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2000 AMC 12 Problem 11

Below is the professionally curated solution for Problem 11 of the 2000 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 12 solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:algebraic manipulationsubstitution

Difficulty rating: 1530

11.

Two non-zero real numbers, aa and b,b, satisfy ab=a−b.ab = a - b. Find a possible value of ab+ba−ab.\frac{a}{b} + \frac{b}{a} - ab.

−2-2

−12-\dfrac{1}{2}

13\dfrac{1}{3}

12\dfrac{1}{2}

22

Solution:

Combining over a common denominator, ab+ba−ab=a2+b2−(ab)2ab. \frac{a}{b} + \frac{b}{a} - ab = \frac{a^2 + b^2 - (ab)^2}{ab}.

Replacing abab with a−ba - b in the numerator, a2+b2−(a−b)2=a2+b2−(a2−2ab+b2)=2ab. a^2 + b^2 - (a - b)^2 = a^2 + b^2 - (a^2 - 2ab + b^2) = 2ab.

Therefore the expression equals 2abab=2.\dfrac{2ab}{ab} = 2.

Thus, the correct answer is E.

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