in-AB-C-prove-that-sin-A-2-a-b-c-

Question Number 175585 by mnjuly1970 last updated on 03/Sep/22

i n A \overset{Δ}{B C} p r o v e t h a t :

s i n (\frac{A}{2}) ⩽ \frac{a}{b + c}

Commented by mahdipoor last updated on 05/Sep/22

\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} = 2 R \Rightarrow

\frac{a}{b + c} = \frac{2 R . s i n A}{2 R (s i n B + s i n C)} = \frac{s i n A}{s i n B + s i n C} =

\frac{2 s i n (A / 2) . c o s (A / 2)}{s i n B + s i n C} =

2 s i n (A / 2) \frac{c o s (\frac{180 - (B + C)}{2})}{s i n B + s i n C} =

2 s i n (A / 2) \frac{s i n (\frac{B + C}{2})}{2 s i n (\frac{B + C}{2}) c o s (\frac{B - C}{2})} =

\frac{s i n (A / 2)}{c o s ((B - C) / 2)}

\Rightarrow\Rightarrow\Rightarrow\Rightarrow

s i n (A / 2) ⩽ \frac{a}{b + c} \Leftrightarrow

s i n (A / 2) ⩽ \frac{s i n (A / 2)}{c o s ((B - C) / 2)} \Leftrightarrow

c o s (\frac{B - C}{2}) ⩽ 1

Commented by mnjuly1970 last updated on 03/Sep/22

z e n d e h b a s h i d o s t a d m a h d i p o o r

m a m n o o n . b e s i a r a l i b o o d .

Commented by mahdipoor last updated on 03/Sep/22

♡

Commented by behi834171 last updated on 03/Sep/22

s i r! s h o u l d u s e : R, i n s i n e r u l e .

r, i s t h e r a d i i o f i n n e r c i r c l e o f A \overset{▴}{B C} .

o s t a d m e h d i p o r! d e q a t l o t f a n .

Commented by mnjuly1970 last updated on 04/Sep/22

e r a d a t m a n d i m o s t a d . .

Commented by mahdipoor last updated on 05/Sep/22

m a m n o o n a m, d o r o s t e s h k a r d a m

Answered by behi834171 last updated on 03/Sep/22

s i n \frac{A}{2} = \sqrt{\frac{(p - b) (p - c)}{b c}} \Rightarrow

\sqrt{\frac{(p - b) (p - c)}{b c}} ⩽ \frac{a}{b + c} \Rightarrow

\frac{\sqrt{p (p - a) (p - b) (p - c)}}{\sqrt{p (p - a) . a b c}} ⩽ \frac{\sqrt{a}}{b + c} \Rightarrow

\frac{S}{\sqrt{p (p - a) 4 R . S}} ⩽ \frac{\sqrt{a}}{b + c} \Rightarrow \frac{\sqrt{r}}{2 \sqrt{(p - a) R}} ⩽ \frac{\sqrt{a}}{b + c} \Rightarrow

\Rightarrow \sqrt{\frac{r}{R}} ⩽ \frac{2 \sqrt{a (p - a)}}{b + c} \Rightarrow \frac{r}{R} ⩽ \frac{4 a (p - a)}{{(b + c)}^{2}}

\frac{4 a (p - a)}{{(b + c)}^{2}} = \frac{2 a (b + c - a)}{{(b + c)}^{2}} = 2 \frac{a}{b + c} - 2 {(\frac{a}{b + c})}^{2}

= 2 t - 2 t^{2}

f (t) = 2 t - 2 t^{2} \Rightarrow \frac{d f}{d t} = 2 (1 - 2 t) = 0 \Rightarrow t = \frac{1}{2}

f_{m i n} = 2 \times \frac{1}{2} - 2 \times \frac{1}{4} = 1 - \frac{1}{2} = \frac{1}{2} \Rightarrow f ⩽ \frac{1}{2} \Rightarrow

\Rightarrow \frac{r}{R} ⩽ \frac{1}{2} \Rightarrow R ⩾ 2 r . t h i s i s t r u e .

[{E u l e r}^{'} s r u l e :

d^{2} = R^{2} - 2 R . r = R (R - 2 r) ⩾ 0 \Rightarrow R ⩾ 2 r]

Commented by mnjuly1970 last updated on 04/Sep/22

z e n d e h b a s h i d o s t a d b o z o r g v a r

k h e i l i a l i b o o d .

Commented by behi834171 last updated on 04/Sep/22

m a m n o n . s h o m a h a m b i n a z i r h a s t e d .

z e n d e b a s h i d .