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Question Number 62861 by ajfour last updated on 26/Jun/19

Commented by mr W last updated on 26/Jun/19

$v e r y h a r d t a s k s i r, m a y b e n o t$ $p o s s i b l e t o s o l v e .$
Commented by ajfour last updated on 26/Jun/19

$s e r i o u s l y S i r, n o m a t t e r w h a t i t r y,$ $i t d o n t s i m p l i f y, b i q u a d r a t i c$ $t r o u b l e .$
Commented by ajfour last updated on 26/Jun/19

$F i n d x_{A} (A b e i n g l o w e r m o s t p o i n t)$ $i n t e r m s o f s .$
Commented by ajfour last updated on 26/Jun/19

$G o o d q u e s t i o n, GREAT s o l u t i o n,$ $S i r . T r u l y a b o v e a l l p r a i s e, t h a n k s a l o t!$
Commented by MJS last updated on 26/Jun/19

${you}^{'} re welcome as always$ $I like your geometrical problems$
Commented by mr W last updated on 26/Jun/19

$t h a n k s t o M J S s i r! i t h i n k i t i s$ $i m p o s s i b l e t o s o l v e t h e p r o b l e m w i t h$ $g i v e n l e n g t h s, s i n c e i t l e a d s a l w a y s$ $t o a q u a r t i c e q u a t i o n w h i c h m a y h a v e$ $m a x . 4 r o o t s . a c t u a l l y i a l s o c o n s i d e r e d$ $a w a y t o s o l v e b y a s s u m i n g t h e$ $p o s i t i o n o f B C a n d t h e n d e t e r m i n e$ $t h e p o i n t A a n d t h e l e n g t h s . i t h i n k$ $w e c a n t h e n a v o i d t h e q u a r t i c e q u a t i o n s .$ $b u t i h a d n o t i m e t o t r y i t o u t t o d a y .$ $i^{'} l l t a k e a t r y t h i s e v e n i n g a n d s e e i f$ $i c a n g e t t h e s a m e r e s u l t a s y o u r s .$
Commented by mr W last updated on 27/Jun/19

$M J S s i r : c a n y o u p l e a s e c h e c k :$ $i {c o u l d n}^{'} t g e t t h e s a m e r e s u l t a s y o u .$ $a c c o r d i n g t o y o u r r e s u l t \frac{\sqrt{3}}{3} < a ⩽ \frac{\sqrt{. . .}}{3}$ $a n d i g o t 0 ⩽ m < \frac{\sqrt{3}}{3} (m = y o u r a) .$ $w e k n o w m = 0 i s p o s s i b l e, b e c a u s e$ $t h e s m a l l e s t e q u i l a t e r a l t r i a n g l e i s$ $s = 2 \sqrt{3} w i t h A (0, 0), B (\sqrt{3}, 3), C (- \sqrt{3}, 3) .$ $i . e . 2 \sqrt{3} ⩽ s < \infty .$ $b u t t h i s c a s e i s n o t i n c l u d e d i n y o u r$ $r e s u l t, s i n c e y o u g o t \frac{\sqrt{3}}{3} < a ⩽ \frac{\sqrt{. . .}}{3} a n d$ $0 ⩽ s < \infty .$
Commented by mr W last updated on 27/Jun/19

Commented by MJS last updated on 27/Jun/19

$I made a mistake, hopefully corrected$ $everything . it now looks as if we can solve$ $it for a = f (s) . it leads to a polynome of$ $3^{rd} degree for a^{2} . please check my work, I^{'} ve$ $been in a hurry . . .$
Commented by mr W last updated on 27/Jun/19

$t h a n k y o u s i r!$ $i c a n n o t g e t y o u r f o r m u l a f o r b,$ $b u t i t s e e m s t o b e c o r r e c t .$
	Answered by ajfour last updated on 26/Jun/19

	$A (a, a^{2}), B (b, b^{2}), C (c, c^{2})$ ${(a - b)}^{2} + {(a^{2} - b^{2})}^{2} = s^{2}$ ${(b - c)}^{2} + {(b^{2} - c^{2})}^{2} = s^{2}$ ${(c - a)}^{2} + {(c^{2} - a^{2})}^{2} = s^{2}$ $. . . . . . . . . . . . . . . . .$
	Answered by mr W last updated on 26/Jun/19

