2005 AMC 10A Problem 2

P1-10~1 minPrint

Functions Set up the equation/formula and compute; no special trick needed

For each pair of real numbers $a \neq b$ , define the operation $\star$ as

$(a \star b) = \frac{a+b}{a-b}.$

What is the value of $((1 \star 2) \star 3)$ ?

Show solution without answering

Key insight

Do the inner operation first: 1 star 2 is -3. Then -3 star 3 has a top of 0 and a bottom of -6, so the answer is 0.

Solution

The new operation is just a rule: add the two numbers on top, subtract them on the bottom. Parentheses tell you which part to do first, so start inside.

1. Inner part. Put $a = 1$ and $b = 2$ into the rule:
$1 \star 2 = \frac{1 + 2}{1 - 2} = \frac{3}{-1} = -3.$
2. Outer part. Now the left number is $-3$ and the right number is $3$ :
$(-3) \star 3 = \frac{-3 + 3}{-3 - 3} = \frac{0}{-6} = 0.$
3. Is anything undefined? A fraction breaks only when its bottom is $0$ . The bottoms were $-1$ and $-6$ , so both steps are fine. A top of $0$ is allowed; it just makes the fraction equal $0$ .

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

Why this works

A made-up symbol like $\star$ is a recipe: plug in the numbers and follow the steps, innermost parentheses first. When a choice says "not defined," ask whether any denominator is actually zero. Here the rule only uses pairs with $a \neq b$ , so the bottom $a - b$ is never zero.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Work from the inside out. First,
$1 \star 2 = \frac{1 + 2}{1 - 2} = \frac{3}{-1} = -3.$
Next, substitute this result as the left input:
$(-3) \star 3 = \frac{-3 + 3}{-3 - 3} = \frac{0}{-6} = 0.$
The denominator is nonzero at every step, so the value is defined and equals $0$ .

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

Review in the AoPS books

Introduction to Algebra, Chapter 16: Functions
Introduction to Algebra, Chapter 17: Graphing Functions (§17.2)
Intermediate Algebra, Chapter 19: Functional Equations (§19.1, §19.2)