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Question Number 35279 by ajfour last updated on 17/May/18

Commented by ajfour last updated on 17/May/18

$G i v e n t w o c i r c l e s o f r a d i i r a n d R .$ $T h e c i r c l e s t o u c h e a c h o t h e r$ $i n t e r n a l l y . T r i a n g l e A B C h a s$ $i t s v e r t e x A a t t h e p o i n t w h e r e$ $t h e c i r c l e s t o u c h . V e r t e x B l i e s$ $o n t h e c i r c u m f e r e n c e o f s m a l l e r$ $c i r c l e o f r a d i u s r w h i l e v e r t e x C$ $l i e s o n c i r c u m f e r e n c e o f c i r c l e$ $o f r a d i u s R .$ $F i n d m a x i m u m a r e a o f △ A B C$ $i n t e r m s o f r, a n d R .$
Commented by ajfour last updated on 17/May/18

Commented by ajfour last updated on 17/May/18

$A (θ, ϕ) = \frac{1}{2} b c \sin (θ + ϕ)$ $b = 2 R \sin ϕ, c = 2 r \sin θ$ $A (θ, ϕ) = 2 r R \sin θ \sin ϕ \sin (θ + ϕ)$ $\frac{\partial A}{\partial θ} = 2 r R \sin ϕ [\cos θ \sin (θ + ϕ)$ $+ \sin θ \cos (θ + ϕ)] = 0$ $\Rightarrow \sin ϕ = 0, π o r$ $\sin (2 θ + ϕ) = 0$ $s i m i l a r l y \frac{\partial A}{\partial ϕ} = 0 \Rightarrow \sin θ = 0, π$ $o r \sin (2 ϕ + θ) = 0$ $f o r o u r c a s e$ $\Rightarrow 2 θ + ϕ = π a n d 2 ϕ + θ = π$ $\Rightarrow θ + ϕ = \frac{2 π}{3}$ $S o θ = ϕ = \frac{π}{3}$ $A = (2 r R) (\frac{3 \sqrt{3}}{8}) = \frac{3 \sqrt{3} r R}{4} .$
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