Question-53910

Tinku Tara June 4, 2023 Geometry 0 Comments

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Question Number 53910 by ajfour last updated on 27/Jan/19

Commented by ajfour last updated on 27/Jan/19

Find maximum area of quadrilateral ABCD.

F i n d m a x i m u m a r e a o f q u a d r i l a t e r a l

A B C D .

Commented by ajfour last updated on 27/Jan/19

Commented by ajfour last updated on 27/Jan/19

At least find θ and φ in terms of R and r such that area ABCDA is maximum.

A t l e a s t f i n d θ a n d ϕ i n t e r m s o f

R a n d r s u c h t h a t a r e a A B C D A i s

m a x i m u m .

Commented by ajfour last updated on 27/Jan/19

shall your result meet with what i calculated, please confirm it, Sir.. tan θ = ((R+r(1+cos φ))/(rsin φ)) tan φ = ((r+R(1+cos θ))/(Rsin θ)) .

s h a l l y o u r r e s u l t m e e t w i t h w h a t

i c a l c u l a t e d, p l e a s e c o n f i r m i t, S i r . .

\tan θ = \frac{R + r (1 + \cos ϕ)}{r \sin ϕ}

\tan ϕ = \frac{r + R (1 + \cos θ)}{R \sin θ} .

Commented by mr W last updated on 27/Jan/19

without calculation i guess the result is when tangent at D is parallel to diagonal AC and tangent at C is parallel to diagonal BD. i.e. CA⊥AD, BD⊥BC.

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

i s w h e n t a n g e n t a t D i s p a r a l l e l t o

d i a g o n a l A C a n d t a n g e n t a t C i s

p a r a l l e l t o d i a g o n a l B D .

i . e . C A ⊥ A D, B D ⊥ B C .

Commented by ajfour last updated on 27/Jan/19

Brilliant professor Sir, but my final cubic solution had error, i am trying to see if Cardano′s fit, please help there.

B r i l l i a n t p r o f e s s o r S i r, b u t m y

f i n a l c u b i c s o l u t i o n h a d e r r o r,

i a m t r y i n g t o s e e i f {C a r d a n o}^{'} s f i t,

p l e a s e h e l p t h e r e .

Commented by mr W last updated on 27/Jan/19

yes, i can confirm that your solution matches that what i guess.

y e s, i c a n c o n f i r m t h a t y o u r s o l u t i o n

m a t c h e s t h a t w h a t i g u e s s .

Commented by mr W last updated on 27/Jan/19

Commented by mr W last updated on 27/Jan/19

∠BAC=(π/2)−θ [R+r(1+cos φ)]tan ((π/2)−θ)=r sin φ ⇒tan θ=((R+r(1+cos φ))/(r sin φ)) similarly: [r+R(1+cos θ)]tan ((π/2)−φ)=R sin θ ⇒tan φ=((r+R(1+cos θ))/(R sin θ))

∠ B A C = \frac{π}{2} - θ

[R + r (1 + \cos ϕ)] \tan (\frac{π}{2} - θ) = r \sin ϕ

\Rightarrow \tan θ = \frac{R + r (1 + \cos ϕ)}{r \sin ϕ}

s i m i l a r l y :

[r + R (1 + \cos θ)] \tan (\frac{π}{2} - ϕ) = R \sin θ

\Rightarrow \tan ϕ = \frac{r + R (1 + \cos θ)}{R \sin θ}

Commented by mr W last updated on 27/Jan/19

this is how i guess the result: if C is fixed on the second circle, i.e. ΔACB is fixed, max. ΔACD is when tangent at D is parallel to AC, for this is the max. altitude over AC. similarly if D is fixed on the first circle, max. ΔBDC is when tangent at C is parallel to BD.

t h i s i s h o w i g u e s s t h e r e s u l t :

i f C i s f i x e d o n t h e s e c o n d c i r c l e, i . e .

Δ A C B i s f i x e d, m a x . Δ A C D i s w h e n

t a n g e n t a t D i s p a r a l l e l t o A C, f o r t h i s

i s t h e m a x . a l t i t u d e o v e r A C .

s i m i l a r l y i f D i s f i x e d o n t h e f i r s t

c i r c l e, m a x . Δ B D C i s w h e n t a n g e n t

a t C i s p a r a l l e l t o B D .

