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Thread: Aime 2006 II #6

  1. #1
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    Aime 2006 II #6

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

    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 $\displaystyle B$ is a vertex for the smaller square as well as a point on $\displaystyle \overline{AE}$, then I get the following equalities.

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

    (1)$\displaystyle x^2+1=y^2$
    (2)$\displaystyle (1-x)^2+(1-x)^2=y^2$

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

    Equating these two expressions and solving for $\displaystyle x$, I get that $\displaystyle x=\frac{4-\sqrt{10}}{2}$, 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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    I made a small error. I should have solved the system and gotten that $\displaystyle x=\frac{4-\sqrt{12}}{2}$, which reduces down to $\displaystyle \frac{2-\sqrt{3}}{1}$, thus leading to the correct answer of 006. Thanks for pointing out that error.
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