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Question Number 89281 by ajfour last updated on 16/Apr/20

Commented by ajfour last updated on 17/Apr/20

$B o t h p a r a b o l a s h a v e s a m e s h a p e .$ $I f e q u a t i o n o f o n e i s y = x^{2},$ $a n d t h e i n s c r i b e d t r i a n g l e i s$ $e q i u i l a t e r a l, f i n d t h e e q . o f t h e$ $o t h e r p a r a b o l a .$ $F i n d a l s o t h e t r i a n g l e s i d e .$
Commented by ajfour last updated on 17/Apr/20

$M j S S i r, m r W S i r, p l e a s e s o l v e .$
	Answered by mr W last updated on 17/Apr/20

	$s a y t h e s i d e l e n g t h o f t h e e q u i l a t e r a l$ $t r i a n g l e i s s .$ $f o r a g i v e n s, t h e c o o r d i n a t e s o f p o i n t s$ $A, B, C c a n b e d e t e r m i n e d a c c o r d i n g$ $t o t h e m e t h o d f r o m a j f o u r s i r i n$ $Q 62938 .$
	Commented by mr W last updated on 24/Apr/20

	$c o o r d i n a t e s o f t h e c e n t r o i d G (h, k) :$ $h = \sqrt{\frac{s^{2}}{48} - \frac{1}{4}}$ $k = \frac{3 s^{2}}{16} - \frac{1}{4}$ $\frac{s^{2}}{48} - \frac{1}{4} ⩾ 0 \Rightarrow s ⩾ 2 \sqrt{3}$ $t h e x - c o o r d i n a t e s o f p o i n t s A, B, C$ $c a n b e c a l c u l a t e d a s r o o t s o f f o l l o w i n g$ $e q u a t i o n :$ $x^{3} + {a x}^{2} + b x + c = 0$ $w i t h$ $a = - 3 h$ $b = \frac{3}{2} (3 h^{2} - k)$ $c = \frac{1}{3 h} (h^{2} + k^{2} - \frac{s^{2}}{3})$ $t h e e q u a t i o n c a n b e r e f o r m e d t o :$ $z^{3} + 3 p z + 2 q = 0$ $w i t h$ $x = z + h$ $p = \frac{h^{2} - k}{2}$ $q = \frac{h (5 h^{2} - 3 k)}{4} + \frac{3 h^{2} + 3 k^{2} - s^{2}}{18 h}$ $t h e r o o t s a r e$ $x_{i} = 2 \sqrt{- p} \sin (\frac{1}{3} \sin^{- 1} \frac{q}{\sqrt{- p^{3}}} + \frac{2 i π}{3}) + h$ $(i = 0, 1, 2 f o r p o i n t A, B, C r e s p e c t i v e l y)$ $x_{A} = x_{0} = 2 \sqrt{- p} \sin (\frac{1}{3} \sin^{- 1} \frac{q}{\sqrt{- p^{3}}}) + h$ $x_{C} = x_{2} = 2 \sqrt{- p} \sin (\frac{1}{3} \sin^{- 1} \frac{q}{\sqrt{- p^{3}}} + \frac{4 π}{3}) + h$
	Commented by ajfour last updated on 18/Apr/20

	$S i r, Q . 89415 p l e a s e, c a n y o u$ $c o n f i r m j u s t t h e a n s w e r . .$
	Commented by mr W last updated on 17/Apr/20

	Commented by mr W last updated on 18/Apr/20

	$t h i s s o l u t i o n w a s o n l y p o s s i b l e d u e t o$ $y o u r e a r l i e r w o n d e r f u l w o r k . t h e r e f o r e$ $b i g t h a n k s t o y o u f o r t h e q u e s t i o n a n d$ $f o r t h e c o n t r i b u t i o n t o t h e s o l u t i o n!$
	Commented by mr W last updated on 17/Apr/20

