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

(→‎Solution 1: Conditional Probability)
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==Solution==
 
==Solution==
  
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===Solution 1: Conditional Probability===
+
===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 and ck and vice versa". WLOG, Assume <cmath>a≥b≥c</cmath>. Scale the triangle so we have side lengths of <cmath>1,{b\a},{c\a}</cmath>. Then it is easilt seen the solution is <math>\boxed{\textbf{(C)}\;1/2.}</math>
 +
 
 +
===Solution 2: Conditional Probability===
  
 
WLOG, let the largest of the three numbers drawn be <math>a>0</math>.  Then the other two numbers are drawn uniformly and independently from the interval <math>[0,a]</math>.  The probability that their sum is greater than <math>a</math> is <math>\boxed{\textbf{(C)}\;1/2.}</math>
 
WLOG, let the largest of the three numbers drawn be <math>a>0</math>.  Then the other two numbers are drawn uniformly and independently from the interval <math>[0,a]</math>.  The probability that their sum is greater than <math>a</math> is <math>\boxed{\textbf{(C)}\;1/2.}</math>
  
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===Solution 2: Calculus===
+
===Solution 3: Calculus===
  
 
When <math>a>b</math>, consider two cases:
 
When <math>a>b</math>, consider two cases:
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<math>a<b</math> is the same. Thus the answer is <math>\frac{1}{2}</math>.
 
<math>a<b</math> is the same. Thus the answer is <math>\frac{1}{2}</math>.
  
−
===Solution 3: Geometry===
+
===Solution 4: Geometry===
  
 
The probability of this occurring is the volume of the corresponding region within a <math>1 \times 1 \times 1</math> cube, where each point <math>(x,y,z)</math> corresponds to a choice of values for each of <math>x, y,</math> and <math>z</math>. The region where, WLOG, side <math>z</math> is too long, <math>z\geq x+y</math>, is a pyramid with a base of area <math>\frac{1}{2}</math> and height <math>1</math>, so its volume is <math>\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}</math>. Accounting for the corresponding cases in <math>x</math> and <math>y</math> multiplies our answer by <math>3</math>, so we have excluded a total volume of <math>\frac{1}{2}</math> from the space of possible probabilities. Subtracting this from <math>1</math> leaves us with a final answer of <math>\frac{1}{2}</math>.
 
The probability of this occurring is the volume of the corresponding region within a <math>1 \times 1 \times 1</math> cube, where each point <math>(x,y,z)</math> corresponds to a choice of values for each of <math>x, y,</math> and <math>z</math>. The region where, WLOG, side <math>z</math> is too long, <math>z\geq x+y</math>, is a pyramid with a base of area <math>\frac{1}{2}</math> and height <math>1</math>, so its volume is <math>\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}</math>. Accounting for the corresponding cases in <math>x</math> and <math>y</math> multiplies our answer by <math>3</math>, so we have excluded a total volume of <math>\frac{1}{2}</math> from the space of possible probabilities. Subtracting this from <math>1</math> leaves us with a final answer of <math>\frac{1}{2}</math>.
  
−
=== Solution 4: More Calculus ===
+
=== Solution 5: More Calculus ===
  
 
The probability of this occurring is the volume of the corresponding region within a <math>1 \times 1 \times 1</math> cube, where each point <math>(x,y,z)</math> corresponds to a choice of values for each of <math>x, y,</math> and <math>z</math>.  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 <math>x + y < z</math>, which has area <math>\frac{z^2}{2}</math> or when <math>x+z<y</math> or <math>y+z<x</math>, which have an area of <math>\frac{(1-z)^2}{2}+\frac{(1-z)^2}{2} = (1-z)^2.</math> Integrating this expression from 0 to 1 in the form
 
The probability of this occurring is the volume of the corresponding region within a <math>1 \times 1 \times 1</math> cube, where each point <math>(x,y,z)</math> corresponds to a choice of values for each of <math>x, y,</math> and <math>z</math>.  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 <math>x + y < z</math>, which has area <math>\frac{z^2}{2}</math> or when <math>x+z<y</math> or <math>y+z<x</math>, which have an area of <math>\frac{(1-z)^2}{2}+\frac{(1-z)^2}{2} = (1-z)^2.</math> Integrating this expression from 0 to 1 in the form
Line 33: Line 37:
 
<math>\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}</math>
 
<math>\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}</math>
  
−
=== Solution 5: Geometry in 2-D ===
+
=== Solution 6: Geometry in 2-D ===
 
WLOG assume that <math>z</math> is the largest number and hence the largest side. Then <math>x,y \leq z</math>. We can set up a square that is <math>z</math> by <math>z</math> in the <math>xy</math> plane. We are wanting all the points within this square that satisfy <math>x+y > z</math>. This happens to be a line dividing the square into 2 equal regions. Thus the answer is <math>\frac{1}{2}</math>.
 
WLOG assume that <math>z</math> is the largest number and hence the largest side. Then <math>x,y \leq z</math>. We can set up a square that is <math>z</math> by <math>z</math> in the <math>xy</math> plane. We are wanting all the points within this square that satisfy <math>x+y > z</math>. This happens to be a line dividing the square into 2 equal regions. Thus the answer is <math>\frac{1}{2}</math>.
  

Revision as of 15:13, 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 and ck and vice versa". WLOG, Assume 

\[a≥b≥c\] (Error compiling LaTeX. Unknown error_msg)

. Scale the triangle so we have side lengths of 

\[1,{b\a},{c\a}\] (Error compiling LaTeX. Unknown error_msg)

. Then it is easilt 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 (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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