If ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x)) then show that ((a + b + c)/(x + y + z)) = ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) .


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Question Number 202290 by MATHEMATICSAM last updated on 24/Dec/23

$If \frac{3 a - b}{x + y} = \frac{3 b - c}{y + z} = \frac{3 c - a}{z + x} then show$ $that \frac{a + b + c}{x + y + z} = \frac{a^{2} + b^{2} + c^{2}}{a x + b y + c z} .$
	Answered by Rasheed.Sindhi last updated on 24/Dec/23
	$If ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x)) then show that ((a + b + c)/(x + y + z)) = ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) ((3a − b)/(x + y)) = ((3b − c)/(y + z)) = ((3c − a)/(z + x)) ⇒((x+y)/(3a − b))=((y+z)/(3b − c))=((z+x)/(3c − a))=k (say) x+y=k(3a − b)...(i) y+z=k(3b − c)...(ii) z+x=k(3c − a)...(iii) (i)+(ii)+(iii): 2(x+y+z)=k{(3a − b)+(3b − c)+(3c − a)} x+y+z=k(a+b+c)...(iv) (iv)−(ii): x=k(a+b+c)−k(3b − c) =k(a−2b+2c) (iv)−(iii): y=k(a+b+c)−k(3c − a) =k(2a+b−2c) (iv)−(i): z=k(a+b+c)−k(3a − b) =k(−2a+2b+c) To prove: ((a + b + c)/(x + y + z)) = ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) lhs: ((a + b + c)/(x + y + z))=((a + b + c)/(k(a+b+c)))=(1/k) rhs: ((a^(2 ) + b^2 + c^2 )/(ax + by + cz)) = ((a^(2 ) + b^2 + c^2 )/(a(k(a−2b+2c)) + b(k(2a+b−2c)) + c(k(−2a+2b+c)))) = ((a^(2 ) + b^2 + c^2 )/(k{a(a−2b+2c)) + b(2a+b−2c)) + c(−2a+2b+c))})) = ((a^(2 ) + b^2 + c^2 )/(k{a^2 −2ab+2ca + 2ab+b^2 −2bc −2ca+2bc+c^2 })) = ((a^(2 ) + b^2 + c^2 )/(k{a^2 +b^2 +c^2 }))=(1/k) ∵ lhs=rhs=1/k ∴ Proved$
	$If \frac{3 a - b}{x + y} = \frac{3 b - c}{y + z} = \frac{3 c - a}{z + x} then show$ $that \frac{a + b + c}{x + y + z} = \frac{a^{2} + b^{2} + c^{2}}{a x + b y + c z}$ $\frac{3 a - b}{x + y} = \frac{3 b - c}{y + z} = \frac{3 c - a}{z + x}$ $\Rightarrow \frac{x + y}{3 a - b} = \frac{y + z}{3 b - c} = \frac{z + x}{3 c - a} = k (s a y)$ $x + y = k (3 a - b) . . . (i)$ $y + z = k (3 b - c) . . . (i i)$ $z + x = k (3 c - a) . . . (i i i)$ $(i) + (i i) + (i i i) :$ $2 (x + y + z) = k {(3 a - b) + (3 b - c) + (3 c - a)}$ $x + y + z = k (a + b + c) . . . (i v)$ $(i v) - (i i) :$ $x = k (a + b + c) - k (3 b - c)$ $= k (a - 2 b + 2 c)$ $(i v) - (i i i) :$ $y = k (a + b + c) - k (3 c - a)$ $= k (2 a + b - 2 c)$ $(i v) - (i) :$ $z = k (a + b + c) - k (3 a - b)$ $= k (- 2 a + 2 b + c)$ $T o p r o v e :$ $\frac{a + b + c}{x + y + z} = \frac{a^{2} + b^{2} + c^{2}}{a x + b y + c z}$ $l h s :$ $\frac{a + b + c}{x + y + z} = \frac{a + b + c}{k (a + b + c)} = \frac{1}{k}$ $r h s :$ $\frac{a^{2} + b^{2} + c^{2}}{a x + b y + c z}$ $= \frac{a^{2} + b^{2} + c^{2}}{a (k (a - 2 b + 2 c)) + b (k (2 a + b - 2 c)) + c (k (- 2 a + 2 b + c))}$ $= \frac{a^{2} + b^{2} + c^{2}}{k {a (a - 2 b + 2 c)) + b (2 a + b - 2 c)) + c (- 2 a + 2 b + c))}}$ $= \frac{a^{2} + b^{2} + c^{2}}{k {a^{2} - 2 a b + 2 c a + 2 a b + b^{2} - 2 b c - 2 c a + 2 b c + c^{2}}}$ $= \frac{a^{2} + b^{2} + c^{2}}{k {a^{2} + b^{2} + c^{2}}} = \frac{1}{k}$ $∵ l h s = r h s = 1 / k$ $∴ P roved$
	Answered by som(math1967) last updated on 24/Dec/23

	$E a c h r a t i o$ $= \frac{3 a - b + 3 b - c + 3 c - a}{x + y + y + z + z + x}$ $= \frac{2 (a + b + c)}{2 (x + y + z)} = \frac{a + b + c}{x + y + z}$ $\Rightarrow \frac{a + b + c}{x + y + z} = \frac{a + b + c - 3 a + b}{x + y + z - x - y}$ $= \frac{a + b + c - 3 b + c}{x + y + z - y - z} = \frac{a + b + c - 3 c + a}{x + y + z - z - x}$ $\Rightarrow \frac{a + b + c}{x + y + z} = \frac{2 b - 2 a + c}{z} = \frac{a - 2 b + 2 c}{x}$ $= \frac{2 a - 2 c + b}{y}$ $\Rightarrow \frac{a + b + c}{x + y + z} = \frac{2 b c - 2 a c + c^{2}}{c z} = \frac{a^{2} - 2 a b + 2 a c}{a x}$ $= \frac{2 a b - 2 b c + b^{2}}{b y}$ $\Rightarrow \frac{a + b + c}{x + y + z}$ $= \frac{2 b c - 2 a c + c^{2} + a^{2} - 2 a b + 2 a c + 2 a b - 2 b c + b^{2}}{c z + a x + b y}$ $\Rightarrow \frac{a + b + c}{x + y + z} = \frac{a^{2} + b^{2} + c^{2}}{a x + b y + c z}$
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