	Commented by mr W last updated on 18/Apr/20

	$s i n c e b o t h p a r a b o l a s h a v e t h e s a m e$ $s h a p e a n d t h e t r i a n g l e i s e q u i l a t e r a l,$ $t h e c o r r e l a t i o n b e t w e e n t h e s e c o n d$ $p a r a b o l a a n d t h e t r i a n g l e i s t h e s a m e$ $a s t h e c o r r e l a t i o n b e t w e e n t h e f i r s t$ $p a r a b o l a a n d t h e t r i a n g l e . t h e s e c o n d$ $p a r a b o l a i s j u s t t h e f i r s t p a r a b o l a$ $a f t e r a r o t a t i o n o f 120 ° a b o u t t h e$ $c e n t r o i d G o f t h e t r i a n g l e . a n d t h e$ $p o i n t A i s r o t a t e d t o t h e p o s i t i o n$ $o f p o i n t C, p o i n t B t o t h e p o s i t i o n o f$ $p o i n t A, p o i n t C t o t h e p o s i t i o n o f$ $p o i n t B .$ $f r o m Q 55613 w e k n o w : w h e n a c u r v e$ $y = f (x) i s r o t a t e d a b o u t t h e p o i n t (h, k)$ $b y a n a n g l e θ (c l o c k w i s e +), i t b e c o m e s$ $t h e c u r c e v = f (u) w i t h$ $u = (x - h) \cos θ - (y - k) \sin θ + h$ $v = (x - h) \sin θ + (y - k) \cos θ + k$ $i n c u r r e n t c a s e :$ $θ = 120 °$ $f r o m y = x^{2} w e g e t t h e n$ $\frac{\sqrt{3}}{2} (x - h) - \frac{1}{2} (y - k) + k =$ ${[- \frac{1}{2} (x - h) - \frac{\sqrt{3}}{2} (y - k) + h]}^{2}$ $o r$ $2 \sqrt{3} (x - h) - 2 y + 3 k = {[x - 2 h + \sqrt{3} (y - k)]}^{2}$ $t h i s i s t h e e q u a t i o n o f t h e s e c o n d$ $p a r a b o l a w i t h$ $h = \sqrt{\frac{s^{2}}{48} - \frac{1}{4}}$ $k = \frac{3 s^{2}}{16} - \frac{1}{4}$ $b u t t h e r e i s a n o t h e r c o n d i t i o n :$ $b o t h p a r a b o l a s s h o u l d t o u c h e a c h o t h e r$ $a t p o i n t C .$
	Commented by mr W last updated on 17/Apr/20

	Commented by ajfour last updated on 18/Apr/20
	hence side of triangle, remains to be obtained.. side of triangle under the tangency condition, must be unique.
	Commented by mr W last updated on 24/Apr/20

	$t h a t m e a n s : w h e n t h e t a n g e n t l i n e$ $o f p a r a b o l a a t p o i n t A i s r o t a t e d b y$ $120 °, i t s h o u l d c o i n c i d e w i t h t h e$ $t a n g e n t l i n e o f t h e p a r a b o l a a t p o i n t C .$ $i . e . θ_{A} = θ_{C} + 120 °$ $a t p o i n t A :$ $\tan θ_{A} = \frac{d y}{d x} = 2 x_{A}$ $a t p o i n t C :$ $\tan θ_{C} = \frac{d y}{d x} = 2 x_{C}$ $\tan θ_{A} = \tan (θ_{C} + 120 °) = \frac{\tan θ_{C} - \sqrt{3}}{1 + \sqrt{3} \tan θ_{C}}$ $\Rightarrow 2 x_{A} = \frac{2 x_{C} - \sqrt{3}}{1 + 2 \sqrt{3} x_{C}} . . . (I)$ $w i t h$ $x_{A} = 2 \sqrt{- p} \sin (\frac{1}{3} \sin^{- 1} \frac{q}{\sqrt{- p^{3}}}) + h$ $x_{C} = 2 \sqrt{- p} \sin (\frac{1}{3} \sin^{- 1} \frac{q}{\sqrt{- p^{3}}} + \frac{4 π}{3}) + h$ $p = \frac{h^{2} - k}{2}$ $q = \frac{h (5 h^{2} - 3 k)}{4} + \frac{3 h^{2} + 3 k^{2} - s^{2}}{18 h}$ $w e g e t f r o m e q n . (I) :$ $s \approx 3.779421$
	Commented by mr W last updated on 18/Apr/20

	Commented by ajfour last updated on 18/Apr/20

	$B u t a f t e r a l l i t s y o u w h o s o l v e d$ $i t r e a l l y, i w a s v e r y p u z z l e d, s o$ $t h a n k y o u e n o r m o u s l y S i r .$
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