2016 AMC 12A Problem 23

Question

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?

\dfrac{1}{6}

\dfrac{1}{3}

\dfrac{1}{2}

\dfrac{2}{3}

\dfrac{5}{6}

The triples

(x,y,z)

fill a unit cube; failure means one number

\ge

the sum of the other two

Each region such as

z\ge x+y

is a tetrahedron of volume

\frac16

The ordered triples

(x,y,z)

fill the unit cube of volume

1.

They fail to form a triangle exactly when one value is at least the sum of the other two.

The region

z\ge x+y

is a tetrahedron with vertices

(0,0,0),(0,0,1),(0,1,1),(1,0,1)

of volume

\frac16.

The analogous regions

x\ge y+z

and

y\ge x+z

also have volume

\frac16

and have disjoint interiors. So the failure probability is

3\cdot\frac16=\frac12,

and the triangle probability is

1-\frac12=\frac12.

Thus, the correct answer is C.

Problem 23 in other years