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

Revision as of 08:28, 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 $\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,dbda=\frac{1}{24}$

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 3: / Line 3: @@
 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]
+<math>\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56</math>
 ==Solution==

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