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

Revision as of 15:54, 4 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?

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

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 $1/2$ .[/quote]

Solution 2: Calculus

When $a>b$ , consider two cases:

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

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

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

@@ Line 1: / Line 1: @@
+==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?
+<math>\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56</math>[/quote]
+==Solution==
+===Solution 1: Logic===
+WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is <math>1/2</math>.
+Thus the answer is <math>1/2</math>.[/quote]
+===Solution 2: Calculus===
+When <math>a>b</math>, consider two cases:
+) <math>0<a<\frac{1}{2}</math>, then
+<math>\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,dbda=\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>a<b</math> is the same. Thus the answer is <math>\frac{1}{2}</math>.

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