Question-165600

Tinku Tara June 4, 2023 Geometry 0 Comments

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Question Number 165600 by mr W last updated on 05/Feb/22

Commented by mr W last updated on 05/Feb/22

attempt to Q165565

a t t e m p t t o Q 165565

Commented by mr W last updated on 05/Feb/22

Commented by mr W last updated on 06/Feb/22

let μ=(b/a)≤1 A((a/(tan θ)), a)=(ma, a) B(b, y_B ) (y_B −a)^2 +((a/(tan θ))−b)^2 =(a+b)^2 (y_B −a)^2 =a^2 (1+(1/(tan θ)))(2μ+1−(1/(tan θ))) let m=(1/(tan θ)) y_B =a+a(√((1+m)(2μ+1−m))) tan θ≥(a/(a+2b))=(1/(2μ+1)) θ≤(π/4) ⇒1≤m≤2μ+1 x_P =(a/(tan θ))+a sin α=a(m+sin α) y_P =a−a cos α=a(1−cos α) x_Q =b−b cos β=aμ(1−cos β) y_Q =y_B +b sin β=a[1+(√((1+m)(2μ+1−m)))+μ sin β] area of triangle OPQ: Δ=((x_Q y_Q )/2)+(((x_P −x_Q )(y_P +y_Q ))/2)−((x_P y_P )/2) 2Δ=x_P y_Q −x_Q y_P 2Δ=a^2 (m+sin α)(1+(√((1+m)(2μ+1−m)))+μ sin β)−a^2 μ(1−cos β)(1−cos α) ((2Δ)/(μa^2 ))=(m+sin α)(((1+(√((1+m)(2μ+1−m))))/μ)+sin β)−(1−cos β)(1−cos α) let Φ=((2Δ)/(μa^2 )), λ=((1+(√((1+m)(2μ+1−m))))/μ) Φ=(m+sin α)(λ+sin β)−(1−cos β)(1−cos α) Φ=mλ−1+λ sin α+cos α+m sin β+cos β−cos (α+β) Φ=mλ−1+(√(1+λ^2 )) cos (α−tan^(−1) λ)+(√(1+m^2 )) cos (β−tan^(−1) m)−cos (α+β) Φ is a function of three variables: m, α, β. i′ll try in geometric way: for maiximum area of triangle OPQ, tangent at Q must be // OP: (1/(tan β))=(y_P /x_P )=((1−cos α)/(m+sin α)) ⇒((cos β)/(sin β))=((1−cos α)/(m+sin α)) m cos β−sin β+sin α cos β+cos α sin β=0 m cos β−sin β+sin (α+β)=0 ⇒(√(1+m^2 )) sin (β−tan^(−1) m)=sin (α+β) (we can get the same also from (∂Φ/∂β)=0) tangent at P must be // OQ: tan α=(y_Q /x_Q )=((1+(√((1+m)(2μ+1−m)))+μ sin β)/(μ(1−cos β))) ⇒((sin α)/(cos α))=((λ+sin β)/(1−cos β)) λ cos α−sin α+cos α sin β+sin α cos β=0 λ cos α−sin α+sin (α+β)=0 ⇒(√(1+λ^2 )) sin (α−tan^(−1) λ)=sin (α+β) (we can get the same also from (∂Φ/∂α)=0) we can see that (π/2)<α+β<π in case of that Δ is maximum. say (√(1+λ^2 )) sin (α−tan^(−1) λ)=(√(1+m^2 )) sin (β−tan^(−1) m)=sin (α+β)=ξ ⇒α=tan^(−1) λ+sin^(−1) (ξ/( (√(1+λ^2 )))) ⇒β=tan^(−1) m+sin^(−1) (ξ/( (√(1+m^2 )))) ⇒α+β=π−sin^(−1) ξ determinant (((sin^(−1) ξ+sin^(−1) (ξ/( (√(1+λ^2 ))))+sin^(−1) (ξ/( (√(1+m^2 ))))=π−tan^(−1) λ−tan^(−1) m))) ...(I) (√(1+λ^2 )) sin (α−tan^(−1) λ)=ξ ⇒cos (α−tan^(−1) λ)=(√(1−(ξ^2 /(1+λ^2 )))) (√(1+m^2 )) sin (β−tan^(−1) m)=ξ ⇒cos (β−tan^(−1) m)=(√(1−(ξ^2 /(1+m^2 )))) sin (α+β)=ξ ⇒cos (α+β)=−(√(1−ξ^2 )) Φ=mλ−1+(√(1+λ^2 )) cos (α−tan^(−1) λ)+(√(1+m^2 )) cos (β−tan^(−1) m)−cos (α+β) determinant (((Φ=mλ−1+(√(1+λ^2 −ξ^2 ))+(√(1+m^2 −ξ^2 )) +(√(1−ξ^2 ))))) ...(II)

