Given that a, b and c are real numbers that stisfy the system of equation above a − (√(b^2 − (1/(16)) )) = (√(c^2 − (1/(16)) )) b − (√(c^2 − (1/(25)))) = (√(a^2 − (1/(25)) )) c − (√(a^2 − (1/(36)) )) = (√(b^2 − (1/(36)))) if a+ b + c = (x/( (√y))) where x, y are positive integers and y is square free, find the value of x + y !


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Question Number 119150 by abdul88 last updated on 22/Oct/20

$G i v e n t h a t a, b a n d c a r e r e a l n u m b e r s$ $t h a t s t i s f y t h e s y s t e m o f e q u a t i o n a b o v e$ $a - \sqrt{b^{2} - \frac{1}{16}} = \sqrt{c^{2} - \frac{1}{16}}$ $b - \sqrt{c^{2} - \frac{1}{25}} = \sqrt{a^{2} - \frac{1}{25}}$ $c - \sqrt{a^{2} - \frac{1}{36}} = \sqrt{b^{2} - \frac{1}{36}}$ $i f a + b + c = \frac{x}{\sqrt{y}} w h e r e x, y a r e p o s i t i v e i n t e g e r s$ $a n d y i s s q u a r e f r e e, f i n d t h e v a l u e o f x + y!$
	Answered by MJS_new last updated on 22/Oct/20

	$the given equations have the shape$ $o - \sqrt{p^{2} - r} = \sqrt{q^{2} - r}$ $o = \sqrt{p^{2} - r} + \sqrt{q^{2} - r}$ $o^{2} = (p^{2} - r) + (q^{2} - r) + 2 \sqrt{p^{2} - r} \sqrt{q^{2} - r}$ $o^{2} - p^{2} - q^{2} + 2 r = 2 \sqrt{p^{2} - r} \sqrt{q^{2} - r}$ $squaring again and transforming gives$ $o^{4} + p^{4} + q^{4} - 2 (o^{2} p^{2} + o^{2} q^{2} + p^{2} q^{2}) + 4 {r o}^{2} = 0$ $we see that o, p, q are interchangeable$ $the only different term is 4 {r o}^{2}$ $\Rightarrow substracting two equations we get$ $\frac{4}{16} a^{2} - \frac{4}{25} b^{2} = 0 \land \frac{4}{16} a^{2} - \frac{4}{36} c^{2} = 0$ $\Rightarrow b = \frac{5}{4} a \land c = \frac{3}{2} a$ $inserting above we get$ $a = \frac{8 \sqrt{7}}{105} \Rightarrow b = \frac{2 \sqrt{7}}{21} \land c = \frac{4 \sqrt{7}}{35}$ $a + b + c = \frac{2 \sqrt{7}}{7} = \frac{2}{\sqrt{7}}$ $\Rightarrow x = 2 \land y = 7 \Rightarrow x + y = 9$
	Commented by PRITHWISH SEN 2 last updated on 22/Oct/20

