2008 AMC 12B Problem 7

Attempt Problem 7 of the 2008 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

For real numbers

a

and

b,

define

a \$ b = (a - b)^2.

What is

(x - y)^2 \$ (y - x)^2?

0

x^2 + y^2

2x^2

2y^2

4xy

Note that

(y - x)^2 = (x - y)^2

So the expression is

t \$ t

with

t = (x - y)^2,

and

t \$ t = (t - t)^2

Since

(y - x)^2 = (x - y)^2,

both inputs to the operation are the same value

t = (x - y)^2.

Therefore

(x - y)^2 \$ (y - x)^2

= t \$ t

= (t - t)^2 = 0.

Thus, the correct answer is A.

Problem 7 in other years