Difference between revisions of "2016 AMC 12A Problems/Problem 23"

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1) <math>0<a<\frac{1}{2}</math>, then  
 
1) <math>0<a<\frac{1}{2}</math>, then  
<math>\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,dbda=\frac{1}{24}</math>
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<math>\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,\text{d}b\,\text{d}a=\frac{1}{24}</math>
  
 
2)<math>\frac{1}{2}<a<1</math>, then  
 
2)<math>\frac{1}{2}<a<1</math>, then  
<math>\int_{\frac{1}{2}}^{1} \left(\int_{0}^{1-a}2b \,db + \int_{1-a}^{a}1+b-a \,db\right)da=\frac{5}{24}</math>
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<math>\int_{\frac{1}{2}}^{1} \left(\int_{0}^{1-a}2b \,\text{d}b + \int_{1-a}^{a}1+b-a \,db\right)\text{d}a=\frac{5}{24}</math>
  
 
<math>a<b</math> is the same. Thus the answer is <math>\frac{1}{2}</math>.
 
<math>a<b</math> is the same. Thus the answer is <math>\frac{1}{2}</math>.
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==See Also==
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{{AMC12 box|year=2016|ab=A|num-b=22|num-a=24}}
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{{MAA Notice}}

Revision as of 12:45, 5 February 2016

Problem

Three numbers in the interval [0,1] are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?

$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$

Solution

Solution 1: Logic

WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$.

Thus the answer is $1/2$.

Solution 2: Calculus

When $a>b$, consider two cases:

1) $0<a<\frac{1}{2}$, then $\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,\text{d}b\,\text{d}a=\frac{1}{24}$

2)$\frac{1}{2}<a<1$, then $\int_{\frac{1}{2}}^{1} \left(\int_{0}^{1-a}2b \,\text{d}b + \int_{1-a}^{a}1+b-a \,db\right)\text{d}a=\frac{5}{24}$

$a<b$ is the same. Thus the answer is $\frac{1}{2}$.


See Also

2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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 12 Problems and Solutions

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