Commented by mr W last updated on 27/Jan/19

i think we got the correct exact solution now.

i t h i n k w e g o t t h e c o r r e c t e x a c t s o l u t i o n

n o w .

Commented by ajfour last updated on 28/Jan/19

I too believed Cardano′s method shall fetch all roots, whether real or complex; MjS sir′s comment once had let me doubt this belief for some time; or may be i misinterpreted it..thanks!

I t o o b e l i e v e d {C a r d a n o}^{'} s m e t h o d

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

o r c o m p l e x; M j S {s i r}^{'} s c o m m e n t

o n c e h a d l e t m e d o u b t t h i s b e l i e f

f o r s o m e t i m e; o r m a y b e i

m i s i n t e r p r e t e d i t . . t h a n k s!

Answered by ajfour last updated on 27/Jan/19

Commented by ajfour last updated on 27/Jan/19

C(R+r+rcos φ, rsin φ) A=Area(△ABC)+Area(△ACD) A=(((R+r)/2))rsin φ+((Rh)/2) Eq. of AD: y=−xtan θ h=[rsin φ+(R+r+rcos φ)tan θ]cos θ =rsin φcos θ+(R+r+rcos φ)sin θ = (R+r)sin θ+rsin (θ+φ) 2A=(R+r)rsin φ+R(R+r)sin θ +Rrsin (θ+φ) let μ=(R/r) ⇒ 2A/r^2 =(μ+1)sin φ+μ(1+μ)sin θ +μsin (θ+φ) ((2A)/((μ+1)r^2 ))=sin φ+μsin θ+λsin (θ+φ) where λ=(μ/(μ+1)) =(R/(R+r)) ((2/((μ+1)r^2 )))((∂(A))/∂θ)=μcos θ+λcos (θ+φ)=0 ⇒ μcos θ = −λcos (θ+φ) ...(i) ((2/((μ+1)r^2 )))((∂(A))/∂φ)= cos φ+λcos (θ+φ)=0 ⇒ cos φ = −λcos (θ+φ) ...(ii) ⇒ cos φ = μcos θ Now using (i) with μ/λ = b=μ+1 bcos θ+cos θcos φ−sin θsin φ=0 cos^2 θ(b+μcos θ)^2 =(1−cos^2 θ)(1−μ^2 cos^2 θ) let cos θ= z ⇒ b^2 z^2 +2μbz^3 =1−μ^2 z^2 −z^2 but b=μ+1 , ⇒ 2(μ^2 +μ+1)z^2 +2μ(μ+1)z^3 −1=0 ⇒ (1/z^3 )−2(μ^2 +μ+1)(1/z)−2μ(μ+1)=0 (1/z)=(1/(cos θ))=sec θ = t ⇒ t^3 +pt+q =0 with p=−2(μ^2 +μ+1) ; q=−2μ(μ+1) (q^2 /4)+(p^3 /(27)) = μ^2 (μ+1)^2 −((8(μ^2 +μ+1)^3 )/(27)) D < 0 perhaps (as μ>1) ....

C (R + r + r \cos ϕ, r \sin ϕ)

A = A r e a (△ A B C) + A r e a (△ A C D)

A = (\frac{R + r}{2}) r \sin ϕ + \frac{R h}{2}

E q . o f A D : y = - x \tan θ

h = [r \sin ϕ + (R + r + r \cos ϕ) \tan θ] \cos θ

= r \sin ϕ \cos θ + (R + r + r \cos ϕ) \sin θ

= (R + r) \sin θ + r \sin (θ + ϕ)

2 A = (R + r) r \sin ϕ + R (R + r) \sin θ

+ R r \sin (θ + ϕ)

l e t μ = \frac{R}{r}

\Rightarrow 2 A / r^{2} = (μ + 1) \sin ϕ + μ (1 + μ) \sin θ

+ μ \sin (θ + ϕ)

\frac{2 A}{(μ + 1) r^{2}} = \sin ϕ + μ \sin θ + λ \sin (θ + ϕ)

w h e r e λ = \frac{μ}{μ + 1} = \frac{R}{R + r}

(\frac{2}{(μ + 1) r^{2}}) \frac{\partial (A)}{\partial θ} = μ \cos θ + λ \cos (θ + ϕ) = 0