	Commented by mr W last updated on 27/Jun/19
	$D(0,h) m=tan θ>0 eqn. of BC: y=h+mx intersection BC with parabola: y=x^2 =h+mx ⇒x^2 −mx−h=0 x_C =((m−(√(m^2 +4h)))/2) x_B =((m+(√(m^2 +4h)))/2) midpoint of BC=M x_M =((x_C +x_B )/2)=(m/2) y_M =((x_C ^2 +x_B ^2 )/2)=(((x_C +x_B )^2 −2x_C x_B )/2)=((m^2 +2h)/2) eqn. of bisector of BC: y−y_M =−(1/m)(x−x_M ) y−((m^2 +2h)/2)=−(1/m)(x−(m/2)) ⇒y=−(x/m)+((m^2 +2h+1)/2) intersection of bisector with parabola: y=x^2 =−(x/m)+((m^2 +2h+1)/2) ⇒x^2 +(x/m)−((m^2 +2h+1)/2)=0 x_A =(1/(2m))[−1+(√(1+2m^2 (m^2 +2h+1)))] y_A =(1/(2m^2 ))[1+m^2 (m^2 +2h+1)−(√(1+2m^2 (m^2 +2h+1)))] ((√3)/2)×BC=MA ⇒((√3)/2)(x_B −x_C )=y_M −y_A ((√3)/2)(√(m^2 +4h))=(1/2)[m^2 +2h−(1/m^2 )−m^2 −2h−1+(1/m)(√((1/m^2 )+2(m^2 +2h+1)))] ⇒m^2 (√(3(m^2 +4h)))=(√(1+2m^2 (m^2 +2h+1)))−(1+m^2 ) ⇒0≤m<((√3)/3) ⇒h≥3 3m^4 (m^2 +4h)=[1+2m^2 (m^2 +2h+1)+1+2m^2 +m^4 −2(1+m^2 )(√(1+2m^2 (m^2 +2h+1)))] 2(1+m^2 )(√(1+2m^2 (m^2 +2h+1)))=2+4m^2 +3m^4 (1−m^2 )+4m^2 (1−3m^2 )h ⇒16m^4 (1−3m^2 )^2 h^2 +8m^2 {(1−3m^2 )(2+4m^2 +3m^4 −3m^6 )−2(1+m^2 )^2 }h+{(2+4m^2 +3m^4 −3m^6 )^2 −4(1+m^2 )^2 (1+2m^2 +2m^4 )}=0 ⇒16(1−3m^2 )^2 h^2 +8(9m^6 −12m^4 −11m^2 −6)h+m^2 (9m^6 −18m^4 −23m^2 −12)=0 ⇒h=(((−9m^6 +12m^4 +11m^2 +6)+(√((9m^6 −12m^4 −11m^2 −6)^2 −m^2 (1−3m^2 )^2 (9m^6 −18m^4 −23m^2 −12))))/(4(1−3m^2 )^2 )) ... i stop here , hope it′s correct, too hard to simplify ... s=(√(1+m^2 ))(x_B −x_C ) ⇒s=(√((1+m^2 )(m^2 +4h))) ⇒s≥2(√3)$
	$D (0, h)$ $m = \tan θ > 0$ $e q n . o f B C :$ $y = h + m x$ $i n t e r s e c t i o n B C w i t h p a r a b o l a :$ $y = x^{2} = h + m x$ $\Rightarrow x^{2} - m x - h = 0$ $x_{C} = \frac{m - \sqrt{m^{2} + 4 h}}{2}$ $x_{B} = \frac{m + \sqrt{m^{2} + 4 h}}{2}$ $m i d p o i n t o f B C = M$ $x_{M} = \frac{x_{C} + x_{B}}{2} = \frac{m}{2}$ $y_{M} = \frac{x_{C}^{2} + x_{B}^{2}}{2} = \frac{{(x_{C} + x_{B})}^{2} - 2 x_{C} x_{B}}{2} = \frac{m^{2} + 2 h}{2}$ $e q n . o f b i s e c t o r o f B C :$ $y - y_{M} = - \frac{1}{m} (x - x_{M})$ $y - \frac{m^{2} + 2 h}{2} = - \frac{1}{m} (x - \frac{m}{2})$ $\Rightarrow y = - \frac{x}{m} + \frac{m^{2} + 2 h + 1}{2}$ $i n t e r s e c t i o n o f b i s e c t o r w i t h p a r a b o l a :$ $y = x^{2} = - \frac{x}{m} + \frac{m^{2} + 2 h + 1}{2}$ $\Rightarrow x^{2} + \frac{x}{m} - \frac{m^{2} + 2 h + 1}{2} = 0$ $x_{A} = \frac{1}{2 m} [- 1 + \sqrt{1 + 2 m^{2} (m^{2} + 2 h + 1)}]$ $y_{A} = \frac{1}{2 m^{2}} [1 + m^{2} (m^{2} + 2 h + 1) - \sqrt{1 + 2 m^{2} (m^{2} + 2 h + 1)}]$ $\frac{\sqrt{3}}{2} \times B C = M A$ $\Rightarrow \frac{\sqrt{3}}{2} (x_{B} - x_{C}) = y_{M} - y_{A}$ $\frac{\sqrt{3}}{2} \sqrt{m^{2} + 4 h} = \frac{1}{2} [m^{2} + 2 h - \frac{1}{m^{2}} - m^{2} - 2 h - 1 + \frac{1}{m} \sqrt{\frac{1}{m^{2}} + 2 (m^{2} + 2 h + 1)}]$ $\Rightarrow m^{2} \sqrt{3 (m^{2} + 4 h)} = \sqrt{1 + 2 m^{2} (m^{2} + 2 h + 1)} - (1 + m^{2})$ $\Rightarrow 0 ⩽ m < \frac{\sqrt{3}}{3}$ $\Rightarrow h ⩾ 3$ $3 m^{4} (m^{2} + 4 h) = [1 + 2 m^{2} (m^{2} + 2 h + 1) + 1 + 2 m^{2} + m^{4} - 2 (1 + m^{2}) \sqrt{1 + 2 m^{2} (m^{2} + 2 h + 1)}]$ $2 (1 + m^{2}) \sqrt{1 + 2 m^{2} (m^{2} + 2 h + 1)} = 2 + 4 m^{2} + 3 m^{4} (1 - m^{2}) + 4 m^{2} (1 - 3 m^{2}) h$ $\Rightarrow 16 m^{4} {(1 - 3 m^{2})}^{2} h^{2} + 8 m^{2} {(1 - 3 m^{2}) (2 + 4 m^{2} + 3 m^{4} - 3 m^{6}) - 2 {(1 + m^{2})}^{2}} h + {{(2 + 4 m^{2} + 3 m^{4} - 3 m^{6})}^{2} - 4 {(1 + m^{2})}^{2} (1 + 2 m^{2} + 2 m^{4})} = 0$ $\Rightarrow 16 {(1 - 3 m^{2})}^{2} h^{2} + 8 (9 m^{6} - 12 m^{4} - 11 m^{2} - 6) h + m^{2} (9 m^{6} - 18 m^{4} - 23 m^{2} - 12) = 0$ $\Rightarrow h = \frac{(- 9 m^{6} + 12 m^{4} + 11 m^{2} + 6) + \sqrt{{(9 m^{6} - 12 m^{4} - 11 m^{2} - 6)}^{2} - m^{2} {(1 - 3 m^{2})}^{2} (9 m^{6} - 18 m^{4} - 23 m^{2} - 12)}}{4 {(1 - 3 m^{2})}^{2}}$ $. . . i s t o p h e r e, h o p e {i t}^{'} s c o r r e c t, t o o h a r d t o s i m p l i f y . . .$ $s = \sqrt{1 + m^{2}} (x_{B} - x_{C})$ $\Rightarrow s = \sqrt{(1 + m^{2}) (m^{2} + 4 h)}$ $\Rightarrow s ⩾ 2 \sqrt{3}$
	Commented by ajfour last updated on 27/Jun/19

	$s i r, y o u h a v e o b t a i n e d h = f (m)$ $b u t w h a t a b o u t m, h o w z t h a t$ $f o u n d ?$ $g r e a t e f f o r t S i r, t h a n k s .$
	Commented by mr W last updated on 27/Jun/19

	$i t r i e d t o u s e m a s b a s e p a r a m e t e r,$ $t h e n f i n d h (m) a n d s (m) . t h i s m e t h o d$ $i s e a s i e r, s i n c e w e o n l y n e e d t o s o l v e$ $q u a d r a t i c e q u a t i o n s . w e f i n d i n t h i s$ $w a y : s ⩾ 2 \sqrt{3} a n d 0 ⩽ m < \frac{1}{\sqrt{3}} .$
	Commented by mr W last updated on 27/Jun/19

	$h = h (m) c o u l d b e m o r e s i m p l i f i e d a s$ $M J S s i r d i d . b u t i {h a v n}^{'} t r e a c h e d t h a t .$
	Answered by MJS last updated on 27/Jun/19

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