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

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== Problem ==
 
== Problem ==
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The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </math> where <math> a, b, </math> and <math> c </math> are [[positive]] [[integer]]s. Find <math>abc</math>.
  
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When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face <math> F </math> is greater than 1/6, the probability of obtaining the face opposite is less than 1/6, the probability of obtaining any one of the other four faces is 1/6, and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is <math>47/288</math>. Given that the [[probability]] of obtaining face <math> F </math>  is <math> m/n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] positive integers, find <math> m+n. </math>
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== Solution 1 ==
 +
We begin by [[equate | equating]] the two expressions:
  
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== Solution ==
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<cmath> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</cmath>
  
−
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>:
+
Squaring both sides yields:  
  
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<math>A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}</math>
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<cmath> 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 </cmath>  
  
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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, we can match coefficients:
  
−
:<math>A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}</math>
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<cmath>  
−
:<math>A(1)\cdot A(6)=\frac{5}{192}</math>
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\begin{align*}
 +
2ab\sqrt{6} &= 104\sqrt{6} \\
 +
2ac\sqrt{10} &=468\sqrt{10} \\
 +
2bc\sqrt{15} &=144\sqrt{15}\\
 +
2a^2 + 3b^2 + 5c^2 &=2006
 +
\end{align*}
 +
</cmath>  
  
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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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Solving the first three equations gives:
 +
<cmath>\begin{eqnarray*}ab &=& 52\\
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ac &=& 234\\
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bc &=& 72 \end{eqnarray*}</cmath>  
  
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:<math>A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6}</math>
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Multiplying these equations gives <math> (abc)^2 = 52 \cdot 234 \cdot 72 = 2^63^413^2 \Longrightarrow abc = \boxed{936}</math>.
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:<math>A(1)+A(6)=\frac{1}{3}</math>
 
  
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Combining the equations:
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<!--
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Since this is the AIME and you do not have a calculator solving <math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math> might prove difficult.
 +
So instead use the three equations given above.
  
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:<math>A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}</math>
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<math> ab = 52 </math>  
  
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:<math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math>
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<math> ac = 234 </math>  
  
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:<math>A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0</math>
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<math> bc = 72 </math>  
  
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:<math>192 A(6)^2 - 64 A(6) + 5 = 0</math>
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Thus <math> a = 52/b = 234/c </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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<math> 52c = 234b </math>
  
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:<math>A(6)=\frac{64\pm16}{384}</math>
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<math> c = 234/52b </math>
  
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:<math>A(6)=\frac{5}{24}, \frac{1}{8}</math>
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<math> c = 9/2b </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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Plugging into last equation leads to:
  
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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> 9/2b^2 = 72 </math>
 +
 
 +
<math> b = 4 </math>
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 +
Plugging into others you get
 +
 
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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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 +
Much easier than taking crazy square roots without a calculator!
 +
 
 +
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:
 +
 
 +
<math>a=13</math>
 +
 
 +
<math>b=4</math>
 +
 
 +
<math>c=18</math>
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 +
Which clearly fits the fourth equation:
 +
<math> 2 \cdot 13^2 + 3 \cdot 4^2 + 5 \cdot 18^2 = 2006 </math>
 +
 
 +
<math>abc=\boxed{936}</math>
 +
-->
 +
 
 +
== Solution 2 ==
 +
We realize that the quantity under the largest radical is a perfect square and attempt to rewrite the radicand as a square. Start by setting <math>x=\sqrt{2}</math>, <math>y=\sqrt{3}</math>, and <math>z=\sqrt{5}</math>.  Since
 +
 
 +
<cmath>(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)</cmath>
 +
 
 +
we attempt to rewrite the radicand in this form:
 +
 
 +
<cmath>2006+2(52xy+234xz+72yz)</cmath>
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 +
Factoring, we see that <math>52=13\cdot4</math>, <math>234=13\cdot18</math>, and <math>72=4\cdot18</math>. Setting <math>p=13</math>, <math>q=4</math>, and <math>r=18</math>, we see that
 +
 
 +
<cmath>2006=13^2x^2+4^2y^2+18^2z^2=169\cdot2+16\cdot3+324\cdot5</cmath>
 +
 
 +
so our numbers check. Thus <math>104\sqrt{2}+468\sqrt{3}+144\sqrt{5}+2006=(13\sqrt{2}+4\sqrt{3}+18\sqrt{5})^2</math>. Square rooting gives us <math>13\sqrt{2}+4\sqrt{3}+18\sqrt{5}</math> and our answer is <math>13\cdot4\cdot18=\boxed{936}</math>
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 17:24, 10 March 2015

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 1

We begin by equating the two expressions:

\[a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}\]

Squaring both sides yields:

\[2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006\]

Since $a$, $b$, and $c$ are integers, we can match coefficients:

\begin{align*} 2ab\sqrt{6} &= 104\sqrt{6} \\ 2ac\sqrt{10} &=468\sqrt{10} \\ 2bc\sqrt{15} &=144\sqrt{15}\\ 2a^2 + 3b^2 + 5c^2 &=2006 \end{align*}

Solving the first three equations gives: \begin{eqnarray*}ab &=& 52\\ ac &=& 234\\ bc &=& 72 \end{eqnarray*}

Multiplying these equations gives $(abc)^2 = 52 \cdot 234 \cdot 72 = 2^63^413^2 \Longrightarrow abc = \boxed{936}$.


Solution 2

We realize that the quantity under the largest radical is a perfect square and attempt to rewrite the radicand as a square. Start by setting $x=\sqrt{2}$, $y=\sqrt{3}$, and $z=\sqrt{5}$. Since

\[(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)\]

we attempt to rewrite the radicand in this form:

\[2006+2(52xy+234xz+72yz)\]

Factoring, we see that $52=13\cdot4$, $234=13\cdot18$, and $72=4\cdot18$. Setting $p=13$, $q=4$, and $r=18$, we see that

\[2006=13^2x^2+4^2y^2+18^2z^2=169\cdot2+16\cdot3+324\cdot5\]

so our numbers check. Thus $104\sqrt{2}+468\sqrt{3}+144\sqrt{5}+2006=(13\sqrt{2}+4\sqrt{3}+18\sqrt{5})^2$. Square rooting gives us $13\sqrt{2}+4\sqrt{3}+18\sqrt{5}$ and our answer is $13\cdot4\cdot18=\boxed{936}$

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions

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