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Math Help - A nice equilateral triangle problem

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    A nice equilateral triangle problem

    There are three parallel lines with given distances  x ~,~ y ~, ~ x+y .

    Find the length of the side of the equilateral triangle whose three vertices lie on the parallel lines respectively .
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    Quote Originally Posted by simplependulum View Post
    There are three parallel lines with given distances  x ~,~ y ~, ~ x+y .

    Find the length of the side of the equilateral triangle whose three vertices lie on the parallel lines respectively .
    Dear simplependulum,

    Please refer the attached drawing.

    If~B\hat{D}F=C\hat{D}E=\alpha~and~BF=x,~CE=y

    The length of a side of the traingle ABC= \left(\frac{x}{\sin{\alpha}}+\frac{y}{\sin{\alpha}  }\right)

    Consider the traingle AEC where, C\hat{A}E=\alpha-60

    Hence, \left(\frac{x}{\sin{\alpha}}+\frac{y}{\sin{\alpha}  }\right)\sin{(\alpha-60)}=y

    By simplification we could obtain, \tan{\alpha}=\frac{\sqrt{3}(x+y)}{x-y}\Rightarrow{\sin{\alpha}}=\frac{\sqrt{3}(x+y)}{2  \sqrt{x^2+xy+y^2}}

    Therefore the length of a side of the traingle ABC= \left(\frac{x}{\sin{\alpha}}+\frac{y}{\sin{\alpha}  }\right)=\frac{2\sqrt{x^2+xy+y^2}}{\sqrt{3}}
    Attached Thumbnails Attached Thumbnails A nice equilateral triangle problem-sp.pdf  
    Last edited by Sudharaka; June 20th 2010 at 07:58 AM.
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    Quote Originally Posted by Sudharaka View Post
    Dear simplependulum,

    Please refer the attached drawing.

    AB=AC~;~since~ABC~is~eqilateral

    A\hat{B}C=A\hat{C}B=60^{o}~;~since~ABC~is~eqilater  al

    BD=CD~;~since~BFD~and~CED~traingles~are~congruent

    Hence the traingles ABD and ADC are congruent. Therefore, A\hat{D}B=A\hat{D}C=90^{o}

    Hence all the sides of the traingle ABC is equal to x+y.
    If I interpreted correctly, the side length is x+y if and only if x = y, and the triangles you said are congruent are only similar in general. I have not solved this problem.
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    Quote Originally Posted by undefined View Post
    If I interpreted correctly, the side length is x+y if and only if x = y, and the triangles you said are congruent are only similar in general. I have not solved this problem.
    Dear undefined,

    Thanks. That is a mistake I have'nt seen. I cannot take BD=CD since it is only valid if x=y. Thanks again for pointing that out. I have edited my previous post with a new approach.
    Last edited by Sudharaka; June 20th 2010 at 07:29 AM.
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    Excellent , the key is making two equations with unknowns , the side and the included angle but we have to eliminate the side we are looking for first , that's a trick .
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