2016 AMC 12A Problems/Problem 23

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: Conditional Probability

Let the largest of the three numbers drawn be $a$ . Then the other two numbers are drawn uniformly and independently from the interval $[0,a]$ . The probability that their sum is at least $a$ 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 $\frac{1}{2}$ .

Solution 4: More Calculus

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 $z$ . We take a horizontal cross section of the cube, essentially picking a value for z. The area where the triangle inequality will not hold is when $x + y < z$ , which has area $\frac{z^2}{2}$ or when $x+z<y$ or $y+z<x$ , which have an area of $\frac{(1-z)^2}{2}+\frac{(1-z)^2}{2} = (1-z)^2.$ Integrating this expression from 0 to 1 in the form

$\int_0^1 \frac{z^2}{2} + (1-z)^2 dz = \frac{z^3}{2} - z^2 + z \biggr |_0^1 = \frac{1}{2} -1 + 1 = \frac{1}{2}$

Solution 5: Geometry in 2-D

WLOG assume that $z$ is the largest number and hence the largest side. Then $x,y \leq z$ . We can set up a square that is $z$ by $z$ in the $xy$ plane. We are wanting all the points within this square that satisfy $x+y > z$ . This happens to be a line dividing the square into 2 equal regions. Thus the answer is $\frac{1}{2}$ .

[][] diagram for this problem goes here (z by z square)

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

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