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Question Number 52539 by ajfour last updated on 09/Jan/19

	Answered by mr W last updated on 09/Jan/19

	$p = \frac{\sqrt{[{(b + c)}^{2} - a^{2}] [a^{2} - {(b - c)}^{2}]}}{2 a}$ $r = \frac{\sqrt{[{(b + c)}^{2} - a^{2}] b c}}{b + c}$ $m = \frac{\sqrt{2 (b^{2} + c^{2}) - a^{2}}}{2}$ $\Rightarrow a^{2} = 2 (b^{2} + c^{2}) - 4 m^{2} . . . (i i i)$ $\Rightarrow (r^{2} - b c) {(b + c)}^{2} = 2 b c [2 m^{2} + 2 b c - {(b + c)}^{2}]$ $\Rightarrow 8 p^{2} [{(b + c)}^{2} - 2 m^{2} - 2 b c) = [4 m^{2} + 4 b c - {(b + c)}^{2}] [{(b + c)}^{2} - 4 m^{2}]$ $l e t X = b + c, Y = b c$ $\Rightarrow (r^{2} - Y) X^{2} = 2 Y (2 m^{2} + 2 Y - X^{2}) . . . (i)$ $\Rightarrow 8 p^{2} (X^{2} - 2 Y - 2 m^{2}) = (4 m^{2} + 4 Y - X^{2}) (X^{2} - 4 m^{2}) . . . (i i)$ $f r o m (i) :$ $\Rightarrow X^{2} = \frac{4 Y (m^{2} + Y)}{r^{2} + Y}$ $\Rightarrow p^{2} [\frac{2 Y (m^{2} + Y)}{r^{2} + Y} - Y - m^{2}] = [m^{2} + Y - \frac{Y (m^{2} + Y)}{r^{2} + Y}] [\frac{Y (m^{2} + Y)}{r^{2} + Y} - m^{2}]$ $\Rightarrow Y = r^{2} \sqrt{\frac{m^{2} - p^{2}}{r^{2} - p^{2}}} . . . (1)$ $\Rightarrow X = 2 \sqrt{\frac{Y (m^{2} + Y)}{r^{2} + Y}} . . . (2)$ $w i t h p = \sqrt{15}, m = \sqrt{31}, r = 2 \sqrt{6}$ $\Rightarrow Y = 24 \sqrt{\frac{31 - 15}{24 - 15}} = 24 \times \frac{4}{3} = 32 (= b c)$ $\Rightarrow X = 2 \sqrt{\frac{32 (31 + 32)}{24 + 32}} = 12 (= b + c)$ $\Rightarrow b a n d c a r e r o o t s o f$ $z^{2} - 12 z + 32 = 0$ $(z - 8) (z - 4) = 0$ $\Rightarrow b = 8, c = 4 (o r b = 4, c = 8)$ $f r o m (i i i) :$ $\Rightarrow a = 6$
	Commented by ajfour last updated on 10/Jan/19

	$c o r r e c t a n s w e r s s i r . t h a n k y o u .$ $(i h a d c r e a t e d t h e q u e s t i o n m y s e l f,$ $s o l v i n g i s a t o u g h e r t a s k i s e e n o w) .$
	Commented by mr W last updated on 10/Jan/19

	${i t}^{'} s a n i c e q u e s t i o n s i r .$ $a t t h e b e g i n i n g i t h o u g h t t h e s i d e s o f$ $t h e t r i a n g l e c o u l d b e n o t u n i q u e . a n d$ $i t h o u g h t {i t}^{'} s m a y b y i m p o s s i b l e t o d e t e r m i n e$ $t h e s i d e s o f t h e t r i a n g l e w i t h d i r e c t$ $f o r m u l a, t h e n i f o u n d t h i s s o l u t i o n$ $w h i c h c a n s o l v e b + c a n d b c d i r e c t l y .$ $w i t h t h i s m e t h o d, t h e s i d e s o f t h e$ $t r i a n g l e c a n a l l b e c a l c u l a t e d d i r e c t l y$ $a n d e x a c t l y .$
	Commented by mr W last updated on 13/Jan/19

	Answered by behi83417@gmail.com last updated on 09/Jan/19

	$4 m^{2} = 2 (b^{2} + c^{2}) - a^{2} = 124 (i)$ $\Rightarrow b^{2} + c^{2} = 62 + \frac{a^{2}}{2}$ $r = \frac{2 b c}{b + c} c o s \frac{A}{2} = \frac{2 b c}{b + c} \sqrt{\frac{p (p - a)}{b c}}$ $\frac{2}{b + c} \sqrt{b c p (p - a)} = 2 \sqrt{6} \Rightarrow b c p (p - a) = 6 {(b + c)}^{2} (i i)$ $b c (\frac{a + b + c}{2}) (\frac{b + c - a}{2}) = 6 {(b + c)}^{2} \Rightarrow$ $b c [{(b + c)}^{2} - a^{2}] = 24 {(b + c)}^{2}$ $b c [62 + \frac{a^{2}}{2} + 2 b c - a^{2}] = 24 {(b + c)}^{2} = 24 (62 + \frac{a^{2}}{2} + 2 b c)$ $\Rightarrow b c (62 + 2 b c - \frac{a^{2}}{2}) = 24 (62 + \frac{a^{2}}{2} + 2 b c)$ $62 b c + 2 {(b c)}^{2} - b c . \frac{a^{2}}{2} = 24 (62 + \frac{a^{2}}{2} + 2 b c)$ $\Rightarrow {(b c)}^{2} + (7 - \frac{a^{2}}{4}) . b c - 6 a^{2} - 744 = 0$ $. . . . .$
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