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>. | 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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| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/203JJbC2Clg?si=XCoDHzGHfoEDhUN7 | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
==See Also== | ==See Also== | ||
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{{AMC10 box|year=2000|num-b=9|num-a=11}} | {{AMC10 box|year=2000|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 23:42, 14 July 2024
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 
.
Video Solution by Daily Dose of Math
https://youtu.be/203JJbC2Clg?si=XCoDHzGHfoEDhUN7
~Thesmartgreekmathdude
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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