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-

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 \dots . :)

Answered by 1549442205PVT last updated on 23/Oct/20

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 \dots :)

Commented by 1549442205PVT last updated on 25/Oct/20

You are welcome