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Question Number 39177 by ajfour last updated on 03/Jul/18

Commented by ajfour last updated on 03/Jul/18

$F i n d x a n d r a d i u s R o f s e m i x i r c l e$ $i n t e r m s o f l, α, a n d 1 .$
Commented by MJS last updated on 03/Jul/18

$new idea$ $start with the center of the square and the$ $triangle ‘ ‘ {sitting}^{″} on the x - axis$ $now {it}^{'} s possible to choose l and α and then$ $construct / calculate the rest$ $first the circle through the vericles and with$ $its center on the x - axis, then find x (which$ $I^{'} d like to call z instead)$
	Answered by MrW3 last updated on 04/Jul/18

	Commented by MrW3 last updated on 04/Jul/18

	$l e t a = s i d e l e n g t h o f s q u a r e = 1$ $C = m i d p o i n t o f A B$ $∠ A B F = \frac{π}{4} - α$ $A D = A F - D F = l \sin (\frac{π}{4} - α) - \frac{a}{\sqrt{2}}$ $= \frac{l (\cos α - \sin α) - a}{\sqrt{2}}$ $\frac{M A}{\sin \frac{π}{4}} = \frac{A D}{\sin α}$ $\Rightarrow M A = \frac{A D}{\sqrt{2} \sin α} = \frac{l (\cos α - \sin α) - a}{2 \sin α}$ $O C = M C \tan α = (M A + \frac{l}{2}) \tan α$ $= [\frac{l (\cos α - \sin α) - a}{2 \sin α} + \frac{l}{2}] \tan α$ $= \frac{l \cos α - a}{2 \cos α} = \frac{l}{2} - \frac{a}{2 \cos α}$ $R = O B = \sqrt{{O C}^{2} + {C B}^{2}}$ $= \sqrt{{(\frac{l}{2} - \frac{a}{2 \cos α})}^{2} + {(\frac{l}{2})}^{2}}$ $= \sqrt{\frac{l^{2}}{2} - \frac{l a}{2 \cos α} + \frac{a^{2}}{4 \cos^{2} α}}$ $\Rightarrow R = \frac{\sqrt{a^{2} + 2 l \cos α (l \cos α - a)}}{2 \cos α}$ $E B = F B - F E = l \cos (\frac{π}{4} - α) - \frac{a}{\sqrt{2}}$ $= \frac{l (\cos α + \sin α) - a}{\sqrt{2}}$ $\frac{G B}{E B} = \frac{F B}{A B} = \frac{l (\cos α + \sin α)}{\sqrt{2} l} = \frac{\cos α + \sin α}{\sqrt{2}}$ $\Rightarrow G B = \frac{\cos α + \sin α}{\sqrt{2}} \times \frac{l (\cos α + \sin α) - a}{\sqrt{2}}$ $= \frac{l (1 + \sin 2 α) - a (\cos α + \sin α)}{2}$ $E O = \frac{G C}{\cos α} = (G B - C B) \frac{1}{\cos α}$ $= [\frac{l (1 + \sin 2 α) - a (\cos α + \sin α)}{2} - \frac{l}{2}] \frac{1}{\cos α}$ $= \frac{l \sin 2 α - a (\cos α + \sin α)}{2 \cos α}$ $= l \sin α - \frac{a}{2} (1 + \tan α)$ $x = R - E O - D E$ $= R - l \sin α + \frac{a}{2} (1 + \tan α) - a$ $\Rightarrow x = R - l \sin α - \frac{a}{2} (1 - \tan α)$
	Commented by ajfour last updated on 04/Jul/18

	$O v e r w h e l m i n g, T h a n k y o u S i r .$
	Answered by MJS last updated on 04/Jul/18

	$center of square$ $C = (\begin{matrix} 0 \\ \frac{1}{2} \end{matrix})$ $A on y = \frac{1}{2} + x; B on y = \frac{1}{2} - x$ $A = (\begin{matrix} a \\ \frac{1}{2} + a \end{matrix}) B = (\begin{matrix} b \\ \frac{1}{2} - b \end{matrix})$ $line l : y = d - x \tan α$ $∣ A B ∣^{2} = l^{2} = 2 a^{2} + 2 b^{2} \Rightarrow a = - \frac{\sqrt{2 l^{2} - 4 b^{2}}}{2}$ $A on l : \frac{1}{2} + a = d - a \tan α \Rightarrow d = \frac{1}{2} + a (1 + \tan α)$ $d = \frac{1}{2} - \frac{\sqrt{2 l^{2} - 4 b^{2}}}{2} (1 + \tan α)$ $B on l : \frac{1}{2} - b = d - b \tan α \Rightarrow b = \frac{1 - 2 d}{2 (1 - \tan α)}$ $b = \frac{l (1 + \tan α)}{2 \sqrt{1 + \tan^{2} α}} = \frac{l}{2} (\sin α + \cos α) \Rightarrow$ $\Rightarrow a = - \frac{l (1 - \tan α)}{2 \sqrt{1 + \tan^{2} α}} = \frac{l}{2} (\sin α - \cos α)$ $A = (\begin{matrix} \frac{l}{2} (\sin α - \cos α) \\ \frac{1}{2} + \frac{l}{2} (\sin α - \cos α) \end{matrix})$ $B = (\begin{matrix} \frac{l}{2} (\sin α + \cos α) \\ \frac{1}{2} - \frac{l}{2} (\sin α + \cos α) \end{matrix})$ $center of circle : M = (\begin{matrix} m \\ 0 \end{matrix})$ ${(x - m)}^{2} + y^{2} = r^{2}$ $A on circle, B on circle \Rightarrow$ $\Rightarrow m = (\frac{l}{\sqrt{1 + \tan^{2} α}} - \frac{1}{2}) \tan α = l \sin α - \frac{1}{2} \tan α$ $r = \frac{1}{2} \sqrt{2 l^{2} + 1 + \tan^{2} α - 2 l \sqrt{1 + \tan^{2} α}} =$ $= \sqrt{\frac{l^{2}}{2} - \frac{l}{2 \cos α} + \frac{1}{{4 \cos}^{2} α}}$ $z = r - m - \frac{1}{2}$
	Commented by ajfour last updated on 04/Jul/18

	$T h a n k y o u S i r . p u r e c o o r d i n a t e$ $m e t h o d!$
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