2000 AMC 10 Problems/Problem 10

Problem

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 $|x-y|$ ?

$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 10$

Solution

Since $6$ and $4$ are fixed sides, the smallest possible side has to be larger than $6-4=2$ and the largest possible side has to be smaller than $6+4=10$ . This gives us the triangle inequality $2<x<10$ and $2<y<10$ . $7$ can be attained by letting $x=9.1$ and $y=2.1$ . However, $8=10-2$ cannot be attained. Thus, the answer is $\boxed{\bold{D}}$ .

Video Solution by Daily Dose of Math

https://youtu.be/203JJbC2Clg?si=XCoDHzGHfoEDhUN7

~Thesmartgreekmathdude

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 10 Problems and Solutions

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2000 AMC 10 Problems/Problem 10

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Video Solution by Daily Dose of Math

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