Question-52190

Question Number 52190 by ajfour last updated on 04/Jan/19

Commented by ajfour last updated on 04/Jan/19

O r e l s e f i n d j u s t θ .

Commented by MJS last updated on 04/Jan/19

putting R = 1 I get θ = \frac{π}{4} + \arcsin \frac{\sqrt{2}}{4} \approx 65.7048 °

maximum perimeter = 1 + \sqrt{3 + \sqrt{7}} + \frac{1}{2} \sqrt{10 + 2 \sqrt{7}} \approx 5.33130

Commented by ajfour last updated on 04/Jan/19

w h a t s t h e e q u a t i o n, s i r ?

Answered by mr W last updated on 04/Jan/19

a s s h o w n i n Q 52061, c a l c u l u s m e t h o d

i s n o t t h e e a s i e s t w a y t o s o l v e s u c h

q u e s t i o n s . t h e b e s t w a y i s t o f o l l o w

t h e n a t u r e d i r e c t l y . t h e o p t i m u m i n

m a t h e m a t i c s i s u s u a l l y t h a t w h a t t h e

n a t u r e d o e s . s y m m e t r y i s e . g . n a t u r e .

i f p o i n t s A a n d C b o t h l i e o n t h e c i r c l e,

w e k n o w w i t h o u t “ {c a l c u l a t i o n}^{″} w h e r e

t h e p o i n t B l i e s s u c h t h a t t h e l e n g t h

o f p a t h A - B - C i s m a x i m u m .

w e c a n s a y {i t}^{'} s t h e s y m m e t r y, b u t w e

a l s o k n o w {i t}^{'} s t h e w a y a l i g h t r a y i n

t h e n a t u r e w i l l f o l l o w .

t h e r e f o r e, w i t h o u t c a l c u l u s, t h e p a t h

A - B - C h a s i t s m a x i m u m l e n g t h,

w h e n O B i s t h e b i s e c t o r o f ∠ B .

l e t α = ∠ O B A = ∠ O B C

\Rightarrow ∠ B = 2 α, ∠ C = \frac{π}{2} + α, ∠ A = π - 2 α - (\frac{π}{2} + α) = \frac{π}{2} - 3 α

B C = 2 R \cos α

\frac{B C}{\sin A} = \frac{C A}{\sin B}

\Rightarrow \frac{2 R \cos α}{\cos 3 α} = \frac{R}{\sin 2 α}

\Rightarrow \cos 3 α = 2 \sin 2 α \cos α

\Rightarrow 4 \cos^{2} α - 3 = 2 \sin 2 α

\Rightarrow 2 \cos 2 α - 1 = 2 \sin 2 α

\Rightarrow \cos 2 α - \sin 2 α = \frac{1}{2}

\Rightarrow \sin \frac{π}{4} \cos 2 α - \cos \frac{π}{4} \sin 2 α = \frac{\sqrt{2}}{4}

\Rightarrow \sin (\frac{π}{4} - 2 α) = \frac{\sqrt{2}}{4}

\Rightarrow \frac{π}{4} - 2 α = \sin^{- 1} \frac{\sqrt{2}}{4}

θ = \frac{π}{2} - 2 α = \frac{π}{4} + (\frac{π}{4} - 2 α)

\Rightarrow θ = \frac{π}{4} + \sin^{- 1} \frac{\sqrt{2}}{4}

Commented by mr W last updated on 04/Jan/19

t h i s r e s u l t c o i n c i d e s w i t h t h e r e s u l t

b y c a l c u l u s m e t h o d a s M J S s i r h a s

s h o w n .

Answered by MJS last updated on 04/Jan/19

R = 1

A = (\begin{matrix} 1 \\ 0 \end{matrix}) C = (\begin{matrix} 0 \\ 0 \end{matrix}) B = (\begin{matrix} - \cos θ \\ 1 + \sin θ \end{matrix})

b =∣ A C ∣= 1

c =∣ A B ∣= \sqrt{3 + 2 \cos θ + 2 \sin θ}

a =∣ B C ∣= \sqrt{2 + 2 \sin θ}

p (θ) = 1 + \sqrt{2 + 2 \sin θ} + \sqrt{3 + 2 \cos θ + 2 \sin θ}

p^{'} (θ) = 0

\frac{\cos θ \sqrt{6 + 4 \cos θ + 4 \sin θ} + 2 (\cos θ - \sin θ) \sqrt{1 + \sin θ}}{2 \sqrt{1 + \sin θ} \sqrt{3 + 2 \cos θ + 2 \sin θ}} = 0

\cos θ = c

\sin θ = s

\dots

s^{3} - (2 c - 1) s^{2} - 2 c s - \frac{c^{2} (2 c + 1)}{2} = 0

2 c (2 s^{2} + 2 s + c^{2}) = 2 s^{3} + 2 s^{2} - c^{2}

c^{2} = 1 - s^{2}

2 c (s^{2} + 2 s + 1) = 2 s^{3} + 3 s^{2} - 1

2 c {(s + 1)}^{2} = {(s + 1)}^{2} (2 s - 1)

2 c = 2 s - 1

1 + 2 \cos θ - 2 \sin θ = 0

1 + 2 \sqrt{2} \cos (\frac{π}{4} + θ) = 0

θ = - \frac{π}{4} + \arccos (- \frac{\sqrt{2}}{4}) = \frac{π}{4} + \arcsin \frac{\sqrt{2}}{4}

Commented by ajfour last updated on 04/Jan/19

G r e a t B e a u t y S i r!

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