Printable

2008 AMC 10B Problem 5

Problem only Back to the problem

2008 AMC 10B Problem 5AMC 10 Step by Step

Name: Date: September 13, 2026

Functions · difficulty 1 of 5 · about 1 min · position P1-10

For real numbers $a$ and $b$ , define $a \$b$ $=(a-b)^2$ . What is $(x-y)^2\$(y-x)^2$ ?

A)
$0$
B)
$x^2+y^2$
C)
$2x^2$
D)
$2y^2$
E)
$4xy$

Answer

Key insight

Since (x-y)^2 and (y-x)^2 are the same number, the operation is applied to two equal inputs and gives (0)^2 = 0.

Solution

Because $y - x = -(x - y)$ and squaring erases the sign, $(y-x)^2 = (x-y)^2$ . So the two inputs to the operation are equal; call them both $t$ . Then
$t \$t = (t - t)^2 = 0^2 = 0.$

The answer is $\boxed{\textbf{(A)}\ 0}$ .

Why this works

A defined operation is just a formula; substitute and simplify. The only content here is noticing that $(x-y)^2$ and $(y-x)^2$ are identical, which makes the difference inside the outer square vanish before any expansion. Look for equal or opposite inputs before doing algebra.

Alternative approach

Plug in numbers: with $x = 3$ , $y = 1$ , both inputs equal $4$ , and $4 \$4 = 0$ . Choices (B) through (E) give $10, 18, 2, 12$ , so only (A) fits.

The trap

Expanding both squares and making a sign error, or assuming (y-x)^2 equals -(x-y)^2.

Common mistakes

Expanding both squares and making a sign error, or assuming (y-x)^2 equals -(x-y)^2.
Squaring the difference of the squares fully by hand and losing track of terms, when the difference is $0$ from the start.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta