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**+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

CPhill Jan 25, 2018