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Question Number 198647 by ajfour last updated on 22/Oct/23

Commented by ajfour last updated on 22/Oct/23

$Q i s c e n t e r o f s m a l l e r c i r c l e o f$ $r a d i u s a . F i n d m a x i m u m a r e a o f$ $△ A B C i n t e r m s o f a, b .$
	Answered by mr W last updated on 22/Oct/23

	Commented by mr W last updated on 23/Oct/23

	$l e t μ = \frac{b}{a}$ $\frac{\sin δ}{a} = \frac{\sin (α - δ)}{b} = \frac{\sin α}{P A}$ $\frac{\sin α}{\tan δ} - \cos α = \frac{b}{a} = μ$ $\Rightarrow \tan δ = \frac{\sin α}{\cos α + μ}$ $P A = \frac{a \sin α}{\sin δ}$ $Δ = \frac{b^{2} \sin (β + γ)}{2} + \frac{b P A [\sin (π + δ - β) + \sin (π - δ - γ)]}{2}$ $Δ = \frac{b^{2} \sin (β + γ)}{2} + \frac{b a \sin α [\sin (β - δ) + \sin (γ + δ)]}{2 \sin δ}$ $Δ = \frac{b^{2} \sin (β + γ)}{2} + \frac{b a \sin α [\sin β \cos δ - \cos β \sin δ + \sin γ \cos δ + \cos γ \sin δ]}{2 \sin δ}$ $Δ = \frac{b^{2} \sin (β + γ)}{2} + \frac{b a \sin α [(\sin β + \sin γ) \cot δ - \cos β + \cos γ]}{2}$ $Δ = \frac{b^{2} \sin (β + γ)}{2} + \frac{b a \sin α [(\sin β + \sin γ) \frac{\cos α + μ}{\sin α} - \cos β + \cos γ]}{2}$ $Φ = \frac{2 Δ}{a b} = μ \sin (β + γ) + (\sin β + \sin γ) (\cos α + μ) - \sin α (\cos β - \cos γ)$ $s u c h t h a t a r e a Δ i s m a x i m u m,$ $\frac{\partial Φ}{\partial α} = (\sin β + \sin γ) (- \sin α) - \cos α (\cos β - \cos γ) = 0$ $\sin α (\sin β + \sin γ) + \cos α (\cos β - \cos γ) = 0 . . . (i i i)$ $\Rightarrow \tan α = \frac{\cos γ - \cos β}{\sin β + \sin γ}$ $(\Rightarrow t h i s p r o v e s t h a t Q A ⊥ B C .)$ $\frac{\partial Φ}{\partial β} = μ \cos (β + γ) + \cos β (\cos α + μ) + \sin α \sin β = 0 . . . (i)$ $\frac{\partial Φ}{\partial γ} = μ \cos (β + γ) + \cos γ (\cos α + μ) - \sin α \sin γ = 0 . . . (i i)$ $(i) - (i i) :$ $(\cos β - \cos γ) (\cos α + μ) + \sin α (\sin β + \sin γ) = 0$ $μ (\cos β - \cos γ) + \cos α (\cos β - \cos γ) + \sin α (\sin β + \sin γ) = 0$ $d u e t o (i i i) :$ $μ (\cos β - \cos γ) = 0$ $\Rightarrow \cos β = \cos γ$ $\Rightarrow β = γ$ $i . e . p o i n t s B a n d C a r e s y m m e t r i c a b o u t x - a x i s!$ $\tan α = \frac{\cos γ - \cos β}{\sin β + \sin γ} = 0$ $\Rightarrow α = 0$ $i . e . p o i n t A l i e s o n x - a x i s!$ $i n s e r t i n t o (i) :$ $μ \cos 2 β + \cos β (1 + μ) = 0$ $μ (2 \cos^{2} β - 1) + \cos β (1 + μ) = 0$ $2 μ \cos^{2} β + (1 + μ) \cos β - μ = 0$ $\Rightarrow \cos β = \frac{- (1 + μ) + \sqrt{9 μ^{2} + 2 μ + 1}}{4 μ}$ $i . e . \cos β = \frac{- (a + b) + \sqrt{a^{2} + 9 b^{2} + 2 a b}}{4 b}$ $◼$
	Answered by mr W last updated on 23/Oct/23

