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2000 AMC 10 Problem 15

2000 AMC 10 Problem 15AMC 10 Step by Step

Name: Date: September 13, 2026

Algebraic Manipulation · difficulty 3 of 5 · about 4 min · position P11-15

Two non-zero real numbers, $a$ and $b,$ satisfy $ab = a - b$ . Which of the following is a possible value of $\frac {a}{b} + \frac {b}{a} - ab$ ?

A)
$- 2$
B)
$\frac { -1 }{2}$
C)
$\frac {1}{3}$
D)
$\frac {1}{2}$
E)
$2$

Answer

Key insight

Combine a/b + b/a into (a^2+b^2)/(ab), then use a^2 + b^2 = (a-b)^2 + 2ab and the given ab = a - b to make everything cancel.

Solution

Write the two fractions over a common denominator:
$\frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}.$
Since $a^2 + b^2 = (a-b)^2 + 2ab$ and we are given $a - b = ab$ , the numerator equals $(ab)^2 + 2ab$ . Therefore
$\frac{a^2+b^2}{ab} = \frac{(ab)^2 + 2ab}{ab} = ab + 2.$
Subtracting $ab$ leaves $2$ . So the expression equals $2$ for every pair satisfying the condition.

The answer is $\boxed{\textbf{(E)}\ 2}$ .

Why this works

Symmetric expressions like $\frac ab + \frac ba$ are functions of $a+b$ , $a-b$ and $ab$ . Rewriting them in those terms lets a single given relation collapse the whole expression. The phrase "a possible value" hints that the value is actually forced.

Alternative approach

Pick a concrete pair: with $a = 1$ , the condition $b = 1 - b$ gives $b = \tfrac12$ . Then $\frac{1}{1/2} + \frac{1/2}{1} - \tfrac12 = 2 + \tfrac12 - \tfrac12 = 2$ . One valid example settles a "which is possible" question.

The trap

Trying to solve for a and b individually; the expression has the same value for every valid pair, so the identity is the whole problem.

Common mistakes

Trying to solve for a and b individually; the expression has the same value for every valid pair, so the identity is the whole problem.
Sign error when expanding $(a-b)^2$ , giving $ab - 2$ and choice (A).

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Test small/specific values or special cases to find or verify the answer