What is the relationship

Algebra Level pending

{ A = 2015 x 2 + 2015 y 2 + 6 z 2 B = 2 ( 2012 x y + 3 y z + 3 z x ) \begin{cases} A = 2015x^2 + 2015y^2 + 6z^2 \\ B = 2(2012xy + 3yz + 3zx) \end{cases}

If x x , y y and z z are distinct numbers, which of the following relationships holds between A A and B B ?


Based on NMTC Sub-Junior Question, Final Level.
B > A B > A Relationship cannot be determined A = B A = B A > B A > B

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Star Light
Mar 15, 2017

Way to approach, rather than method, which can be very short.

First look:
Relationship cannot be determined looks appealing.

Try substituting 0, 1, and -1:
A > B
Rules out A = B and A < B

To eliminate relationship cannot be determined, we need to solve algebraically:
2 ( 2012 x y + 3 y z + 3 z x ) = 4024 x y + 6 y z + 6 z x 2(2012xy + 3yz + 3zx) = 4024xy + 6yz + 6zx

The connection between x^2, y^2 and xy is given by the formula for (x + y)^2:
2015 x 2 + 2015 y 2 = 2015 ( x 2 + y 2 ) = 2015 ( x 2 + y 2 2 x y + 2 x y ) = 2015 [ ( x y ) 2 + 2 x y ] = 2015 ( x y ) 2 + 4030 x y 2015x^2 + 2015y^2 \\ = 2015(x^2 + y^2) \\ = 2015(x^2 + y^2 - 2xy + 2xy) \\ = 2015[(x - y)^2 + 2xy] \\ = 2015(x - y)^2 + 4030xy

First term is positive:
2012 ( x y ) 2 + 4024 x y > 4024 x y 2012(x - y)^2 + 4024xy > 4024xy

Similar process for remaining two terms
This makes A > B correct.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...