Difference between revisions of "2000 AMC 10 Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | From the triangle inequality, <math>2<x<10</math> and <math>2<y<10</math>. | + | From the triangle inequality, <math>2<x<10</math> and <math>2<y<10</math>. <math>7</math> can be attain 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>. |
| − | <math>\boxed{\ | ||
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==See Also== | ==See Also== | ||
Revision as of 21:46, 3 March 2017
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
From the triangle inequality, 
and 
. 
can be attain 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
