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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 6]] |
| − | Let <math> \mathcal{S} </math> be the set of real numbers that can be represented as repeating decimals of the form <math> 0.\overline{abc} </math> where <math> a, b, c </math> are distinct digits. Find the sum of the elements of <math> \mathcal{S}. </math>
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| − | == Solution ==
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| − | {{image}}
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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>\displaystyle a + b + c = 3 + 3 + 6 = 012</math>.
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| − | == See also ==
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| − | {{AIME box|year=2006|n=II|num-b=5|num-a=7}}
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| − | [[Category:Intermediate Geometry Problems]]
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