Find-the-maximum-area-of-a-triangle-inscribed-in-an-ellipse-with-parameters-a-and-b-

Question Number 50674 by mr W last updated on 18/Dec/18

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

i n s c r i b e d i n a n e l l i p s e w i t h p a r a m e t e r s

a a n d b .

Commented by ajfour last updated on 18/Dec/18

Commented by MJS last updated on 18/Dec/18

all triangles created as follows have the

same (maximum) area

1 . draw a circle with radius a

2 . insert an equilateral triangle with 0 ⩽ α < \frac{2 π}{3}

A = (\begin{matrix} a \cos α \\ a \sin α \end{matrix}) B = (\begin{matrix} a \cos (α + \frac{2 π}{3}) \\ a \sin (α + \frac{2 π}{3}) \end{matrix}) C = (\begin{matrix} a \cos (α - \frac{2 π}{3}) \\ a \sin (α - \frac{2 π}{3}) \end{matrix})

3 . compress (\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = (\begin{matrix} x \\ \frac{b}{a} y \end{matrix})

Commented by mr W last updated on 19/Dec/18

t h a n k y o u s i r!

Answered by ajfour last updated on 18/Dec/18

l e t a > b, B (a \cos θ, b \sin θ)

A = a b \cos θ (1 - \sin θ)

\frac{d A}{d θ} = 0 \Rightarrow \cos^{2} θ = \sin^{2} θ - \sin θ

\Rightarrow 2 \sin^{2} θ - \sin θ - 1 = 0

{(\sin θ - \frac{1}{4})}^{2} = \frac{9}{16}

\Rightarrow \sin θ = - \frac{1}{2} \Rightarrow θ = - \frac{π}{6}

A_{m a x} = \frac{3 \sqrt{3} a b}{4} .

Commented by mr W last updated on 18/Dec/18

t h a n k s! b u t w h y m u s t A b e o n t h e a x i s ?

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Dec/18

△ A B C A (a c o s α, b c o s α) B (a c o s β, b s i n β)

C (a c o s γ, b s i n γ)

\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]

= \frac{1}{2} a b [c o s α (s i n β - s i n γ) + c o s β (s i n γ - s i n α) + c o s γ (s i n α - s i n β)]

= \frac{a b}{2} [s i n (β - α) + s i n (γ - β) + s i n (α - γ)]

= \frac{a b}{2} [- 2 s i n (\frac{α - β}{2}) c o s (\frac{α - β}{2}) + 2 s i n (\frac{α - β}{2}) c o s (\frac{2 γ - α - β}{2})]

= \frac{a b}{2} \times 2 s i n (\frac{α - β}{2}) [2 s i n (\frac{γ - β}{2}) s i n (\frac{γ - α}{2})]

= - 2 a b s i n (\frac{α - β}{2}) s i n (\frac{β - γ}{2}) s i n (\frac{γ - α}{2})

s o a r e a o f △ A B C i s

2 a b s i n (\frac{α - β}{2}) s i n (\frac{β - γ}{2}) s i n (\frac{γ - α}{2})

w a i t p l s m a x i m i s i n g t h e a r e a \dots

n o w l e t P c o r e s p o n d t o A i n a u x i l a r y c i r c l e

Q \to B a n d R \to C

a r e a o f △ P Q R = 2 a^{2} s i n (\frac{α - β}{2}) s i n (\frac{β - γ}{2}) s i n (\frac{γ - α}{2})

\frac{a r e a △ A B C}{a r e a △ P Q R} = \frac{2 a b}{2 a^{2}} = \frac{b}{a}

m a x a r e a △ A B C = \frac{b}{a} \times m a x a r e a △ P Q R

△ P Q R a c h i v e m a x a r e a w h e n i t i s e a u i l a t e r a l

t r i a n g l e . h e n c e △ A B C i s a l s o e q u i l a t e r a l

t r i a n g l e \dots

a r e a o f e q u i l a t e r a l t r i a n g l e i n s c r i b e d i n c i r c l e

o f r s d i u s a i s

= \frac{1}{2} (a c o s 30^{o} + a c o s 30^{o}) \times (a + a s i n 30^{o})

= \frac{a^{2}}{2} (2 \times \frac{\sqrt{3}}{2}) (1 + s i n 30^{o}) = \frac{3 \sqrt{3}}{4} a^{2}

s o r e q u i r e d a r e a o f t r i z n g l e △ A B C = \frac{b}{a} \times \frac{3 \sqrt{3}}{4} a^{2}

= \frac{3 \sqrt{3}}{4} a b

Commented by mr W last updated on 20/Dec/18

t h a n k s s i r!

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