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Question Number 54160 by ajfour last updated on 30/Jan/19

Commented by ajfour last updated on 30/Jan/19

$F i n d m a x i m u m a r e a o f △ A B C i n$ $t e r m s o f R a n d r .$
	Answered by mr W last updated on 30/Jan/19

	Commented by mr W last updated on 30/Jan/19

	$α = \frac{π}{2} - \frac{β}{2}$ $\Rightarrow 2 α = π - β . . (i)$ $β = \frac{π}{2} - \frac{α}{2}$ $\Rightarrow 2 β = π - α . . (i i)$ $\Rightarrow α = β = \frac{π}{3} = 60 °$ $A C = 2 R \cos \frac{α}{2} = \sqrt{3} R$ $A B = 2 r \cos \frac{β}{2} = \sqrt{3} r$ $∠ C A B = 180 ° - \frac{α}{2} - \frac{β}{2} = 120 °$ $Δ_{m a x} = \frac{1}{2} \times \sqrt{3} R \times \sqrt{3} r \times \sin 120 °$ $\Rightarrow Δ_{m a x} = \frac{3 \sqrt{3} R r}{4}$
	Commented by ajfour last updated on 30/Jan/19

	$y e s S i r, s t r a i g h t a n d c l e a r, t h a n k s$ $a g a i n (i a m q u i t e d u m b a t t h i s) .$
	Commented by mr W last updated on 30/Jan/19
	$i tried above to find the maximum triangle only using logic. but we can get the same result using calculus: Δ=(1/2)×2R cos (α/2)×2r cos (β/2)×sin ((α/2)+(β/2)) =2Rr cos (α/2) cos (β/2) sin ((α/2)+(β/2)) ((∂(Δ))/∂α)=2Rrcos (β/2){−(1/2)sin (α/2) sin ((α/2)+(β/2))+(1/2)cos (α/2) cos ((α/2)+(β/2))}=0 sin (α/2) sin ((α/2)+(β/2))=cos (α/2) cos ((α/2)+(β/2)) ⇒tan (α/2) tan ((α/2)+(β/2))=1 ...(i) ((∂(Δ))/∂β)=2Rrcos (α/2){−(1/2)sin (β/2) sin ((α/2)+(β/2))+(1/2)cos (β/2) cos ((α/2)+(β/2))}=0 ⇒tan (β/2) tan ((α/2)+(β/2))=1 ...(ii) ⇒tan (α/2)=tan (β/2) ⇒α=β ⇒tan (α/2) tan α=1 ⇒tan (α/2) ((2 tan (α/2))/(1−tan^2 (α/2)))=1 ⇒3 tan^2 (α/2)=1 ⇒tan (α/2)=(1/(√3)) ⇒(α/2)=30° ⇒α=β=60°$
	$i t r i e d a b o v e t o f i n d t h e m a x i m u m t r i a n g l e$ $o n l y u s i n g l o g i c . b u t w e c a n g e t t h e$ $s a m e r e s u l t u s i n g c a l c u l u s :$ $Δ = \frac{1}{2} \times 2 R \cos \frac{α}{2} \times 2 r \cos \frac{β}{2} \times \sin (\frac{α}{2} + \frac{β}{2})$ $= 2 R r \cos \frac{α}{2} \cos \frac{β}{2} \sin (\frac{α}{2} + \frac{β}{2})$ $\frac{\partial (Δ)}{\partial α} = 2 R r \cos \frac{β}{2} {- \frac{1}{2} \sin \frac{α}{2} \sin (\frac{α}{2} + \frac{β}{2}) + \frac{1}{2} \cos \frac{α}{2} \cos (\frac{α}{2} + \frac{β}{2})} = 0$ $\sin \frac{α}{2} \sin (\frac{α}{2} + \frac{β}{2}) = \cos \frac{α}{2} \cos (\frac{α}{2} + \frac{β}{2})$ $\Rightarrow \tan \frac{α}{2} \tan (\frac{α}{2} + \frac{β}{2}) = 1 . . . (i)$ $\frac{\partial (Δ)}{\partial β} = 2 R r \cos \frac{α}{2} {- \frac{1}{2} \sin \frac{β}{2} \sin (\frac{α}{2} + \frac{β}{2}) + \frac{1}{2} \cos \frac{β}{2} \cos (\frac{α}{2} + \frac{β}{2})} = 0$ $\Rightarrow \tan \frac{β}{2} \tan (\frac{α}{2} + \frac{β}{2}) = 1 . . . (i i)$ $\Rightarrow \tan \frac{α}{2} = \tan \frac{β}{2}$ $\Rightarrow α = β$ $\Rightarrow \tan \frac{α}{2} \tan α = 1$ $\Rightarrow \tan \frac{α}{2} \frac{2 \tan \frac{α}{2}}{1 - \tan^{2} \frac{α}{2}} = 1$ $\Rightarrow 3 \tan^{2} \frac{α}{2} = 1$ $\Rightarrow \tan \frac{α}{2} = \frac{1}{\sqrt{3}}$ $\Rightarrow \frac{α}{2} = 30 °$ $\Rightarrow α = β = 60 °$
	Commented by ajfour last updated on 30/Jan/19

	$u s i n g c a l c u l u s i^{'} d g o t t h e s a m e,$ $b u t w o u l d y o u p l e a s e e x p l a i n y o u r$ $l o g i c t o m e i n s o m e m o r e d e t a i l, S i r ?$
	Commented by mr W last updated on 30/Jan/19

	$t h e l o g i c i s :$ $w i t h m a x i m u m t r i a n g l e$ $t a n g e n t a t C i s p a r a l l e l t o A B, i . e .$ $C P ⊥ A B .$ $t a n g e n t a t B i s p a r a l l e l t o A C, i . e .$ $B Q ⊥ A C .$
	Commented by mr W last updated on 30/Jan/19

	$t h e l o g i c i s t h e s a m e a s i n Q 53910 .$
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