Difference between revisions of "2000 AMC 10 Problems/Problem 10"

Revision as of 14:36, 19 April 2021

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 $|x-y|$ ?

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

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 $\boxed{\bold{D}}$ .

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 {{AMC10 box|year=2000|num-b=9|num-a=11}}
 {{MAA Notice}}
+[[Category:Introductory Geometry Problems]]

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Difference between revisions of "2000 AMC 10 Problems/Problem 10"

Revision as of 14:36, 19 April 2021

Problem

Solution

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