l e t μ = \frac{b}{a} ⩽ 1

A (\frac{a}{\tan θ}, a) = (m a, a)

B (b, y_{B})

{(y_{B} - a)}^{2} + {(\frac{a}{\tan θ} - b)}^{2} = {(a + b)}^{2}

{(y_{B} - a)}^{2} = a^{2} (1 + \frac{1}{\tan θ}) (2 μ + 1 - \frac{1}{\tan θ})

l e t m = \frac{1}{\tan θ}

y_{B} = a + a \sqrt{(1 + m) (2 μ + 1 - m)}

\tan θ ⩾ \frac{a}{a + 2 b} = \frac{1}{2 μ + 1}

θ ⩽ \frac{π}{4}

\Rightarrow 1 ⩽ m ⩽ 2 μ + 1

x_{P} = \frac{a}{\tan θ} + a \sin α = a (m + \sin α)

y_{P} = a - a \cos α = a (1 - \cos α)

x_{Q} = b - b \cos β = a μ (1 - \cos β)

y_{Q} = y_{B} + b \sin β = a [1 + \sqrt{(1 + m) (2 μ + 1 - m)} + μ \sin β]

a r e a o f t r i a n g l e O P Q :

Δ = \frac{x_{Q} y_{Q}}{2} + \frac{(x_{P} - x_{Q}) (y_{P} + y_{Q})}{2} - \frac{x_{P} y_{P}}{2}

2 Δ = x_{P} y_{Q} - x_{Q} y_{P}

2 Δ = a^{2} (m + \sin α) (1 + \sqrt{(1 + m) (2 μ + 1 - m)} + μ \sin β) - a^{2} μ (1 - \cos β) (1 - \cos α)

\frac{2 Δ}{μ a^{2}} = (m + \sin α) (\frac{1 + \sqrt{(1 + m) (2 μ + 1 - m)}}{μ} + \sin β) - (1 - \cos β) (1 - \cos α)

l e t Φ = \frac{2 Δ}{μ a^{2}}, λ = \frac{1 + \sqrt{(1 + m) (2 μ + 1 - m)}}{μ}

Φ = (m + \sin α) (λ + \sin β) - (1 - \cos β) (1 - \cos α)

Φ = m λ - 1 + λ \sin α + \cos α + m \sin β + \cos β - \cos (α + β)

Φ = m λ - 1 + \sqrt{1 + λ^{2}} \cos (α - \tan^{- 1} λ) + \sqrt{1 + m^{2}} \cos (β - \tan^{- 1} m) - \cos (α + β)

Φ i s a f u n c t i o n o f t h r e e v a r i a b l e s :

m, α, β .

i^{'} l l t r y i n g e o m e t r i c w a y :

f o r m a i x i m u m a r e a o f t r i a n g l e O P Q,

t a n g e n t a t Q m u s t b e / / O P :

\frac{1}{\tan β} = \frac{y_{P}}{x_{P}} = \frac{1 - \cos α}{m + \sin α}

\Rightarrow \frac{\cos β}{\sin β} = \frac{1 - \cos α}{m + \sin α}

m \cos β - \sin β + \sin α \cos β + \cos α \sin β = 0

m \cos β - \sin β + \sin (α + β) = 0

\Rightarrow \sqrt{1 + m^{2}} \sin (β - \tan^{- 1} m) = \sin (α + β)

(w e c a n g e t t h e s a m e a l s o f r o m \frac{\partial Φ}{\partial β} = 0)

t a n g e n t a t P m u s t b e / / O Q :

\tan α = \frac{y_{Q}}{x_{Q}} = \frac{1 + \sqrt{(1 + m) (2 μ + 1 - m)} + μ \sin β}{μ (1 - \cos β)}

