2016 AMC 12A Problems/Problem 23

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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: Super WLOG

WLOG assume $a$ is the largest. Scale the triangle to $1,{b}/{a},{c}/{a}$ or $1,x,y$ with $x,y \in [0,1]$ and equally likely to be any pair of numbers. Then, $x+y>1$, 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(A--C,dashed); fill(A--D--C--cycle,gray);  label("$0$",A,SW); label("$1$",B,S); label("$1$",D,W); label("$y$",(0,.5),W); label("$x$",(.5,0),S); label("$x+y>1$",(2/7,5/7)); [/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 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 $\sqrt(2)$ and the other has 3 sides of length $\sqrt(2)$ and 3 sides of length $1.$ The volume of this region is $\frac 1 2$. Hence our solution is $C.$

Solution 9: Cheap Solution

Pretend that the values of $x$, $y$, and $z$ are integers ranging from $[1,n]$. Test out the probability of the first few values of $n$ (for example, $P(2) = \frac{5}{8}, P(3) = \frac{15}{27}, P(4) = \frac{34}{64}, P(5) = \frac{65}{125}$). Since real numbers contain infinite increments, the answer is the limit as $n$ approaches infinity of $P(n)$, which is easily hypothesized as $\frac{1}{2}$, or $C.$ -solution by fidgetboss_4000 get rect

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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