\Rightarrow μ \cos θ = - λ \cos (θ + ϕ) \dots (i)

(\frac{2}{(μ + 1) r^{2}}) \frac{\partial (A)}{\partial ϕ} = \cos ϕ + λ \cos (θ + ϕ) = 0

\Rightarrow \cos ϕ = - λ \cos (θ + ϕ) \dots (i i)

\Rightarrow \cos ϕ = μ \cos θ

N o w u s i n g (i) w i t h μ / λ = b = μ + 1

b \cos θ + \cos θ \cos ϕ - \sin θ \sin ϕ = 0

\cos^{2} θ {(b + μ \cos θ)}^{2} = (1 - \cos^{2} θ) (1 - μ^{2} \cos^{2} θ)

l e t \cos θ = z

\Rightarrow b^{2} z^{2} + 2 μ {b z}^{3} = 1 - μ^{2} z^{2} - z^{2}

b u t b = μ + 1, \Rightarrow

2 (μ^{2} + μ + 1) z^{2} + 2 μ (μ + 1) z^{3} - 1 = 0

\Rightarrow \frac{1}{z^{3}} - 2 (μ^{2} + μ + 1) \frac{1}{z} - 2 μ (μ + 1) = 0

\frac{1}{z} = \frac{1}{\cos θ} = \sec θ = t

\Rightarrow t^{3} + p t + q = 0

w i t h p = - 2 (μ^{2} + μ + 1); q = - 2 μ (μ + 1)

\frac{q^{2}}{4} + \frac{p^{3}}{27} = μ^{2} {(μ + 1)}^{2} - \frac{8 {(μ^{2} + μ + 1)}^{3}}{27}

D < 0 p e r h a p s (a s μ > 1)

\dots .

Commented by ajfour last updated on 27/Jan/19

Thanks a lot Sir!

T h a n k s a l o t S i r!

Commented by mr W last updated on 27/Jan/19

Commented by mr W last updated on 28/Jan/19

t^3 −pt−q =0 with p=2(μ^2 +μ+1) ; q=2μ(μ+1) (u+v)^3 −3uv(u+v)−(u^3 +v^3 )=0 let t=u+v u^3 v^3 =((p/3))^3 =(8/(27))(μ^2 +μ+1)^3 u^3 +v^3 =q=2μ(μ+1) u^3 and v^3 are roots of x^2 −2μ(μ+1)x+(8/(27))(μ^2 +μ+1)^3 =0 D=4[μ^2 (μ+1)^2 −(8/(27))(μ^2 +μ+1)^3 ]<0 u^3 (v^3 )=μ(μ+1)±i(√((8/(27))(μ^2 +μ+1)^3 −μ^2 (μ+1)^2 ))=A±iB with A=μ(μ+1), B=(√((8/(27))(μ^2 +μ+1)^3 −μ^2 (μ+1)^2 )) A+iB=(√(A^2 +B^2 ))((A/( (√(A^2 +B^2 ))))+i(B/( (√(A^2 +B^2 )))))=(√(A^2 +B^2 ))(cos α+i sin α) with α=tan^(−1) (B/A)=tan^(−1) (√(((8(μ^2 +μ+1)^3 )/(27μ^2 (μ+1)^2 ))−1)) A^2 +B^2 =(8/(27))(μ^2 +μ+1)^3 u^3 (v^3 )=(√(A^2 +B^2 ))(cos α±i sin α) ⇒u=((A^2 +B^2 ))^(1/6) [cos (((α+2nπ)/3))+i sin (((α+2nπ)/3))] ⇒v=((A^2 +B^2 ))^(1/6) [cos (((α+2nπ)/3))−i sin (((α+2nπ)/3))] ⇒t=u+v=2((A^2 +B^2 ))^(1/6) cos (((α+2nπ)/3)), n=0,1,2 ⇒t=sec θ=2(√((2(μ^2 +μ+1))/3)) cos ((1/3)tan^(−1) (√(((8(μ^2 +μ+1)^3 )/(27μ^2 (μ+1)^2 ))−1)))

t^{3} - p t - q = 0

w i t h p = 2 (μ^{2} + μ + 1); q = 2 μ (μ + 1)

