Difference between revisions of "2000 AMC 10 Problems/Problem 10"
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The sides of a triangle with positive area have lengths <math>4</math>, <math>6</math>, and <math>x</math>. The sides of a second triangle with positive area have lengths <math>4</math>, <math>6</math>, and <math>y</math>. What is the smallest positive number that is '''not''' a possible value of <math>|x-y|</math>? | The sides of a triangle with positive area have lengths <math>4</math>, <math>6</math>, and <math>x</math>. The sides of a second triangle with positive area have lengths <math>4</math>, <math>6</math>, and <math>y</math>. What is the smallest positive number that is '''not''' a possible value of <math>|x-y|</math>? | ||
− | <math>\ | + | <math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 10</math> |
==Solution== | ==Solution== | ||
− | + | 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{D}}</math>. | |
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− | <math> | ||
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− | <math>8</math> | ||
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− | <math>\boxed{\ | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=9|num-a=11}} | {{AMC10 box|year=2000|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 08:06, 8 November 2021
Problem
The sides of a triangle with positive area have lengths , , and . The sides of a second triangle with positive area have lengths , , and . What is the smallest positive number that is not a possible value of ?
Solution
Since and are fixed sides, the smallest possible side has to be larger than and the largest possible side has to be smaller than . This gives us the triangle inequality and . can be attained by letting and . However, cannot be attained. Thus, the answer is .
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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