+0  
 
0
123
1
avatar+638 

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
Sort: 

1+0 Answers

 #1
avatar+85757 
+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

32 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details