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Let x = 2012(a - b), y = 2012(b - c) and z = 2012(c - a) where a,b,c are real numbers, and assume xy + yz + zx  ≠ 0. Compute {x^2 + y^2 + z^2}/{xy + yz + zx}

waffles  Jan 25, 2018
 #1
avatar+92808 
+1

x^2  =  2012^2(a - b)^2 

y^2 =  2012^2 (b - c)^2

z^2 =  2012^2 (c - a)^2

xy = 2012^2(a - b) (b - c)

yz  = 2012^2 (b -c) (c - a)

xz =  2012^2 (a - b)(c - a)

 

So

 

 {x^2 + y^2 + z^2}/{xy + yz + zx}  =

 

[  2012^2 [  ( a- b)^2  + (b - c)^2 + ( c - a)^2 ] ]/ [ 2012^2 [ (a-b)(b -c) +(b -c)(c - a) +(a -b) (c-a)] ]

 

[ ( a - b)^2 + (b - c)^2  + (c - a)^2) ]  /  [  (b - c) ( a - b + c - a)  + (a - b)(c - a) ]

 

[ ( a - b)^2 + (b  - c)^2 + (c - a)^2 ]  /  [  (b -c) (c - b) + ( a - b)(c - a) ] 

 

[  a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ac + a^2]  / [  - ( b - c)^2  + (a - b)(c - a)]

 

[  2 [ a^2 + b^2 + c^2]  - 2 [ ab + bc + ac ] ]/ [ -b^2 + 2bc - c^2   + ac - bc - a^2 + ab ]

 

[ 2 [ a^2 + b^2 + c^2] - 2 [ ab + bc + ac] ] / [ - [ a^2 + b^2 + c^2]  + [ ab + bc + ac] ]

 

[ 2  [ a^2 + b^2 + c^2 - ab - bc - ac ] ]  /  [ - 1 [ a^2 + b^2 + c^2 - ab + bc + ac ] ]

 

=

 

2 / -1  =

 

-2

 

 

cool cool cool

CPhill  Jan 25, 2018

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