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Question Number 158924 by ajfour last updated on 10/Nov/21

Commented by ajfour last updated on 10/Nov/21

$a, b, c a r e s l a n t e d g e s o f o u t e r$ $t e t r a h e d r o n . F i n d m i n i m u m$ $v o l u m e o f i n n e r t e t r a h e d r o n$ $w h o s e v e r t i c e s l i e o n f a c e s o f$ $o u t e r o n e w h e n t h e o u t e r o n e$ $h a s a m a x i m u m v o l u m e .$
Commented by ajfour last updated on 10/Nov/21

$t h a n k s, S i r; i d i d b l u n d e r!$
Commented by mr W last updated on 10/Nov/21

$c e n t e r o f s p h e r e (r, r, r)$ $p l a n e A B C : \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ $r = \pm \frac{\frac{r}{a} + \frac{r}{b} + \frac{r}{c} - 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} (w e t a k e -)$ $r = \frac{1}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}}$
	Answered by mr W last updated on 10/Nov/21

	$P a r t I$ $s a y θ = a n g l e b e t w e e n O A a n d O B$ $s a y φ = a n g l e b e t w . O C a n d p l a n e O A B$ $t h e n t h e v o l u m e o f t h e o u t e r$ $t e t r a h e d r o n i s$ $V_{O} = \frac{1}{3} \times \frac{a b \sin θ}{2} \times c \sin φ = \frac{a b c \sin θ \sin φ}{6}$ ${i t}^{'} s c l e a r V_{O} i s m a x i m u m w h e n$ $θ = φ = 90 °, i . e . a, b, c a r e p e r p e n d i c u l a r$ $t o e a c h o t h e r .$
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