2000 AMC 10 Problem 10

Below is the professionally curated solution for Problem 10 of the 2000 AMC 10, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 10 solutions, or check the answer key.

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Concepts:triangle inequality bounding to limit cases

Difficulty rating: 1370

10.

The sides of a triangle with positive area have lengths $4,$ $6,$ and $x.$ The sides of a second triangle with positive area have lengths $4,$ $6,$ and $y.$ What is the smallest positive number that is not a possible value of $|x - y|?$

$2$

$4$

$6$

$8$

$10$

Solution:

By the triangle inequality, each of $x$ and $y$ can be any number strictly between $6 - 4 = 2$ and $6 + 4 = 10.$

Then $|x - y|$ can take any value with $0 \le |x - y| \lt 8.$

The smallest positive number not attainable is $10 - 2 = 8.$

Thus, the correct answer is D.

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