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

Revision as of 14:28, 5 February 2016

Problem

Three numbers in the interval $\left[0,1\right]$ 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 $\frac{1}{2}$ .

Solution 3: Geometry

The probability of this occurring is the volume of the corresponding region within a $1 \times 1 \times 1$ cube, where each point $(x,y,z)$ corresponds to a choice of values for each of $x, y,$ and $z$ . The region where, WLOG, side $z$ is too long, $z\geq x+y$ , is a pyramid with a base of area $\frac{1}{2}$ and height $1$ , so its volume is $\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}$ . Accounting for the corresponding cases in $x$ and $y$ multiplies our answer by $3$ , so we have excluded a total volume of $\frac{1}{2}$ from the space of possible probabilities. Subtracting this from $1$ leaves us with a final answer of $\frac{1}{2}$ .

@@ Line 25: / Line 25: @@
 <math>a<b</math> is the same. Thus the answer is <math>\frac{1}{2}</math>.
+===Solution 3: Geometry===
+The probability of this occurring is the volume of the corresponding region within a <math>1 \times 1 \times 1</math> cube, where each point <math>(x,y,z)</math> corresponds to a choice of values for each of <math>x, y,</math> and <math>z</math>. The region where, WLOG, side <math>z</math> is too long, <math>z\geq x+y</math>, is a pyramid with a base of area <math>\frac{1}{2}</math> and height <math>1</math>, so its volume is <math>\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}</math>. Accounting for the corresponding cases in <math>x</math> and <math>y</math> multiplies our answer by <math>3</math>, so we have excluded a total volume of <math>\frac{1}{2}</math> from the space of possible probabilities. Subtracting this from <math>1</math> leaves us with a final answer of <math>\frac{1}{2}</math>.
 ==See Also==
 {{AMC12 box|year=2016|ab=A|num-b=22|num-a=24}}
 {{MAA Notice}}

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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