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Difference between revisions of "2006 AIME I Problems/Problem 6"

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== Problem ==
 
== Problem ==
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Square <math> ABCD </math> has sides of length 1. Points <math> E </math> and <math> F </math> are on <math> \overline{BC} </math> and <math> \overline{CD}, </math> respectively, so that <math> \triangle AEF </math> is equilateral. A square with vertex <math> B </math> has sides that are parallel to those of <math> ABCD </math> and a vertex on <math> \overline{AE}. </math> The length of a side of this smaller square is <math>\frac{a-\sqrt{b}}{c}, </math> where <math> a, b, </math> and <math> c </math> are positive integers and <math> b</math> is not divisible by the square of any prime. Find <math> a+b+c. </math>
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Let <math> \mathcal{S} </math> be the set of [[real number]]s that can be represented as repeating [[Decimal| decimals]] of the form <math> 0.\overline{abc} </math> where <math> a, b, c </math> are distinct [[digit]]s. Find the sum of the elements of <math> \mathcal{S}. </math>
  
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== Solution ==
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== Solution 1 ==
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{{image}}
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Numbers of the form <math>0.\overline{abc}</math> can be written as <math>\frac{abc}{999}</math>. There are <math>10\times9\times8=720</math> such numbers. Each digit will appear in each place value <math>\frac{720}{10}=72</math> times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is <math>\frac{45\times72\times111}{999}= \boxed{360} </math>.
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Call the vertices of the new square A', B', C', and D', in relation to the vertices of <math>ABCD</math>, and define <math>s</math> to be one of the sides of that square. Since the sides are [[parallel]], by [[corresponding angles]] and AA~ we know that triangles <math>AA'D'</math> and <math>D'C'E</math> are similar. Thus, the sides are proportional: <math>\frac{AA'}{A'D'} = \frac{D'C'}{C'E} \Longrightarrow \frac{1 - s}{s} = \frac{s}{1 - s - CE}</math>. Simplifying, we get that <math>s^2 = (1 - s)(1 - s - CE)</math>.
 
  
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<math>\angle EAF</math> is <math>60</math> degrees, so <math>\angle BAE = \frac{90 - 60}{2} = 15</math>. Thus, <math>\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}</math>, so <math>AE = \frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \sqrt{6} - \sqrt{2}</math>. Since <math>\triangle AEF</math> is [[equilateral]], <math>EF = AE = \sqrt{6} - \sqrt{2}</math>. <math>\triangle CEF</math> is a <math>45-45-90 \triangle</math>, so <math>CE = \frac{AE}{\sqrt{2}} = \sqrt{3} - 1</math>. Substituting back into the equation from the beginning, we get <math>s^2 = (1 - s)(2 - \sqrt{3} - s)</math>, so <math>(3 - \sqrt{3})s = 2 - \sqrt{3}</math>. Therefore, <math>s = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 - \sqrt{3}}{6}</math>, and <math>a + b + c = 3 + 3 + 6 = 012</math>.
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== Solution 2 ==
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Alternatively, for every number, <math>0.\overline{abc}</math>, there will be exactly one other number, such that when they are added together, the sum is <math>0.\overline{999}</math>, or, more precisely, 1. As an example, <math>.\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1</math>.
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Thus, the solution can be determined by dividing the total number of [[permutation]]s by 2. The answer is <math>\frac{10 \cdot 9 \cdot 8}{2} = \frac{720}{2}= \boxed{360}</math>.
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Another method, albeit a little risky, that can be used is to note that the numbers between 1 and 999 with distinct digits average out to <math>\frac{999}{2}</math>. Then the total sum becomes <math>\frac{\frac{999}{2}\times720}{999}</math> which reduces to <math>\boxed{360}</math>
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 18:14, 20 November 2023

Problem

Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits. Find the sum of the elements of $\mathcal{S}.$

Solution 1

Numbers of the form $0.\overline{abc}$ can be written as $\frac{abc}{999}$. There are $10\times9\times8=720$ such numbers. Each digit will appear in each place value $\frac{720}{10}=72$ times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is $\frac{45\times72\times111}{999}= \boxed{360}$.

Solution 2

Alternatively, for every number, $0.\overline{abc}$, there will be exactly one other number, such that when they are added together, the sum is $0.\overline{999}$, or, more precisely, 1. As an example, $.\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1$.

Thus, the solution can be determined by dividing the total number of permutations by 2. The answer is $\frac{10 \cdot 9 \cdot 8}{2} = \frac{720}{2}= \boxed{360}$.

Another method, albeit a little risky, that can be used is to note that the numbers between 1 and 999 with distinct digits average out to $\frac{999}{2}$. Then the total sum becomes $\frac{\frac{999}{2}\times720}{999}$ which reduces to $\boxed{360}$

See also

2006 AIME I (Problems • Answer Key • Resources)
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Problem 7
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