Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 211145 by mr W last updated on 29/Aug/24

Commented by mr W last updated on 30/Aug/24

$t h a n k s! b u t i t h i n k p e o p l e k n o w$ $w h a t i e x a c t l y m e a n .$
Commented by mr W last updated on 29/Aug/24

$a c o r n e r o f a s o l i d c u b e w i t h e d g e$ $l e n g t h d i s c h o p p e d o f f a s s h o w n .$ $f i n d t h e r a d i u s R o f t h e l a r g e s t$ $s o l i d s p h e r e w h i c h c a n b e c u t f r o m$ $t h e r e m a i n i n g o b j e c t .$ $a s s u m e 0 < a, b, c ⩽ d .$
Commented by MathematicalUser2357 last updated on 30/Aug/24
$0<a,b,c≤d is not an equation or logic you should say: “∧_(l∈{a,b,c}) (0<l∧l≤d)”$
$0 < a, b, c ⩽ d is not an equation or logic$ $you should say : ‘ ‘ \underset{l \in {a, b, c}}{\land} {(0 < l \land l ⩽ d)}^{″}$
	Answered by mr W last updated on 04/Sep/24

	$s a y a, b, c a r e t h e x, y, x a x e s .$ $t h e e q u a t i o n o f t h e c r o s s s e c t i o n i s$ $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ $i f n o c o r n e r i s c h o p p e d o f f, t h e n$ $t h e l a r g e s t s p h e r e h a s$ $c e n t e r a t (\frac{d}{2}, \frac{d}{2}, \frac{d}{2}) a n d$ $r a d i u s R = R_{0} = \frac{d}{2}$ $i f$ $\frac{\frac{d}{2 a} + \frac{d}{2 b} + \frac{d}{2 c} - 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} ⩾ R_{0} = \frac{d}{2}$ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} ⩾ \frac{2}{d}$ $R = R_{0} = \frac{d}{2}$ $o t h e r w i s e$ $c a s e 1 : c e n t e r (R_{1}, R_{1}, d - R_{1}), r a d i u s R_{1}$ $\frac{\frac{R_{1}}{a} + \frac{R_{1}}{b} + \frac{d - R_{1}}{c} - 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} = R_{1}$ $\Rightarrow \frac{R_{1}}{d} = \frac{\frac{1}{c} - \frac{1}{d}}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} - \frac{1}{a} - \frac{1}{b} + \frac{1}{c}}$ $c a s e 2 : c e n t e r (d - R_{2}, d - R_{2}, R_{2}), r a d i u s R_{2}$ $\frac{\frac{d - R_{2}}{a} + \frac{d - R_{2}}{b} + \frac{R_{2}}{c} - 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} = R_{2}$ $\Rightarrow \frac{R_{2}}{d} = \frac{\frac{1}{a} + \frac{1}{b} - \frac{1}{d}}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} + \frac{1}{a} + \frac{1}{b} - \frac{1}{c}}$ $c a s e 3 : c e n t e r (d - R_{3}, d - R_{3}, d - R_{3}), r a d i u s R_{3}$ $\frac{\frac{d - R_{3}}{a} + \frac{d - R_{3}}{b} + \frac{d - R_{3}}{c} - 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} = R_{3}$ $\Rightarrow \frac{R_{3}}{d} = \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{1}{d}}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$ $w e c a n s h o w t h a t R_{1} < R_{3}, R_{2} < R_{3}$ $t h a t m e a n s t h e r a d i u s o f m a x i m u m$ $s p h e r e f r o m c a s e 1 t o c a s e 3 i s R_{3} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum