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Question Number 62938 by ajfour last updated on 27/Jun/19

Commented by ajfour last updated on 27/Jun/19

$Q u e s t i o n 62861 (r e v i s i t)$
	Answered by ajfour last updated on 27/Jun/19

	$l e t A (x_{1}, x_{1}^{2}), B {(x_{2}, x_{2})}^{2}, C (x_{3}, x_{3}^{2})$ $c i c u m c e n t r e o f △ b e (h, k)$ $R = \frac{s}{\sqrt{3}}$ ${(x - h)}^{2} + {(x^{2} - k)}^{2} = s^{2} / 3$ $T h i s h a s f o u r r o o t s x_{1}, x_{2}, x_{3}, x_{4}$ $x^{4} + (1 - 2 k) x^{2} - 2 h x + (h^{2} + k^{2} - \frac{s^{2}}{3}) = 0$ $x_{1} x_{2} x_{3} x_{4} = h^{2} + k^{2} - s^{2} / 3 . . . . (i)$ $x_{1} x_{2} x_{3} + x_{4} (x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1}) = 2 h . . . (i i)$ $(x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1})$ $+ x_{4} (x_{1} + x_{2} + x_{3}) = 1 - 2 k . . (i i i)$ $x_{1} + x_{2} + x_{3} + x_{4} = 0 . . . (i v)$ $B u t (h, k) a l s o b e i n g c e n t r o i d$ $o f △ A B C, w e h a v e$ $x_{1} + x_{2} + x_{3} = 3 h . . . (v)$ $x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 3 k . . . . (v i)$ $f r o m (i v) & (v)$ $x_{1} + x_{2} + x_{3} = 3 h, x_{4} = - 3 h$ ${(x_{1} + x_{2} + x_{3})}^{2} = x_{1}^{2} + x_{2}^{2} + x_{3}^{2}$ $+ 2 (x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1})$ $\Rightarrow x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1} = \frac{9 h^{2} - 3 k}{2}$ $N o w f r o m (i i)$ $x_{1} x_{2} x_{3} = 2 h + 3 h (\frac{9 h^{2} - 3 k}{2})$ $A n d u s i n g (i)$ $2 h + 3 h (\frac{9 h^{2} - 3 k}{2}) + \frac{1}{3 h} (h^{2} + k^{2} - \frac{s^{2}}{3}) = 0$ $. . . . . (I)$ $A n d u s i n g (i i i)$ $\frac{9 h^{2} - 3 k}{2} - 9 h^{2} = 1 - 2 k$ $\Rightarrow k = 9 h^{2} + 2 . . . . . . (I I)$ $u s i n g t h i s i n (I)$ $12 h^{2} + 81 h^{4} - 27 h^{2} (9 h^{2} + 2) + 2 h^{2}$ $+ 2 {(9 h^{2} + 2)}^{2} - \frac{2 s^{2}}{3} = 0$ $\Rightarrow 32 h^{2} = \frac{2 s^{2}}{3} - 8$ $\Rightarrow h^{2} = \frac{s^{2} - 12}{48} = \frac{s^{2}}{48} - \frac{1}{4}$ $k = \frac{3 s^{2}}{16} - \frac{1}{4}$ $N o w e q . b e l o w h a s f o u r r o o t s,$ $x_{4} = - 3 h = - \frac{\sqrt{3 (s^{2} - 12)}}{4}$ $x_{A} i s t h e s m a l l e r p o s i t i v e o n e$ $a m o n g x_{1}, x_{2}, a n d x_{3} .$ $x^{4} + (1 - 2 k) x^{2} - 2 h x + (h^{2} + k^{2} - \frac{s^{2}}{3}) = 0 .$ $___________________________$ $b u t a s$ $x_{1} x_{2} x_{3} = - (\frac{h^{2} + k^{2} - s^{2} / 3}{3 h}) = - q$ $x_{1} x_{2} + x_{2} x_{3} + x_{3} x_{1} = \frac{9 h^{2} - 3 k}{2} = p$ $a n d x_{1} + x_{2} + x_{3} = 3 h$ $w e c a n s o l v e r a t h e r a c u b i c$ $x^{3} - 3 {h x}^{2} + p x + q = 0$ $w i t h h = \frac{\sqrt{s^{2} - 12}}{4 \sqrt{3}}, k = \frac{3 s^{2}}{16} - \frac{1}{4},$ $q = \frac{h^{2} + k^{2} - s^{2} / 3}{3 h}, p = \frac{9 h^{2} - 3 k}{2} ◼$
	Commented by mr W last updated on 27/Jun/19

	$a c h e c k w i t h s = 2 \sqrt{3} :$ $h = 0, k = 2$ $q = 0, p = - 3$ $x^{3} - 3 x = 0$ $\Rightarrow x = 0 (p o i n t A)$ $\Rightarrow x = \pm \sqrt{3} (p o i n t B a n d C)$ $\Rightarrow c o r r e c t!$
	Commented by mr W last updated on 27/Jun/19

	$g r e a t s o l u t i o n s i r! y o u h a v e f o u n d$ $a w a y t o g e t x_{A, B, C} f r o m s!$
	Commented by ajfour last updated on 27/Jun/19

	$s a y i f s = 2 \sqrt{15}$ $\Rightarrow h = 1, k = 11, p = - 12, q = 34$ $\Rightarrow x^{3} - 3 x^{2} - 12 x + 34 = 0$ $x_{A} = x_{1} \approx 2.61323$ $x_{B} = x_{2} \approx 3.8056$ $x_{C} = x_{3} \approx - 3.41883 .$
	Commented by ajfour last updated on 27/Jun/19
	Thank you sir, it only needed our focus on center of triangle which is centroid as well as circumcenter.
	Commented by mr W last updated on 28/Jun/19

	$a s t r o k e o f g e n i u s!$
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