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Math Help - determine equilateral triangle's length using a point from the inside

  1. #1
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    determine equilateral triangle's length using a point from the inside

    Problem:
    A point P within an equilateral triangle has a distance of 4, 5, and 6 units respectively from the vertices. Determine the lengths of the sides of the equilateral triangle.

    I was able to break it down into 3 triangles (coming from point P). So what I did was to use Cosine Law on those triangles and was able to get

    2 sin 30 degrees x sin (60 - 2 theta) = -2/a

    Am I on the right track. It seems that if I continue, I'd be getting 3 unknowns, 2 equations which is not feasible, unless I'm good at guessing. Do you have any other ways to solve this ?

    Thank you in advance.

    Regards,
    Bibbo
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  2. #2
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    Hello bibbonacci
    Quote Originally Posted by bibbonacci View Post
    Problem:
    A point P within an equilateral triangle has a distance of 4, 5, and 6 units respectively from the vertices. Determine the lengths of the sides of the equilateral triangle.

    I was able to break it down into 3 triangles (coming from point P). So what I did was to use Cosine Law on those triangles and was able to get

    2 sin 30 degrees x sin (60 - 2 theta) = -2/a

    Am I on the right track. It seems that if I continue, I'd be getting 3 unknowns, 2 equations which is not feasible, unless I'm good at guessing. Do you have any other ways to solve this ?

    Thank you in advance.

    Regards,
    Bibbo
    If the angles at the point P are \theta, \phi, \psi and the length of the side of the triangle is \displaystyle a, then, using the Cosine Rule on all three triangles gives:
    \displaystyle \cos\theta=\frac{41-a^2}{40}

    \displaystyle \cos\phi = \frac{61-a^2}{60}

    \displaystyle \cos\psi = \frac{52-a^2}{48}=\cos(\theta+\phi), since \psi = 2\pi-(\theta+\phi)
    Eliminating \displaystyle \theta and \displaystyle \phi gives a pretty messy equation for \displaystyle a, which I have solved using a spreadsheet (but perhaps that's cheating!), giving the answer a \approx 8.536, to 3 d.p.

    Perhaps someone can find an elegant algebraic solution?

    Grandad
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  3. #3
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    Opalg's Avatar
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    Quote Originally Posted by Grandad View Post
    If the angles at the point P are \theta, \phi, \psi and the length of the side of the triangle is \displaystyle a, then, using the Cosine Rule on all three triangles gives:
    \displaystyle \cos\theta=\frac{41-a^2}{40}

    \displaystyle \cos\phi = \frac{61-a^2}{60}

    \displaystyle \cos\psi = \frac{52-a^2}{48}=\cos(\theta+\phi), since \psi = 2\pi-(\theta+\phi)
    Eliminating \displaystyle \theta and \displaystyle \phi gives a pretty messy equation for \displaystyle a, which I have solved using a spreadsheet (but perhaps that's cheating!), giving the answer a \approx 8.536, to 3 d.p.

    Perhaps someone can find an elegant algebraic solution?

    Grandad
    I agree with that answer. See this thread. But I don't think that an "elegant algebraic solution" is possible.
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