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Question Number 17743 by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17

	Answered by mrW1 last updated on 10/Jul/17

	$let k = \frac{\sqrt{2}}{2} l$ $A (0, k, 0)$ $B (k, 0, 0)$ $C (0, - k, 0)$ $D (- k, 0, 0)$ $M (p, q, r)$ $\Rightarrow a^{2} = p^{2} + {(q - k)}^{2} + r^{2} . . . (i)$ $\Rightarrow b^{2} = {(p - k)}^{2} + q^{2} + r^{2} . . . (ii)$ $\Rightarrow c^{2} = p^{2} + {(q + k)}^{2} + r^{2} . . . (iii)$ $\Rightarrow d^{2} = {(p + k)}^{2} + q^{2} + r^{2} . . . (iv)$ $a^{2} + c^{2} = {2 p}^{2} + {2 q}^{2} + {2 r}^{2} + {2 k}^{2} = 2 (p^{2} + q^{2} + r^{2} + k^{2})$ $b^{2} + d^{2} = {2 p}^{2} + {2 q}^{2} + {2 r}^{2} + {2 k}^{2} = 2 (p^{2} + q^{2} + r^{2} + k^{2})$ $\Rightarrow a^{2} + c^{2} = b^{2} + d^{2}$ $(iii) - (i) :$ $c^{2} - a^{2} = 4 qk$ $\Rightarrow q = \frac{c^{2} - a^{2}}{4 k}$ $(iv) - (ii) :$ $d^{2} - b^{2} = 4 pk$ $\Rightarrow p = \frac{d^{2} - b^{2}}{4 k}$ $from (i) :$ $r^{2} = - [p^{2} + {(q - k)}^{2} - a^{2}]$ $= - [\frac{{(d^{2} - b^{2})}^{2}}{{16 k}^{2}} + {(\frac{c^{2} - a^{2}}{4 k} - k)}^{2} - a^{2}]$ $= - [\frac{{(d^{2} - b^{2})}^{2} + {(c^{2} - a^{2} - {4 k}^{2})}^{2} - {16 a}^{2} k^{2}}{{16 k}^{2}}]$ $= - [\frac{d^{4} + b^{4} - {2 d}^{2} b^{2} + c^{4} + a^{4} + {16 k}^{4} - {2 a}^{2} c^{2} - {8 k}^{2} c^{2} + {8 k}^{2} a^{2} - {16 a}^{2} k^{2}}{{16 k}^{2}}]$ $= - [\frac{a^{4} + b^{4} + c^{4} + d^{4} - 2 (a^{2} c^{2} + d^{2} b^{2}) + {16 k}^{4} - {8 k}^{2} (a^{2} + c^{2})}{{16 k}^{2}}]$ $= - [\frac{{(a^{2} - c^{2})}^{2} + {(b^{2} - d^{2})}^{2} - {8 k}^{2} (a^{2} + c^{2}) + {16 k}^{4}}{{16 k}^{2}}]$ $= - [\frac{{(a^{2} - c^{2})}^{2} + {(b^{2} - d^{2})}^{2} - {8 k}^{2} (a^{2} + c^{2} - {2 k}^{2})}{{16 k}^{2}}]$ $= - [\frac{{(a^{2} - c^{2})}^{2} + {(b^{2} - d^{2})}^{2} - 4 l^{2} (a^{2} + c^{2} - l^{2})}{8 l^{2}}]$ $\Rightarrow h =∣ r ∣= \frac{\sqrt{4 l^{2} (a^{2} + c^{2} - l^{2}) - {(a^{2} - c^{2})}^{2} - {(b^{2} - d^{2})}^{2}}}{2 \sqrt{2} l}$
	Commented by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17

	$t h a n k y o u d e a r m r W 1 .$ $f i r s t a n d a l w a y s t h e b e s t .$
	Answered by ajfour last updated on 10/Jul/17

	Commented by ajfour last updated on 10/Jul/17

	${AP}^{2} = a^{2} = h^{2} + {(x - k)}^{2} + y^{2}$ ${BP}^{2} = b^{2} = h^{2} + x^{2} + {(y - k)}^{2}$ ${CP}^{2} = c^{2} = h^{2} + {(x + k)}^{2} + y^{2}$ ${DP}^{2} = d^{2} = h^{2} + x^{2} + {(y + k)}^{2}$ $a^{2} + c^{2} = b^{2} + d^{2} = 2 (h^{2} + x^{2} + y^{2} + k^{2})$ $. . . . . (i)$ $c^{2} - a^{2} = 4 kx; d^{2} - b^{2} = 4 ky$ $\Rightarrow {16 k}^{2} (x^{2} + y^{2}) = {(c^{2} - a^{2})}^{2} + {(d^{2} - b^{2})}^{2}$ $a^{2} + b^{2} + c^{2} + d^{2} = 4 (h^{2} + x^{2} + y^{2} + k^{2})$ $so, {16 k}^{2} h^{2} = {4 k}^{2} (a^{2} + b^{2} + c^{2} + d^{2})$ $- {16 k}^{2} (x^{2} + y^{2} + k^{2})$ ${16 k}^{2} h^{2} = {4 k}^{2} (a^{2} + b^{2} + c^{2} + d^{2})$ $- [{(c^{2} - a^{2})}^{2} + {(d^{2} - b^{2})}^{2}] - {16 k}^{4}$ $h = \sqrt{\frac{a^{2} + b^{2} + c^{2} + d^{2}}{4} - \frac{(c^{2} - a^{2}) + {(d^{2} - b^{2})}^{2}}{{16 k}^{2}} - k^{2}}$ $with {2 k}^{2} = l^{2} .$
	Commented by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17

	$t h a n k y o u d e a r m r A j f o u r .$ $y o u r a n s w e r s a r e s p i c i a l a n d d i f f e r e n t$ $a n y t i m e .$
	Commented by ajfour last updated on 10/Jul/17

	$but mrW 1 {Sir}^{'} s answer dont agree$ $with my answer . please find the error .$
	Commented by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17

	$h^{2} = \frac{a^{2} + b^{2} + c^{2} + d^{2}}{4} - \frac{{(a^{2} - c^{2})}^{2} + {(b^{2} - d^{2})}^{2}}{8 l^{2}} - \frac{l^{2}}{2}$
	Commented by mrW1 last updated on 10/Jul/17

	$I had a mistake with ‘ ‘ -^{″} sign . But$ $the result is the same even when we$ $used different expressions, since$ $a^{2} + c^{2} = b^{2} + d^{2} .$
	Commented by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17

	$t h a n k s f o r y o u r c a r e a n d c o r r e c t i o n .$
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