Difference between revisions of "2016 AMC 12A Problems/Problem 23"

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===Solution 1: Super WLOG===
 
===Solution 1: Super WLOG===
  
This lemma will be used for the solution: "If a triangle with side lengths <math>a,b,c</math> exists so does a triangle with side lengths <math>ak,bk,ck</math> and vice versa". WLOG <math>a\geq b\geq c</math> Scale the triangle so we have side lengths of <math>1,{b}/{a},{c}/{a}</math> Then it is easily seen the solution is <math>\boxed{\textbf{(C)}\;1/2}</math>
+
This lemma will be used for the solution: "If a triangle with side lengths <math>a,b,c</math> exists so does a triangle with side lengths <math>ak,bk,ck</math> and vice versa". WLOG <math>a\geq b\geq c</math>, scale the triangle so we have side lengths of <math>1,{b}/{a},{c}/{a}</math> Then it is easily seen the solution is <math>\boxed{\textbf{(C)}\;1/2}</math>
  
 
===Solution 2: Conditional Probability===
 
===Solution 2: Conditional Probability===

Revision as of 16:20, 21 March 2016

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

This lemma will be used for the solution: "If a triangle with side lengths $a,b,c$ exists so does a triangle with side lengths $ak,bk,ck$ and vice versa". WLOG $a\geq b\geq c$, scale the triangle so we have side lengths of $1,{b}/{a},{c}/{a}$ Then it is easily seen the solution is $\boxed{\textbf{(C)}\;1/2}$

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

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

Solution 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)

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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