Difference between revisions of "2000 AMC 10 Problems/Problem 10"
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From the triangle inequality, <math>2<x<10</math> and <math>2<y<10</math>. The smallest positive number not possible is <math>10-2</math>, which is <math>8</math>. | From the triangle inequality, <math>2<x<10</math> and <math>2<y<10</math>. The smallest positive number not possible is <math>10-2</math>, which is <math>8</math>. | ||
<math>\boxed{\text{D}}</math> | <math>\boxed{\text{D}}</math> | ||
| + | |||
| + | 7 is the correct answer, but it is not listen here. | ||
==See Also== | ==See Also== | ||
Revision as of 17:16, 8 February 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 
. The smallest positive number not possible is 
, which is 
.
7 is the correct answer, but it is not listen here.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