{(u + v)}^{3} - 3 u v (u + v) - (u^{3} + v^{3}) = 0

l e t t = u + v

u^{3} v^{3} = {(\frac{p}{3})}^{3} = \frac{8}{27} {(μ^{2} + μ + 1)}^{3}

u^{3} + v^{3} = q = 2 μ (μ + 1)

u^{3} a n d v^{3} a r e r o o t s o f

x^{2} - 2 μ (μ + 1) x + \frac{8}{27} {(μ^{2} + μ + 1)}^{3} = 0

D = 4 [μ^{2} {(μ + 1)}^{2} - \frac{8}{27} {(μ^{2} + μ + 1)}^{3}] < 0

u^{3} (v^{3}) = μ (μ + 1) \pm i \sqrt{\frac{8}{27} {(μ^{2} + μ + 1)}^{3} - μ^{2} {(μ + 1)}^{2}} = A \pm i B

w i t h A = μ (μ + 1), B = \sqrt{\frac{8}{27} {(μ^{2} + μ + 1)}^{3} - μ^{2} {(μ + 1)}^{2}}

A + i B = \sqrt{A^{2} + B^{2}} (\frac{A}{\sqrt{A^{2} + B^{2}}} + i \frac{B}{\sqrt{A^{2} + B^{2}}}) = \sqrt{A^{2} + B^{2}} (\cos α + i \sin α)

w i t h α = \tan^{- 1} \frac{B}{A} = \tan^{- 1} \sqrt{\frac{8 {(μ^{2} + μ + 1)}^{3}}{27 μ^{2} {(μ + 1)}^{2}} - 1}

A^{2} + B^{2} = \frac{8}{27} {(μ^{2} + μ + 1)}^{3}

u^{3} (v^{3}) = \sqrt{A^{2} + B^{2}} (\cos α \pm i \sin α)

\Rightarrow u = \sqrt[6]{A^{2} + B^{2}} [\cos (\frac{α + 2 n π}{3}) + i \sin (\frac{α + 2 n π}{3})]

\Rightarrow v = \sqrt[6]{A^{2} + B^{2}} [\cos (\frac{α + 2 n π}{3}) - i \sin (\frac{α + 2 n π}{3})]

\Rightarrow t = u + v = 2 \sqrt[6]{A^{2} + B^{2}} \cos (\frac{α + 2 n π}{3}), n = 0, 1, 2

\Rightarrow t = \sec θ = 2 \sqrt{\frac{2 (μ^{2} + μ + 1)}{3}} \cos (\frac{1}{3} \tan^{- 1} \sqrt{\frac{8 {(μ^{2} + μ + 1)}^{3}}{27 μ^{2} {(μ + 1)}^{2}} - 1})

Commented by ajfour last updated on 27/Jan/19

EXCELLENT Sir, but wont any of the other two solutions might fit?

EXCELLENT S i r, b u t w o n t a n y o f

t h e o t h e r t w o s o l u t i o n s m i g h t f i t ?

Commented by mr W last updated on 27/Jan/19

no, because we expect that 0<θ<(π/2).

n o, b e c a u s e w e e x p e c t t h a t 0 < θ < \frac{π}{2} .

Commented by mr W last updated on 27/Jan/19

example: μ=(R/r)=1, θ=φ=68.53° check: from tan θ=((R+r(1+cos φ))/(r sin φ)) ⇒tan θ=((2+cos θ)/(sin θ)) 2cos^2 θ+2cos θ−1=0 cos θ=(((√3)−1)/2) ⇒θ=cos^(−1) (((√3)−1)/2)=68.53°

e x a m p l e :

μ = \frac{R}{r} = 1, θ = ϕ = 68.53 °

c h e c k :

f r o m \tan θ = \frac{R + r (1 + \cos ϕ)}{r \sin ϕ}

\Rightarrow \tan θ = \frac{2 + \cos θ}{\sin θ}

{2 \cos}^{2} θ + 2 \cos θ - 1 = 0

\cos θ = \frac{\sqrt{3} - 1}{2}

\Rightarrow θ = \cos^{- 1} \frac{\sqrt{3} - 1}{2} = 68.53 °

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