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Difference between revisions of "2000 AMC 10 Problems/Problem 10"

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==Problem==
 
==Problem==
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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>? 
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<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 10</math>
  
 
==Solution==
 
==Solution==
  
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The largest possible value for <math>x</math> is <math>9</math>. The smallest is <math>3</math>.
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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>9-3=6</math>.
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==Video Solution by Daily Dose of Math==
  
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<math>8</math> is the smallest that cannot be made (of the choices listed)
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https://youtu.be/203JJbC2Clg?si=XCoDHzGHfoEDhUN7
  
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<math>\boxed{\text{D}}</math>
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~Thesmartgreekmathdude
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2000|num-b=9|num-a=11}}
 
{{AMC10 box|year=2000|num-b=9|num-a=11}}
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{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 23:42, 14 July 2024

Problem

The sides of a triangle with positive area have lengths $4$, $6$, and $x$. The sides of a second triangle with positive area have lengths $4$, $6$, and $y$. What is the smallest positive number that is not a possible value of $|x-y|$?

$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 10$

Solution

Since $6$ and $4$ are fixed sides, the smallest possible side has to be larger than $6-4=2$ and the largest possible side has to be smaller than $6+4=10$. This gives us the triangle inequality $2<x<10$ and $2<y<10$. $7$ can be attained by letting $x=9.1$ and $y=2.1$. However, $8=10-2$ cannot be attained. Thus, the answer is $\boxed{\bold{D}}$.

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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