Prove-in-AB-C-cosA-sin-2-A-cosB-sin-2-B-cosC-sin-2-C-r-R-r-incircle-radius-R-circumcircle-radius-

Question Number 211276 by mnjuly1970 last updated on 04/Sep/24

P r o v e, i n A \overset{Δ}{B C} :

\frac{c o s A}{{s i n}^{2} A} + \frac{c o s B}{{s i n}^{2} B} + \frac{c o s C}{{s i n}^{2} C} ⩾ \frac{r}{R}

r : i n c i r c l e r a d i u s

R : c i r c u m c i r c l e r a d i u s

Answered by Ghisom last updated on 04/Sep/24

r = \frac{a}{\cot \frac{β}{2} + \cot \frac{γ}{2}}; cyclic for a, b, c & α, β, γ

\Rightarrow sum these up and use γ = π - (α + β)

and let u = \tan α \land v = \tan β to get

3 r = \frac{(1 - u v) v}{1 + v^{2}} a + \frac{(1 - u v) u}{1 + u^{2}} b + \frac{u v}{u + v} c

R = \frac{a}{2 \sin α}; cyclic for a, b, c & α, β, γ

again sum these up to get

3 R = \frac{1 + u^{2}}{4 u} a + \frac{1 + v^{2}}{4 u} b + \frac{(1 + u^{2}) (1 + v^{2})}{4 (u + v) (1 - u v)} c

\Rightarrow

\frac{r}{R} = \frac{4 (1 - u v) u v}{(1 + u^{2}) (1 + v^{2})}

\Rightarrow

\frac{r}{R} = \cos α + \cos β \underset{= \cos γ}{\underset{⏟}{- \cos (α + β)}} - 1

\Rightarrow \frac{r}{R} = \cos α + \cos β + \cos γ - 1

[0 < \frac{r}{R} ⩽ \frac{1}{2}]

in any triangle 0 < θ < π \Rightarrow 0 < \sin θ < 1 \Rightarrow

\frac{\cos θ}{\sin^{2} θ} ⩾ \cos θ

\Rightarrow Σ \frac{\cos θ}{\sin^{2} θ} ⩾ (Σ \cos θ) - 1

[

{what}^{'} s left to prove :

2 ⩽ \frac{\cos α}{\sin^{2} α} + \frac{\cos β}{\sin^{2} β} - \frac{\cos (α + β)}{\sin^{2} (α + β)}

because of symmetry there should be a

\min / \max at α = β = \frac{π}{3}

]

Commented by mnjuly1970 last updated on 04/Sep/24

t h a n k s a l o t s i r G h i s o m

g r a t e f u \underset{―}{\underset{⏟}{⋚}}

Commented by Ghisom last updated on 04/Sep/24

I found another way to end my proof \dots

Commented by Frix last updated on 04/Sep/24

Nice .

Answered by mahdipoor last updated on 05/Sep/24

h (A, B, C) = A + B + C = 180

f (A, B, C) = \frac{c o s A}{{s i n}^{2} A} + \frac{c o s B}{{s i n}^{2} B} + \frac{c o s C}{{s i n}^{2} C}

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

S = r \frac{a + b + c}{2} = \frac{b a}{2} s i n C \Rightarrow

r \frac{a + \frac{s i n B}{s i n A} a + \frac{s i n C}{s i n A} a}{2} = \frac{\frac{s i n B}{s i n A} a \times a}{2} s i n C \Rightarrow

r = \frac{a \times s i n B \times s i n C}{s i n A + s i n B + s i n C}

g (A, B, C) = \frac{r}{R} = 2 \times \frac{s i n A \times s i n B \times s i n C}{s i n A + s i n B + s i n C}

{\begin{cases} \frac{\partial f}{\partial x_{i}} = λ \frac{\partial f}{\partial x_{i}} \\ h = 180 \end{cases} \Rightarrow f_{m i n} = f (60, 60, 60) = 2 \Rightarrow 2 ⩽ f ⩽ \infty

{\begin{cases} \frac{\partial g}{\partial x_{i}} = λ \frac{\partial g}{\partial x_{i}} \\ h = 180 \end{cases} \Rightarrow g_{m a x} = g (60, 60, 60) = \frac{1}{2} \Rightarrow 0 ⩽ g ⩽ \frac{1}{2}

\Rightarrow g ⩽ \frac{1}{2} < 2 ⩽ f \Rightarrow g < f

Commented by mnjuly1970 last updated on 04/Sep/24

t h a n k s a l o t s i r . .

Commented by Frix last updated on 04/Sep/24

Error : R = \frac{a}{2 \sin A} etc .

Commented by Ghisom last updated on 04/Sep/24

line 3 : R = \frac{a}{2 \sin α} = \frac{b}{2 \sin β} = \frac{c}{2 \sin γ}

\Rightarrow \min = \frac{1}{2} \neq \frac{1}{4}

Commented by mahdipoor last updated on 05/Sep/24

t h a n k y o u, i e d i t e d

Commented by Frix last updated on 05/Sep/24

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