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