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Question Number 56289 by mr W last updated on 13/Mar/19

Commented by mr W last updated on 13/Mar/19

$S e e Q 56264$ $F i n d t h e m a x . a n d m i n . c i r c u m s c r i b i n g$ $e q u i l a t e r a l t r i a n g l e o f a g i v e n t r i a n g l e$ $w i t h s i d e l e n g t h e s a, b a n d c .$
Commented by ajfour last updated on 13/Mar/19

Commented by MJS last updated on 13/Mar/19

$solution method :$ $(1) coordinates for A, B, C$ $(2) lines p, q, r through A, B, C with slopes$ $θ - 60 °, θ, θ + 60 °$ $(3) intersections P, Q, R (always forming an$ $equilateral triangle)$ $(4) calculate s (θ)$ $(5) \frac{d}{d θ} s (θ) = 0 \Rightarrow θ = \arctan t which leads to$ $a 2^{nd} degree polynome we can always solve$ $sorry I^{'} ve got no time to type at the moment$
Commented by mr W last updated on 14/Mar/19

$p e r f e c t l y d o n e!$
Commented by ajfour last updated on 14/Mar/19

$t h a n k y o u S i r .$
Commented by mr W last updated on 15/Mar/19

$s i r, c a n y o u p l e a s e c h e c k f o l l o w i n g i s s u e :$ $t h e s_{m a x} s h o u l d b e s y m m e t r i c a b o u t$ $a, b, c . i . e . i f w e e x c h a n g e a a n d b, s_{m a x}$ $s h o u l d b e t h e s a m e . b u t t h e f o r m u l a$ $a b o v e i s n o t s y m m e t r i c f o r a, b, c .$ $e . g .$ $a = 5, b = 4, c = 3 {d o n}^{'} t g i v e t h e s a m e s_{m a x}$ $a s a = 4, b = 5, c = 3 .$
Commented by ajfour last updated on 15/Mar/19

$v e r y s t r a n g e, n o w a y o u t y e t! S i r .$
	Answered by mr W last updated on 13/Mar/19

	$s = e d g e l e n g t h o f e q u i l a t e r a l t r i a n g l e$ $\sin β = \frac{x}{a} \sin \frac{π}{3} = \frac{\sqrt{3} x}{2 a}$ $α = π - (β + \frac{π}{3})$ $P B = \frac{\sin (β + \frac{π}{3})}{\sin \frac{π}{3}} a = (\frac{\sin β}{\tan \frac{π}{3}} + \cos β) b = \frac{x}{2} + a \sqrt{1 - \frac{3 x^{2}}{4 a^{2}}}$ $\sin φ = \frac{s - x}{b} \sin \frac{π}{3} = \frac{\sqrt{3} (s - x)}{2 b}$ $γ = π - (φ + \frac{π}{3})$ $Q A = \frac{\sin (φ + \frac{π}{3})}{\sin \frac{π}{3}} b = (\frac{\sin φ}{\tan \frac{π}{3}} + \cos φ) b = \frac{s - x}{2} + b \sqrt{1 - \frac{3 {(s - x)}^{2}}{4 b^{2}}}$ $R B = s - \frac{x}{2} - a \sqrt{1 - \frac{3 x^{2}}{4 a^{2}}}$ $R A = s - \frac{s - x}{2} - b \sqrt{1 - \frac{3 {(s - x)}^{2}}{4 b^{2}}} = \frac{s + x}{2} - b \sqrt{1 - \frac{3 {(s - x)}^{2}}{4 b^{2}}}$ $c^{2} = {(\frac{2 s - x}{2} - \sqrt{a^{2} - \frac{3 x^{2}}{4}})}^{2} + {(\frac{s + x}{2} - \sqrt{b^{2} - \frac{3 {(s - x)}^{2}}{4}})}^{2} - (\frac{2 s - x}{2} - \sqrt{a^{2} - \frac{3 x^{2}}{4}}) (\frac{s + x}{2} - \sqrt{b^{2} - \frac{3 {(s - x)}^{2}}{4}})$ $4 c^{2} = {(2 s - x - \sqrt{4 a^{2} - 3 x^{2}})}^{2} + {(s + x - \sqrt{4 b^{2} - 3 {(s - x)}^{2}})}^{2} - (2 s - x - \sqrt{4 a^{2} - 3 x^{2}}) (s + x - \sqrt{4 b^{2} - 3 {(s - x)}^{2}})$ $. . . . . .$
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