	$superb \overset{}{!}$
	Commented by MJS_new last updated on 22/Oct/20

	$thank you$
	Commented by abdul88 last updated on 23/Oct/20

	$t h a n k y o u S i r . . . . :)$
	Answered by 1549442205PVT last updated on 23/Oct/20
	$We need the condition a≥(1/5),b≥(1/4),c≥(1/4) for the equalities are defined a − (√(b^2 − (1/(16)) )) = (√(c^2 − (1/(16)) )) ⇔a= (√(b^2 − (1/(16)) )) + (√(c^2 − (1/(16)) )) ⇔a^2 =b^2 +c^2 −(1/8)+2(√(b^2 c^2 −((b^2 +c^2 )/(16))+(1/(256)))) ⇔a^2 −b^2 −c^2 +(1/8)=2(√(b^2 c^2 −((b^2 +c^2 )/(16))+(1/(256)))) a^4 +b^4 +c^4 +(1/(64))−2a^2 b^2 −2a^2 c^2 +(a^2 /4)−((b^2 +c^2 )/4)+2b^2 c^2 =4b^2 c^2 −((b^2 +c^2 )/4)+(1/(64))⇔ a^4 +b^4 +c^4 −2(a^2 b^2 +a^2 c^2 +b^2 c^2 )+(a^2 /4)=0(1) Somilarly,we have b − (√(c^2 − (1/(25)))) = (√(a^2 − (1/(25)) ))⇔ a^4 +b^4 +c^4 −2(a^2 b^2 +a^2 c^2 +b^2 c^2 )+((4b^2 )/(25))=0(2) c − (√(a^2 − (1/(36)) )) = (√(b^2 − (1/(36)))) ⇔ a^4 +b^4 +c^4 −2(a^2 b^2 +a^2 c^2 +b^2 c^2 )+((4c^2 )/(36))=0(3) Substracting(1) from(2),(3) we get { (((a^2 /4)−((4b^2 )/(25))=0)),(((a^2 /4)−(c^2 /9)=0)) :}⇔ { ((b^2 =((25a^2 )/(16)))),((c^2 =((9a^2 )/4))) :}⇒ { ((b=((5a)/4))),((c=((3a)/2))) :} Replace into (3)we get a^4 +((625a^2 )/(256))+((81a^4 )/(16))−2(((25a^4 )/(16))+((9a^4 )/4)+((225a^4 )/(64)))+(a^2 /4)=0 ⇔−((1575)/(256))a^4 +(a^2 /4)=0⇔1575a^2 =64 ⇒a=(8/(15(√7))),b=((10)/( 15(√7))),c=((12)/( 15(√7))) a+b+c=((8+10+12)/(15(√7)))=(2/( (√7)))=(x/( (√y))) ⇒y=7,x=2⇒(x+y)!=9!=362880$
	$We need the condition a ⩾ \frac{1}{5}, b ⩾ \frac{1}{4}, c ⩾ \frac{1}{4}$ $for the equalities are defined$ $a - \sqrt{b^{2} - \frac{1}{16}} = \sqrt{c^{2} - \frac{1}{16}}$ $\Leftrightarrow a = \sqrt{b^{2} - \frac{1}{16}} + \sqrt{c^{2} - \frac{1}{16}}$ $\Leftrightarrow a^{2} = b^{2} + c^{2} - \frac{1}{8} + 2 \sqrt{b^{2} c^{2} - \frac{b^{2} + c^{2}}{16} + \frac{1}{256}}$ $\Leftrightarrow a^{2} - b^{2} - c^{2} + \frac{1}{8} = 2 \sqrt{b^{2} c^{2} - \frac{b^{2} + c^{2}}{16} + \frac{1}{256}}$ $a^{4} + b^{4} + c^{4} + \frac{1}{64} - 2 {\overset{2}{a b}}^{2} - {2 a}^{2} c^{2} + \frac{a^{2}}{4} - \frac{b^{2} + c^{2}}{4} + {2 b}^{2} c^{2}$ $= {4 b}^{2} c^{2} - \frac{b^{2} + c^{2}}{4} + \frac{1}{64} \Leftrightarrow$ $a^{4} + b^{4} + c^{4} - 2 (a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2}) + \frac{a^{2}}{4} = 0 (1)$ $Somilarly, we have$ $b - \sqrt{c^{2} - \frac{1}{25}} = \sqrt{a^{2} - \frac{1}{25}} \Leftrightarrow$ $a^{4} + b^{4} + c^{4} - 2 (a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2}) + \frac{{4 b}^{2}}{25} = 0 (2)$ $c - \sqrt{a^{2} - \frac{1}{36}} = \sqrt{b^{2} - \frac{1}{36}} \Leftrightarrow$ $a^{4} + b^{4} + c^{4} - 2 (a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2}) + \frac{{4 c}^{2}}{36} = 0 (3)$ $Substracting (1) from (2), (3) we get$ ${\begin{cases} \frac{a^{2}}{4} - \frac{{4 b}^{2}}{25} = 0 \\ \frac{a^{2}}{4} - \frac{c^{2}}{9} = 0 \end{cases} \Leftrightarrow {\begin{cases} b^{2} = \frac{{25 a}^{2}}{16} \\ c^{2} = \frac{{9 a}^{2}}{4} \end{cases} \Rightarrow {\begin{cases} b = \frac{5 a}{4} \\ c = \frac{3 a}{2} \end{cases}$ $Replace into (3) we get$ $a^{4} + \frac{{625 a}^{2}}{256} + \frac{{81 a}^{4}}{16} - 2 (\frac{{25 a}^{4}}{16} + \frac{{9 a}^{4}}{4} + \frac{{225 a}^{4}}{64}) + \frac{a^{2}}{4} = 0$ $\Leftrightarrow - \frac{1575}{256} a^{4} + \frac{a^{2}}{4} = 0 \Leftrightarrow {1575 a}^{2} = 64$ $\Rightarrow a = \frac{8}{15 \sqrt{7}}, b = \frac{10}{15 \sqrt{7}}, c = \frac{12}{15 \sqrt{7}}$ $a + b + c = \frac{8 + 10 + 12}{15 \sqrt{7}} = \frac{2}{\sqrt{7}} = \frac{x}{\sqrt{y}}$ $\Rightarrow y = 7, x = 2 \Rightarrow (x + y)! = 9! = 362880$
	Commented by abdul88 last updated on 23/Oct/20

	$T h a n k Y o u S i r . . . :)$
	Commented by 1549442205PVT last updated on 25/Oct/20

	$You are welcome$
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