\Rightarrow \frac{\sin α}{\cos α} = \frac{λ + \sin β}{1 - \cos β}

λ \cos α - \sin α + \cos α \sin β + \sin α \cos β = 0

λ \cos α - \sin α + \sin (α + β) = 0

\Rightarrow \sqrt{1 + λ^{2}} \sin (α - \tan^{- 1} λ) = \sin (α + β)

(w e c a n g e t t h e s a m e a l s o f r o m \frac{\partial Φ}{\partial α} = 0)

w e c a n s e e t h a t \frac{π}{2} < α + β < π i n c a s e o f

t h a t Δ i s m a x i m u m .

s a y \sqrt{1 + λ^{2}} \sin (α - \tan^{- 1} λ) = \sqrt{1 + m^{2}} \sin (β - \tan^{- 1} m) = \sin (α + β) = ξ

\Rightarrow α = \tan^{- 1} λ + \sin^{- 1} \frac{ξ}{\sqrt{1 + λ^{2}}}

\Rightarrow β = \tan^{- 1} m + \sin^{- 1} \frac{ξ}{\sqrt{1 + m^{2}}}

\Rightarrow α + β = π - \sin^{- 1} ξ

\begin{matrix} \sin^{- 1} ξ + \sin^{- 1} \frac{ξ}{\sqrt{1 + λ^{2}}} + \sin^{- 1} \frac{ξ}{\sqrt{1 + m^{2}}} = π - \tan^{- 1} λ - \tan^{- 1} m \end{matrix} \dots (I)

\sqrt{1 + λ^{2}} \sin (α - \tan^{- 1} λ) = ξ

\Rightarrow \cos (α - \tan^{- 1} λ) = \sqrt{1 - \frac{ξ^{2}}{1 + λ^{2}}}

\sqrt{1 + m^{2}} \sin (β - \tan^{- 1} m) = ξ

\Rightarrow \cos (β - \tan^{- 1} m) = \sqrt{1 - \frac{ξ^{2}}{1 + m^{2}}}

\sin (α + β) = ξ

\Rightarrow \cos (α + β) = - \sqrt{1 - ξ^{2}}

Φ = m λ - 1 + \sqrt{1 + λ^{2}} \cos (α - \tan^{- 1} λ) + \sqrt{1 + m^{2}} \cos (β - \tan^{- 1} m) - \cos (α + β)

\begin{matrix} Φ = m λ - 1 + \sqrt{1 + λ^{2} - ξ^{2}} + \sqrt{1 + m^{2} - ξ^{2}} + \sqrt{1 - ξ^{2}} \end{matrix} \dots (I I)

Commented by mr W last updated on 05/Feb/22

Commented by mr W last updated on 05/Feb/22

Commented by mr W last updated on 05/Feb/22

Commented by mr W last updated on 05/Feb/22

the positions of the circles are defined by the parameter m=(1/(tan θ)) with 1≤m≤2μ+1. for a given m we can determine the parameter ξ which defines the positions of points P and Q on the circles, i.e. α and β, such that Δ_(OPQ) is maximum relatively to the given m. but since the equation for ξ can not be solved analytically, i have not found a way to determine the m for the absolute maximum Δ_(OPQ) .

t h e p o s i t i o n s o f t h e c i r c l e s a r e d e f i n e d

b y t h e p a r a m e t e r m = \frac{1}{\tan θ} w i t h

1 ⩽ m ⩽ 2 μ + 1 .

f o r a g i v e n m w e c a n d e t e r m i n e t h e

p a r a m e t e r ξ w h i c h d e f i n e s t h e

p o s i t i o n s o f p o i n t s P a n d Q o n t h e

c i r c l e s, i . e . α a n d β, s u c h t h a t Δ_{O P Q} i s

m a x i m u m r e l a t i v e l y t o t h e g i v e n m .

b u t s i n c e t h e e q u a t i o n f o r ξ c a n n o t b e

s o l v e d a n a l y t i c a l l y, i h a v e n o t f o u n d

a w a y t o d e t e r m i n e t h e m f o r t h e

a b s o l u t e m a x i m u m Δ_{O P Q} .

Commented by aleks041103 last updated on 05/Feb/22

In a very cute vectorial way you can prove that for max area OP⊥QB and OQ⊥PA It is possible to prove that for max area (x_P /y_Q )=((y_B −y_A )/(x_A −x_B )) later i′ll post a proof! I hope that helps!

