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Question Number 19964 by ajfour last updated on 19/Aug/17

Commented by ajfour last updated on 19/Aug/17

$I n s c r i b e d i n a r i g h t c i r c u l a r c o n e$ $i s a s p h e r e w h o s e s u r f a c e a r e a i s$ $e q u a l t o t h e a r e a o f t h e b a s e i f t h e$ $c o n e . I n w h a t r a t i o i s t h e l a t e r a l$ $s i d e o f t h e c o n e d i v i d e d b y t h e l i n e$ $o f t a n g e n c y o f t h e s p h e r e a n d c o n e .$
	Answered by ajfour last updated on 20/Aug/17

	$4 π r^{2} = π R^{2} \Rightarrow R = 2 r$ $B T = O B = R = 2 r$ $h = {(\sqrt{{A T}^{2} + r^{2}} + r)}^{2} = {(A T + B T)}^{2} - {O B}^{2}$ $\Rightarrow {A T}^{2} + r^{2} + r^{2} + 2 r \sqrt{{A T}^{2} + r^{2}}$ $= {A T}^{2} + 4 r^{2} + 4 r (A T) - 4 r^{2}$ $\Rightarrow \sqrt{{A T}^{2} + r^{2}} = - r + 2 (A T)$ $\Rightarrow {A T}^{2} + r^{2} = r^{2} + 4 {(A T)}^{2} - 4 r (A T)$ $\Rightarrow 3 A T = 4 r o r A T = \frac{4 r}{3}$ $H e n c e \frac{A T}{B T} = \frac{4 r / 3}{2 r} = \frac{2}{3} .$
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