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

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

Revision as of 01:12, 3 February 2018

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 $\boxed{\bold{C}}$ since 8 cannot be attained.

See Also

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

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