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Question Number 79943 by TawaTawa last updated on 29/Jan/20

Commented by TawaTawa last updated on 29/Jan/20

$Equilateral triangle not equilibrium triangle$
Commented by TawaTawa last updated on 29/Jan/20

$I cannot show it . the b part$
Commented by TawaTawa last updated on 29/Jan/20

$Alright sir, help me please . God bless you sir .$
Commented by mr W last updated on 29/Jan/20

${w h a t}^{'} s y o u r d i f f i c u l t y ?$
Commented by mr W last updated on 30/Jan/20

$q u e s t i o n h a s t e r r i b l y s o m e e r r o r s .$ $i a n s w e r e d t h e q u e s t i o n i n t h e w a y$ $h o w i t h i n k i t s h o u l d b e . e . g . E B$ $c a n n e v e r b e h o r i z o n t a l . b u t E A c a n$ $b e h o r i z o n t a l .$
	Answered by mr W last updated on 30/Jan/20

	Commented by mr W last updated on 30/Jan/20

	$P A R T A$ $r e d s h a d e d : c u t o f f p a r t$ $w i t h w e i g h t G_{2} a n d c e n t r o i d a t M_{2}$ $b l u e s h a d e d : r e m a i n i n g p a r t$ $w i t h w e i g h t G_{1} a n d c e n t r o i d a t M_{1}$ $b l a c k t r i a n g l e : w h o l e l a m i n a$ $w i t h w e i g h t G a n d c e n t r o i d a t M$ $A B = B C = C A = 2 a$ $O C = \sqrt{3} a$ $C M = \frac{2}{3} O C = \frac{2 \sqrt{3} a}{3}$ $G = a r e a Δ A B C = \frac{2 a}{2} \times \sqrt{3} a = \sqrt{3} a^{2}$ $E D = D C = C E = 2 x$ $F C = \sqrt{3} x$ ${C M}_{2} = \frac{2}{3} F C = \frac{2 \sqrt{3} x}{3}$ $G_{2} = a r e a Δ E D C = \frac{2 x}{2} \times \sqrt{3} x = \sqrt{3} x^{2}$ $G_{1} = G - G_{2} = \sqrt{3} (a^{2} - x^{2})$ ${M M}_{2} = C M - {C M}_{2} = \frac{2 \sqrt{3}}{3} (a - x)$ $G_{1} \times M_{1} M = G_{2} \times {M M}_{2}$ $\Rightarrow M_{1} M = \frac{G_{2}}{G_{1}} \times {M M}_{2} = \frac{\sqrt{3} x^{2}}{\sqrt{3} (a^{2} - x^{2})} \times \frac{2 \sqrt{3}}{3} (a - x)$ $\Rightarrow M_{1} M = \frac{2 \sqrt{3} x^{2}}{3 (a + x)}$ $M_{1} F = M_{1} M + M C - F C = \frac{2 \sqrt{3} x^{2}}{3 (a + x)} + \frac{2 \sqrt{3} a}{3} - \sqrt{3} x$ $M_{1} F = \frac{2 x^{2} + 2 (a^{2} + a x) - 3 (a x + x^{2})}{\sqrt{3} (a + x)}$ $M_{1} F = \frac{2 a^{2} - a x - x^{2}}{\sqrt{3} (a + x)}$ $\Rightarrow M_{1} F = \frac{(a - x) (2 a + x)}{\sqrt{3} (a + x)}$
	Commented by mr W last updated on 30/Jan/20

	Commented by mr W last updated on 30/Jan/20

	$P A R T B$ $θ = 120 ° - α$ $\tan α = \frac{M_{1} F}{E F} = \frac{(a - x) (2 a + x)}{\sqrt{3} (a + x) x}$ ${l e t}^{'} s m a k e α = 30 °, t h e n$ $\tan α = \frac{1}{\sqrt{3}}$ $\frac{(a - x) (2 a + x)}{\sqrt{3} (a + x) x} = \frac{1}{\sqrt{3}}$ $(a - x) (2 a + x) = (a + x) x$ $x^{2} + a x - a^{2} = 0$ $\Rightarrow x = \frac{(\sqrt{5} - 1) a}{2} \approx 0.618 a (g o l d e n r a t i o!)$ $t h a t m e a n s i f x = \frac{(\sqrt{5} - 1) a}{2}, w e w i l l$ $g e t α = 30 ° a n d θ = 120 - 30 = 90 ° .$ $i n t h i s c a s e, w h e n t h e r e m a i n i n g$ $l a m i n a i s f r e e l y s u s p e n d e d a t p o i n t$ $E, {E M}_{1} w i l l b e v e r t i c a l a n d E A w i l l$ $b e h o r i z o n t a l .$
	Commented by mr W last updated on 30/Jan/20

	Commented by TawaTawa last updated on 30/Jan/20

	$Wow, God bless you sir .$ $I will study it and understand .$ $Thanks for your time always .$
	Commented by mr W last updated on 30/Jan/20

	$E x t r a t a s k :$ $P r o v e w i t h o u t c a l c u l a t i o n t h a t i t$ $i s i m p o s s i b l e t h a t E B i s v e r t i c a l$ $w h e n A B D E i s f r e e l y s u s p e n d e d$ $f r o m E, r e g a r d l e s s h o w l a r g e t h e c u t$ $o f f p a r t C D E i s .$
	Commented by mr W last updated on 30/Jan/20

	Commented by mr W last updated on 30/Jan/20

	$E X T R A T A S K$ $O F = s y m m e t r y a x i s$ $S = m i d p o i n t o f O F$ $S i n c e t h e p a r t l e f t f r o m S i s l a r g e r$ $t h a n t h e p a r t r i g h t f r o m S, t h e c e n t r o i d$ $M_{1} m u s t l i e o n t h e l e f t s i d e o f S .$ $O n t h e o t h e r s i d e, l e t N b e t h e$ $i n t e r s e c t i o n p o i n t f r o m E B a n d O F,$ $s i n c e E D i s s m a l l e r t h a n A B, N m u s t$ $l i e o n t h e r i g h t s i d e o f S .$ $t h a t i s t o s a y, M_{1} c a n n e v e r c o i n c i d e$ $w i t h N .$ $W h e n A B D E i s s u s p e n d e t f r o m p o i n t$ $E, {E M}_{1} w i l l b e v e r t i c a l . S i n c e M_{1}$ $c a n n e v e r l i e o n t h e l i n e E B, i t m e a n s$ $t h a t E B c a n n e v e r b e v e r t i c a l, w h e n$ $A B D E i s s u s p e n d e d o n p o i n t E .$
	Commented by mr W last updated on 30/Jan/20

	Commented by TawaTawa last updated on 30/Jan/20

	$I really appreciate your effort sir .$ $Am preparing for the exam . God bless you more sir .$ $you helped me alot .$
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