I n a v e r y c u t e v e c t o r i a l w a y y o u c a n

p r o v e t h a t f o r m a x a r e a

O P ⊥ Q B a n d O Q ⊥ P A

I t i s p o s s i b l e t o p r o v e t h a t f o r m a x a r e a

\frac{x_{P}}{y_{Q}} = \frac{y_{B} - y_{A}}{x_{A} - x_{B}}

l a t e r i^{'} l l p o s t a p r o o f! I h o p e t h a t h e l p s!

Commented by mr W last updated on 05/Feb/22

that′s right sir. thanks! this is also that what my diagram at the beginning shows: tangent at Q // AP, i.e. BQ⊥OP tangent at P // AQ, i.e. AP⊥OQ

{t h a t}^{'} s r i g h t s i r . t h a n k s!

t h i s i s a l s o t h a t w h a t m y d i a g r a m

a t t h e b e g i n n i n g s h o w s :

t a n g e n t a t Q / / A P, i . e . B Q ⊥ O P

t a n g e n t a t P / / A Q, i . e . A P ⊥ O Q

Commented by Tawa11 last updated on 05/Feb/22

Great sir

Great sir

Commented by mr W last updated on 06/Feb/22

Now i have found a way to find the absolute maximum area of the triangle and the corresponding parameter m. The relationship between the paramters ξ and m is the equation I: determinant (((sin^(−1) ξ+sin^(−1) (ξ/( (√(1+λ^2 ))))+sin^(−1) (ξ/( (√(1+m^2 ))))=π−tan^(−1) λ−tan^(−1) m))) the area of the triangle is expressed by the equation II: determinant (((Φ=mλ−1+(√(1+λ^2 −ξ^2 ))+(√(1+m^2 −ξ^2 )) +(√(1−ξ^2 ))))) with Φ=((2Δ)/(μa^2 )). the maximum Φ is when the curves of both equations just touch each other, as following diagram shows. example: μ=(b/a)=0.75 Φ_(max) =11.125757 or Δ_(max) =((μa^2 Φ_(max) )/2)=4.172a^2 at m=1.9858 or θ=tan^(−1) ((1/m))

N o w i h a v e f o u n d a w a y t o f i n d

t h e a b s o l u t e m a x i m u m a r e a o f t h e

t r i a n g l e a n d t h e c o r r e s p o n d i n g

p a r a m e t e r m .

T h e r e l a t i o n s h i p b e t w e e n t h e p a r a m t e r s

ξ a n d m i s t h e e q u a t i o n I :

\begin{matrix} \sin^{- 1} ξ + \sin^{- 1} \frac{ξ}{\sqrt{1 + λ^{2}}} + \sin^{- 1} \frac{ξ}{\sqrt{1 + m^{2}}} = π - \tan^{- 1} λ - \tan^{- 1} m \end{matrix}

t h e a r e a o f t h e t r i a n g l e i s e x p r e s s e d

b y t h e e q u a t i o n I I :

\begin{matrix} Φ = m λ - 1 + \sqrt{1 + λ^{2} - ξ^{2}} + \sqrt{1 + m^{2} - ξ^{2}} + \sqrt{1 - ξ^{2}} \end{matrix}

w i t h Φ = \frac{2 Δ}{μ a^{2}} .

t h e m a x i m u m Φ i s w h e n t h e c u r v e s

o f b o t h e q u a t i o n s j u s t t o u c h e a c h

o t h e r, a s f o l l o w i n g d i a g r a m s h o w s .

\underset{―}{e x a m p l e :}

μ = \frac{b}{a} = 0.75

Φ_{m a x} = 11.125757 o r

Δ_{m a x} = \frac{μ a^{2} Φ_{m a x}}{2} = 4.172 a^{2}

a t m = 1.9858 o r θ = \tan^{- 1} (\frac{1}{m})

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

further examples: μ=(b/a)=0.4 Φ_(max) =13.51418 or Δ_(max) =((μa^2 Φ_(max) )/2)=2.703a^2 at m=1.3871

f u r t h e r e x a m p l e s :

μ = \frac{b}{a} = 0.4

Φ_{m a x} = 13.51418 o r

Δ_{m a x} = \frac{μ a^{2} Φ_{m a x}}{2} = 2.703 a^{2}

a t m = 1.3871

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