2016 AMC 12A Problems/Problem 23

Revision as of 15:28, 5 February 2016 by RavenclawPrefect (talk | contribs) (Added a third geometry-based solution.)


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 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 $\frac{1}{2}$.

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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