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Difference between revisions of "2006 AIME I Problems/Problem 5"

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== Solution ==
 
== Solution ==
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We begin by [[equate | equating]] the two expressions:
 
  
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<math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math>
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For now, assume that face <math>F</math> has a 6, so the opposite face has a 1.  Let <math>A(n)</math> be the probability of rolling a number <math>n</math> on one die and let <math>B(n)</math> be the probability of rolling a number <math>n</math> on the other die. 7 can be obtained by rolling a <math>A(n)=2</math> and <math>B(n)=5</math>, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of <math>\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}</math>, totaling <math>4 \cdot \frac{1}{36} = \frac{1}{9}</math>. Subtracting all these probabilities from <math>\frac{47}{288}</math> leaves <math>\frac{15}{288}=\frac{5}{96}</math> chance of getting a 1 on die <math>A</math> and a 6 on die <math>B</math> or a 6 on die <math>A</math> and a 1 on die <math>B</math>:
  
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Squaring both sides yeilds:
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<math>A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}</math>
  
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<math> 2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006 </math>  
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Since the two dice are identical, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math> so
  
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Since <math>a</math>, <math>b</math>, and <math>c</math> are integers:
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:<math>A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}</math>
 +
:<math>A(1)\cdot A(6)=\frac{5}{192}</math>
  
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1: <math> 2ab\sqrt{6} = 104\sqrt{6} </math>  
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Also, we know that <math>A(2)=A(3)=A(4)=A(5)=\frac{1}{6}</math> and that the total probability must be <math>1</math>, so:
  
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2: <math> 2ac\sqrt{10} = 468\sqrt{10} </math>
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:<math>A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6}</math>
 +
:<math>A(1)+A(6)=\frac{1}{3}</math>
  
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3: <math> 2bc\sqrt{15} = 144\sqrt{15} </math>
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Combining the equations:
  
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4: <math> 2a^2 + 3b^2 + 5c^2 = 2006 </math>  
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:<math>A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}</math>
  
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Solving the first three equations gives:  
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:<math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math>
  
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<math> ab = 52 </math>  
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:<math>A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0</math>
  
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<math> ac = 234 </math>  
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:<math>192 A(6)^2 - 64 A(6) + 5 = 0</math>
  
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<math> bc = 72 </math>  
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:<math>A(6)=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192}</math>
  
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Multiplying these equations gives:  
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:<math>A(6)=\frac{64\pm16}{384}</math>
  
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<math> (abc)^2 = 52 \cdot 234 \cdot 72</math>
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:<math>A(6)=\frac{5}{24}, \frac{1}{8}</math>
  
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<math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math>
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We know that <math>A(6)>\frac{1}{6}</math>, so it can't be <math>\frac{1}{8}</math>.  Therefore, it has to be <math>\frac{5}{24}</math> and the answer is <math>5+24=29</math>.
  
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If it was required to solve for each variable, dividing the product of the three variables by the product of any two variables would yield the third variable. Doing so yields:
+
Note also that the initial assumption that face <math>F</math> was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was.
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<math>a=13</math>
 
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<math>b=4</math>
 
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<math>c=18</math>
 
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Which clearly fits the fourth equation:
 
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<math> 2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006 </math>
 
  
 
== See also ==
 
== See also ==

Revision as of 13:08, 25 September 2007

Problem

The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers. Find $abc$.



Solution

For now, assume that face $F$ has a 6, so the opposite face has a 1. Let $A(n)$ be the probability of rolling a number $n$ on one die and let $B(n)$ be the probability of rolling a number $n$ on the other die. 7 can be obtained by rolling a $A(n)=2$ and $B(n)=5$, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$, totaling $4 \cdot \frac{1}{36} = \frac{1}{9}$. Subtracting all these probabilities from $\frac{47}{288}$ leaves $\frac{15}{288}=\frac{5}{96}$ chance of getting a 1 on die $A$ and a 6 on die $B$ or a 6 on die $A$ and a 1 on die $B$:

$A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}$

Since the two dice are identical, $B(1)=A(1)$ and $B(6)=A(6)$ so

$A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}$
$A(1)\cdot A(6)=\frac{5}{192}$

Also, we know that $A(2)=A(3)=A(4)=A(5)=\frac{1}{6}$ and that the total probability must be $1$, so:

$A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6}$
$A(1)+A(6)=\frac{1}{3}$

Combining the equations:

$A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}$
$\frac{A(6)}{3}-A(6)^2=\frac{5}{192}$
$A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0$
$192 A(6)^2 - 64 A(6) + 5 = 0$
$A(6)=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192}$
$A(6)=\frac{64\pm16}{384}$
$A(6)=\frac{5}{24}, \frac{1}{8}$

We know that $A(6)>\frac{1}{6}$, so it can't be $\frac{1}{8}$. Therefore, it has to be $\frac{5}{24}$ and the answer is $5+24=29$.

Note also that the initial assumption that face $F$ was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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