)}{(\frac{2 b c + b^{2} + c^{2} - a^{2}}{2 b c})} \sqrt{\frac{2 b c - b^{2} - c^{2} + a^{2}}{2 b c + b^{2} + c^{2} - a^{2}}}

r = \frac{2 b c}{(a + b + c)} \sqrt{\frac{(a + b - c) (a + c - b)}{(b + c + a) (b + c - a)}} .

Commented by mr W last updated on 15/Jan/19

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