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Math Help - Aime 2006 II #6

  1. #1
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    Aime 2006 II #6

    "Square ABCD has sides of length 1. Points E and F are on \overline{BC} and \overline{CD}, respectively, so that \triangle AEF is equilateral. A square with vertex B has sides that are parallel to those of ABCD and a vertex on \overline{AE}. The length of a side of this smaller square is \displaystyle \frac{a-\sqrt{b}}{c}, where a, b, and c are positive integers and bis not divisible by the square of any prime. Find a+b+c."

    I've tried to draw my version of this diagram, but I'm not so good at paint, so I'll try to do my best in explaining how I interpret this problem.

    There is this square with an inscribed equilateral triangle, and a smaller square. Because the smaller square's sides are parallel to that of the larger one, and B is a vertex for the smaller square as well as a point on \overline{AE}, then I get the following equalities.

    Let x be one side of the smaller square and let y be one side of the inscribed triangle. Then it follows that:

    (1) x^2+1=y^2
    (2) (1-x)^2+(1-x)^2=y^2

    Equating these two expressions and solving for x, I get that x=\frac{4-\sqrt{10}}{2}, which is incorrect. Thoughts?
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  2. #2
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    Quote Originally Posted by Jameson
    ...
    (1) x^2+1=y^2
    (2) (1-x)^2+(1-x)^2=y^2

    Equating these two expressions and solving for x, I get that x=\frac{4-\sqrt{10}}{2}, which is incorrect. Thoughts?
    Hello,

    using your system of equations I get x= 2+sqrt(3) or x = 2- sqrt(3). Both solutions are positive. Which one fits best to your problem I don't know.

    Greetings

    EB
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  3. #3
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    I made a small error. I should have solved the system and gotten that x=\frac{4-\sqrt{12}}{2}, which reduces down to \frac{2-\sqrt{3}}{1}, thus leading to the correct answer of 006. Thanks for pointing out that error.
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