Question-38775

Question Number 38775 by ajfour last updated on 29/Jun/18

Commented by ajfour last updated on 29/Jun/18

O n a s e g m e n t o f l e n g t h c, i s k e p t

a u n i t s q u a r e, o v e r h a n g i n g a l i t t l e;

s u c h t h a t (s e e d i a g r a m) .

F i n d t h e l e n g t h o f s e g m e n t c

u n d e r t h e b o t t o m e d g e o f s q u a r e .

Commented by MrW3 last updated on 29/Jun/18

I t h i n k y o u h a v e t o g i v e a n a d d i t i o n a l

c o n d i t i o n, e . g . t h e r a d i u s o f t h e c i r c l e,

o t h e r w i s e t h e s e g m e n t c i s n o t u n i q u e .

Answered by MrW3 last updated on 30/Jun/18

l e t R = r a d i u s o f t h e c i r c l e

l e t a = s i d e o f t h e s q u a r e (a = 1)

c = 2 R - a + x

\frac{a}{h} = \frac{h}{2 R - a} \Rightarrow h = \sqrt{a (2 R - a)}

\frac{x}{a} = \frac{a}{h} \Rightarrow x = \frac{a^{2}}{h} = \frac{a^{2}}{\sqrt{a (2 R - a)}}

\Rightarrow c = 2 R - a + \frac{a^{2}}{\sqrt{a (2 R - a)}}

Commented by MrW3 last updated on 30/Jun/18

Commented by ajfour last updated on 30/Jun/18

l e t \tan θ = \frac{x}{a}

\tan θ = \frac{x}{a} = \frac{a}{h} = \frac{h}{c - x}

\Rightarrow \frac{x^{2}}{a^{2}} = \frac{a}{c - x} a n d a s a = 1

x^{3} - {c x}^{2} + 1 = 0

L e t y = \frac{1}{x}

\Rightarrow y^{3} - c y + 1 = 0

l e t y = u + v

\Rightarrow u^{3} + v^{3} + 3 u v (u + v) - c (u + v) + 1 = 0

o r u^{3} + v^{3} + (u + v) (3 u v - c) + 1 = 0

l e t 3 u v = c \Rightarrow u^{3} v^{3} = \frac{c^{3}}{27}

\Rightarrow u^{3} + v^{3} = - 1

u^{3}, v^{3} a r e r o o t s o f

z^{2} + z + \frac{c^{3}}{27} = 0

u^{3}, v^{3} = \frac{- 1 \pm \sqrt{1 - \frac{4 c^{3}}{27}}}{2}

= \frac{- 1}{2} \pm \sqrt{\frac{1}{4} - \frac{c^{3}}{27}}

y = \sqrt[3]{- \frac{1}{2} - \sqrt{\frac{1}{4} - \frac{c^{3}}{27}}}

+ \sqrt[3]{- \frac{1}{2} + \sqrt{\frac{1}{4} - \frac{c^{3}}{27}}}

x = \frac{1}{y} .

Commented by MJS last updated on 30/Jun/18

a = 1 \Rightarrow c = \frac{1 + \sqrt{{(2 R - 1)}^{3}}}{\sqrt{2 R - 1}}; R > \frac{1}{2}

\frac{d c}{d R} = 2 - \frac{1}{\sqrt{{(2 R - 1)}^{3}}}

\Rightarrow c is minimal for R = \frac{1}{2} + \frac{\sqrt[3]{2}}{4}; c = \frac{3 \sqrt[3]{2}}{2}

Commented by MJS last updated on 01/Jul/18

x^{3} - {c x}^{2} + 1 = 0

x = z + \frac{c}{3}

{(z + \frac{c}{3})}^{3} - c {(z + \frac{c}{3})}^{2} + 1 = 0

z^{3} - \frac{c^{2}}{3} z + \frac{27 - 2 c^{3}}{27} = 0

u = \sqrt[3]{\frac{1}{27} c^{3} - \frac{1}{2} + \frac{\sqrt{3}}{18} \sqrt{27 - 4 c^{3}}}; v = \sqrt[3]{\frac{1}{27} c^{3} - \frac{1}{2} - \frac{\sqrt{3}}{18} \sqrt{27 - 4 c^{3}}}

z_{1} = u + v; x_{1} = z_{1} + \frac{c}{3}

z_{2} = (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) v; x_{2} = z_{2} + \frac{c}{3}

z_{3} = (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) u + (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) v; x_{3} = z_{3} + \frac{c}{3}

I think this is easier than to find x = \frac{1}{y} with

y being a sum of two cubic roots . anyway

both methods lead to the same values, if you

use a calculator to get approximate values

the methods are equal \dots

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