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Question Number 49530 by behi83417@gmail.com last updated on 07/Dec/18

Commented by behi83417@gmail.com last updated on 07/Dec/18

$△ a n d △ a r e e q u i l a t e r a l t r i a n g l e s .$ $A B = 4, E G = 2 \sqrt{2} [o r : A B = 6, E G = 2 \sqrt{6}]$ $\Rightarrow A E = ?, ∡ A E I = ?$
Commented by mr W last updated on 07/Dec/18

$n o t p o s s i b l e s i r!$ $E G ⩾ A B / 2 = 6 / 2 = 3$ $\Rightarrow w i t h E G = \sqrt{6} < 3 n o s o l u t i o n!$
Commented by behi83417@gmail.com last updated on 07/Dec/18

$y e s s i r . y o u a r e r i g h t . v a l u e s c h a n g e d .$
	Answered by mr W last updated on 07/Dec/18

	$A E = x$ $4 = x + \frac{x}{2} + \sqrt{{(2 \sqrt{2})}^{2} - {(\frac{\sqrt{3} x}{2})}^{2}}$ $4 - \frac{3 x}{2} = \sqrt{8 - \frac{3 x^{2}}{4}}$ $8 - 3 x = \sqrt{32 - 3 x^{2}}$ $64 - 48 x + 9 x^{2} = 32 - 3 x^{2}$ $8 - 12 x + 3 x^{2} = 0$ $\Rightarrow x = \frac{6 \pm 2 \sqrt{3}}{3} = 0.845 (o r 3.155)$ $\frac{\sin θ}{x} = \frac{\sin 60 °}{2 \sqrt{2}}$ $\Rightarrow \sin θ = \frac{6 - 2 \sqrt{3}}{3} \times \frac{\sqrt{3}}{2 \times 2 \sqrt{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$ $\Rightarrow θ = \sin^{- 1} \frac{\sqrt{6} - \sqrt{2}}{4} = 15 °$ $\Rightarrow ∠ A E I = 180 ° - 60 ° - 15 ° = 105 °$
	Commented by behi83417@gmail.com last updated on 07/Dec/18

	$r i g h t a n s w e r a n d p e r f e c t w a y .$ $t h a n k y o u v e r y m u c h d e a r m a s t e r .$
	Answered by behi83417@gmail.com last updated on 07/Dec/18

	${E G}^{2} = {G B}^{2} + {E B}^{2} - 2 G B . E B . c o s 60$ ${(2 \sqrt{2})}^{2} = x^{2} + {(4 - x)}^{2} - 2 x (4 - x) . \frac{1}{2}$ $3 x^{2} - 12 x + 8 = 0 \Rightarrow x = A E = \frac{12 \pm \sqrt{144 - 96}}{6}$ $A E = \frac{12 \pm 4 \sqrt{3}}{6} = 2 \pm \frac{2}{\sqrt{3}} . [= 0.845 \lor 3.15]$ $c o s A \overset{}{E I} = \frac{{(2 \sqrt{2})}^{2} + x^{2} - {(4 - x)}^{2}}{4 \sqrt{2} x} =$ $= \frac{8 + x^{2} - x^{2} + 8 x - 16}{4 \sqrt{2} x} = \sqrt{2} (1 - \frac{1}{x}) =$ $= \sqrt{2} (1 - \frac{1}{0.845}) = - 0.259 \Rightarrow A \overset{}{E I} = 105^{∙} .$
	Answered by behi83417@gmail.com last updated on 07/Dec/18

	$3 S_{A E I} + S_{E G I} = S_{A B C}$ $\Rightarrow 3 \times \frac{1}{2} x (4 - x) . \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} (4^{2} - 8)$ $\Rightarrow 3 x^{2} - 12 x + 8 = 0 \Rightarrow x = \frac{12 \pm \sqrt{144 - 96}}{6}$ $\Rightarrow x = A E = \frac{12 \pm 4 \sqrt{3}}{6} = 2 \pm \frac{2}{\sqrt{3}} .$
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