	$f o r t h e t r i a n g l e w i t h m a x i m u m a r e a$ $t h e t a n g e n t s a t t h e v e r t e x e s a r e$ $p a r a l l e l t o t h e c o r e s p o n d i n g o p p o s i t e$ $s i d e s .$ $d u e t o s y m m e t r y, t h i s t r i a n g l e m u s t$ $b e s y m m e t r i c a b o u t t h e x - a x i s .$ $t h a t m e a n s f o l l o w i n g r e d t r i a n g l e$ $h a s m a x i m u m a r e a .$
	Commented by mr W last updated on 23/Oct/23

	$b \cos θ = (b \sin θ + b + a) \tan θ$ $b (1 - \sin^{2} θ) = (b \sin θ + b + a) \sin θ$ $2 b \sin^{2} θ + (a + b) \sin θ - b = 0$ $\Rightarrow \sin θ = \frac{- (a + b) + \sqrt{a^{2} + 9 b^{2} + 2 a b}}{4 b}$ $A_{m a x} = (b \sin θ + b + a) b \cos θ$ $= \frac{(3 a + 3 b + \sqrt{a^{2} + 9 b^{2} + 2 a b}) \sqrt{2 (a + 3 b) (b - a) + 2 (a + b) \sqrt{a^{2} + 9 b^{2} + 2 a b}}}{16}$
	Commented by ajfour last updated on 22/Oct/23

	Commented by mr W last updated on 23/Oct/23

	Commented by ajfour last updated on 22/Oct/23

	$S i r, w h i c h (△ o r △) s e e m s t o b e o f$ $g r e a t e r a r e a ?$
	Commented by mr W last updated on 23/Oct/23

	$a c t u a l l y i j u s t b e l i e v e t h a t t h e$ $m a x i m u m t r i a n g l e m u s t a l s o b e$ $s y m m e t r i c a b o u t x - a x i s, b u t {i t}^{'} s n o t$ $p r o v e d .$
	Commented by mr W last updated on 22/Oct/23

	Commented by ajfour last updated on 22/Oct/23

	$h a h a, g u e s s m n o t i n m y s e n s e s, t h a n k s$ $f o r e v e r y t h i n g! i w i l l f o l l o w u r {s o l}^{n} t i l l$ $f i n i s h .$
	Commented by mr W last updated on 23/Oct/23

	$m y h y p o t h e s i s a b o v e i s n o w p r o v e d,$ $s e e g e n e r a l l m e t h o d b e l o w .$ $p l e a s e h e l p c h e c k i n g i f i m a d e a n y$ $m i s t a k e s . t h a n k y o u!$
	Commented by ajfour last updated on 23/Oct/23
	Thanks sir. You were correct indeed, and very detailed everything you still present.
	Answered by ajfour last updated on 23/Oct/23

	Commented by ajfour last updated on 23/Oct/23

	$△ = \frac{1}{2} (2 b \cos θ) (b \sin θ + b \sin ϕ + a \cos δ)$ $\frac{\partial △}{\partial δ} = - b \cos θ \sin δ = 0$ $c l e a r l y w e c h o o s e δ = 0 a s (a < b)$ $\frac{\partial △}{\partial ϕ} = b \cos θ (b \cos ϕ) = 0 \Rightarrow ϕ = \frac{π}{2}$ $\frac{\partial △}{\partial θ} = b \cos θ (b \cos θ + 0)$ $- b \sin θ (b \sin θ + b \sin ϕ + a \cos δ) = 0$ $n o w s i n c e δ = 0, a n d ϕ = \frac{π}{2}$ $w e g e t b^{2} \cos^{2} θ = b \sin θ (b \sin θ + a + b)$ $\cos^{2} θ = \sin θ (\sin θ + 1 + \frac{a}{b})$ $\Rightarrow 2 \sin^{2} θ + (1 + \frac{a}{b}) \sin θ - 1 = 0$ $\sin θ i s o b t a i n e d f r o m h e r e . N o w$ $△ = b^{2} \cos θ (\sin θ + 1 + \frac{a}{b})$
	Commented by mr W last updated on 23/Oct/23
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