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Question Number 111988 by ajfour last updated on 05/Sep/20

Commented by ajfour last updated on 05/Sep/20

$I f △ A B C i s e q u i l a t e r a l w i t h s i d e s,$ $f i n d x - c o o r d i n a t e o f p o i n t A i n$ $t e r m s o f s .$
	Answered by mr W last updated on 06/Sep/20

	$c e n t e r o f e q u i l a t e r a l D (h, k)$ $A (a, 0)$ $R = \frac{2}{3} \times \frac{\sqrt{3} s}{2} = \frac{s}{\sqrt{3}}$ ${(a - h)}^{2} + k^{2} = R^{2} = \frac{s^{2}}{3}$ $\Rightarrow a = h + \sqrt{\frac{s^{2}}{3} - k^{2}}$ $x_{B} = h + (a - h) \cos 120 ° - (0 - k) \sin 120 °$ $= h - \frac{1}{2} (a - h) + \frac{\sqrt{3}}{2} k$ $= \frac{1}{2} (3 h + \sqrt{3} k - a)$ $y_{B} = k + (a - h) \sin 120 ° + (0 - k) \cos 120 °$ $= k + \frac{\sqrt{3}}{2} (a - h) + \frac{1}{2} k$ $= \frac{1}{2} (- \sqrt{3} h + 3 k + \sqrt{3} a)$ $y_{B} = x_{B}^{2}$ $\frac{1}{2} (- \sqrt{3} h + 3 k + \sqrt{3} a) = \frac{1}{4} {(3 h + \sqrt{3} k - a)}^{2}$ $\Rightarrow 2 (- \sqrt{3} h + 3 k + \sqrt{3} a) = {(3 h + \sqrt{3} k - a)}^{2}$ $\Rightarrow 2 (3 k + \sqrt{s^{2} - 3 k^{2}}) = {(2 h + \sqrt{3} k - \sqrt{\frac{s^{2}}{3} - k^{2}})}^{2} . . . (i)$ $x_{C} = h + (a - h) \cos 240 ° - (0 - k) \sin 240 °$ $= h - \frac{1}{2} (a - h) + \frac{\sqrt{3}}{2} (0 - k)$ $= \frac{1}{2} (3 h - \sqrt{3} k - a)$ $y_{C} = k + (a - h) \sin 240 ° + (0 - k) \cos 240 °$ $= k - \frac{\sqrt{3}}{2} (a - h) - \frac{1}{2} (0 - k)$ $= \frac{1}{2} (\sqrt{3} h + 3 k - \sqrt{3} a)$ $y_{C} = x_{C}^{2}$ $\frac{1}{2} (\sqrt{3} h + 3 k - \sqrt{3} a) = \frac{1}{4} {(3 h - \sqrt{3} k - a)}^{2}$ $\Rightarrow 2 (\sqrt{3} h + 3 k - \sqrt{3} a) = {(3 h - \sqrt{3} k - a)}^{2}$ $\Rightarrow 2 (3 k - \sqrt{s^{2} - 3 k^{2}}) = {(2 h - \sqrt{3} k - \sqrt{\frac{s^{2}}{3} - k^{2}})}^{2} . . (i i)$ $w e g e t h a n d k f r o m (i) a n d (i i) .$ $f r o m (i) :$ $2 h = \sqrt{2 (3 k + \sqrt{s^{2} - 3 k^{2}})} - \sqrt{3} k + \sqrt{\frac{s^{2}}{3} - k^{2}}$ $f r o m (i i) :$ $2 h = \sqrt{2 (3 k - \sqrt{s^{2} - 3 k^{2}})} + \sqrt{3} k + \sqrt{\frac{s^{2}}{3} - k^{2}}$ $\sqrt{2 (3 k + \sqrt{s^{2} - 3 k^{2}})} - \sqrt{3} k = \sqrt{2 (3 k - \sqrt{s^{2} - 3 k^{2}})} + \sqrt{3} k$ $\Rightarrow \sqrt{2 (3 k + \sqrt{s^{2} - 3 k^{2}})} - \sqrt{2 (3 k - \sqrt{s^{2} - 3 k^{2}})} = 2 \sqrt{3} k$ $s o l u t i o n p o s s i b l e f o r s ⩽\approx 4.75$
	Commented by mr W last updated on 06/Sep/20

	Commented by mr W last updated on 06/Sep/20

	Commented by mr W last updated on 06/Sep/20

	Answered by ajfour last updated on 06/Sep/20
	$Let s=2p(√3) A(a, 0) (a/p)=3cos θ{1+tan^2 θ−tan θ(√(tan^2 θ−(1/3))) } and θ is obtained from tan θ+(√(tan^2 θ−(1/3)))=2psin θtan θ or (1/(sin^2 θ))+3(2psin θ−1)^2 =4 ★$
	$L e t s = 2 p \sqrt{3}$ $A (a, 0)$ $\frac{a}{p} = 3 \cos θ {1 + \tan^{2} θ - \tan θ \sqrt{\tan^{2} θ - \frac{1}{3}}}$ $a n d θ i s o b t a i n e d f r o m$ $\tan θ + \sqrt{\tan^{2} θ - \frac{1}{3}} = 2 p \sin θ \tan θ$ $o r$ $\frac{1}{\sin^{2} θ} + 3 {(2 p \sin θ - 1)}^{2} = 4 ★$
	Commented by ajfour last updated on 06/Sep/20

	Commented by ajfour last updated on 06/Sep/20

	Commented by ajfour last updated on 06/Sep/20

	$p = 1, s = 2 \sqrt{3}, a = 2 \sqrt{3}, θ = 30 °$
	Commented by mr W last updated on 06/Sep/20

	$n i c e s o l u t i o n s i r!$
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