2016 AMC 12A Problems/Problem 23: Difference between revisions

Revision as of 11:45, 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 $\frac{1}{2}$ .

2016 AMC 12A (Problems • Answer Key • Resources)
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

@@ Line 18: / Line 18: @@
 ) <math>0<a<\frac{1}{2}</math>, then
-<math>\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,dbda=\frac{1}{24}</math>
+<math>\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,\text{d}b\,\text{d}a=\frac{1}{24}</math>
 )<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>
+<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>.
+==See Also==
+{{AMC12 box|year=2016|ab=A|num-b=22|num-a=24}}
+{{MAA Notice}}

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