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: Super WLOG

WLOG assume $a$ is the largest. Then, $b+c>a$ , meaning the solution is $\boxed{\textbf{(C)}\;1/2}$ , as shown in the graph below. $[asy] pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair E = (0,0); draw(A--B--C--D--cycle); draw(B--D,dashed); fill(B--D--C--cycle,gray); label("$0$",A,SW); label("$a$",B,S); label("$a$",D,W); label("$y$",(0,.5),W); label("$x$",(.5,0),S); label("$x+y>a$",(5/7,5/7)); [/asy]$

Solution 2: Conditional Probability

WLOG, let the largest of the three numbers drawn be $a>0$ . Then the other two numbers are drawn uniformly and independently from the interval $[0,a]$ . The probability that their sum is greater than $a$ is $\boxed{\textbf{(C)}\;1/2.}$

Solution 3: 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 4: 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 5: 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

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

Solution 5.1 (better explanation)

This problem is going to require some geometric probability, so let's dive right in. Take three integers $x$ , $y$ , and $z$ . Then for the triangle inequality to hold, the following $3$ inequalities must be true

$x + y > z$ $y + z > x$ $x + z > y$

Now, it would be really easy if these equations only had two variables instead of $3$ , because then we could graph it in a $2$ -dimensional plane instead of a $3$ -dimensional cube. So, we assume $z$ is a constant. We will deal with it later.

Now, since we are graphing, we should probably write these equations in terms of $y$ so they are in slope-intercept form and are easier to graph.

$y > -x + z$ $y > x - z$ $y < x + z$

Now, note that all solutions $(x,y)$ are in a $1$ $x$ $1$ square in the $xy$ -plane because $x$ , $y \in (0, 1]$ .

I recommend drawing the following figure to get an idea of what is going on. The first line is a line with a negative slope that cuts off a $45-45-90$ triangle with side length $z$ of the bottom left corner of the square. The second line is a line with a positive slope that cuts off a $45-45-90$ triangle with side length $1-z$ off the top left corner of the square. The third line also has a positive slope and cuts a $45-45-90$ triangle with side length $1 - z$ off the bottom right corner of the square.

Note: All triangles are $45-45-90$ because the lines have slopes of $1$ or $-1$ .

Using the $>$ and $<$ signs in the lines, we see that the area that satisfies all three inequalities is the area not enclosed in the three triangles. So, our plan of attack will be to

Find the area of the triangles -> Subtract that from the area of the square -> Use probability to get the answer.

Except, now, we have one problem. $z$ is still a variable. But, we want $z$ to be a constant. Well, what if we just took the area over every possible value of $z$ ? Well, that would be a bit hard, if not impossible to do by hand, but there is a handy math tool that will let us do that: the integral!

To find the area of the triangles, our plan of attack will be

Find the area in terms of $z$ -> take the integral from $0$ to $1$ of the expression for the area (this will cover every possible value of $z$

The area of the triangles is

$\frac{z^2}{2} + (1-z)^2 = \frac{1}{2}\left(-3x^2 + 4x\right)$ .

The integral from $0$ to $1$ is

\[\frac{1}{2}\int_0^1\left(-3x^2 + 4x\right)dx = \frac{1}{2}(-x^3 + 2x^2)\left|_0^1\right{} = (\frac{1}{2})(-1 + 2) = \frac{1}{2}\] (Error compiling LaTeX. Unknown error_msg)

The total area of all the possible unit squares is quite obviously

$\int_0^11dx = 1$

Thus, the area not enclosed by the triangles is $1 - \dfrac{1}{2} = \dfrac{1}{2}$ , and the total area of the square is $1$ . Thus, the desired probability is

$\frac{\frac{1}{2}}{1} = \frac{1}{2}$

or $\mathbb{C}$ .

~Extremelysupercooldude

Solution 6: 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)

Solution 7: More WLOG, Complementary Probability

The triangle inequality simplifies to considering only one case: $\text{the smallest side+ the second smallest side} > \text{the largest side}$ . Consider the complement (the same statement, except with a less than or equal to). Assume (WLOG) $a$ is the largest, so on average $a=1/2$ (now equal to becomes a degenerate case with probability $0$ , so we no longer need to consider it). We now want $b+c<1/2$ , so imagine choosing $b+c$ at once rather than independently. But we know that $b+c$ is between $0$ and $2$ . The complement is thus: $(1/2-0)/2=1/4$ . But keep in mind that we choose each $b$ and $c$ randomly and independently, so if there are $k$ ways to choose $b+c$ together, there are $2k$ ways to choose them separately, and therefore the complement actually doubles to match each case (a good example of this is to restrict b and c to integers such that if $b+c=3$ , then we only count this once, but in reality: we have two cases $1+2$ , and $2+1$ ; similar reasoning also generalizes to non-integral values). The complement is then actually $2(1/4)=1/2$ . Therefore, our desired probability is given by $1-\text{complement}=1/2, C$

Solution 8: 3D geometry

We can draw a 3D space where each coordinate is in the range [0,1]. Drawing the lines $x+y>z, x+z>y,$ and $y+z>x,$ We have a 3D space that consists of two tetrahedrons. One is a regular tetrahedron with side length $\sqrt2$ and the other has 3 sides of length $\sqrt2$ and 3 sides of length $1.$ The volume of this region is $\frac 1 2$ . Hence our solution is $C.$

Video Solution by Punxsutawney Phil

https://youtu.be/DDRRP9VLYO8

Video Solution by The Art of Problem Solving

https://www.youtube.com/watch?v=FqRsTNB89ps&list=PLyhPcpM8aMvI7N78mYZyatqveRU30iNcf&index=3

- AMBRIGGS

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