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Math Help - Problem 49

  1. #1
    Grand Panjandrum
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    Problem 49

    (minor variant of a problem due to Roy Barbara)

    Let a,\ b,\ c be three positive real numbers.

    Find necessary and sufficient conditions on a,\ b,\ c for there to exist an interior point P in the equilateral triangle ABC with unit side, such that |PA|=a,\ |PB|=b,\ |PC|=c.

    I have a solution to this, but as I have not looked it up so I cannot tell if it is the originator's solution but as it is not as neat as I would like it is probably clumsy compared to the best solution, so lets see what we can do

    CB
    Last edited by CaptainBlack; October 29th 2008 at 01:12 PM.
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  2. #2
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    Quote Originally Posted by CaptainBlack View Post
    (minor variant of a problem due to Roy Barbara)

    Let a,\ b,\ c be three positive real numbers.

    Find necessary and sufficient conditions on a,\ b,\ c for there to exist an interior point P in the equilateral triangle ABC with unit side, such that |PA|=a,\ |PB|=b,\ |PC|=c.

    I have a solution to this, but as I have not looked it up so I cannot tell if it is the originator's solution but as it is not as neat as I would like it is probably clumsy compared to the best solution, so lets see what we can do

    CB
    Ahh.. maybe its my modification to the question that is the reason my solution is clumsier than I would like??

    I shall have to look into this possibility

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    (minor variant of a problem due to Roy Barbara)

    Let a,\ b,\ c be three positive real numbers.

    Find necessary and sufficient conditions on a,\ b,\ c for there to exist an interior point P in the equilateral triangle ABC with unit side, such that |PA|=a,\ |PB|=b,\ |PC|=c.
    Here is my solution, I let you appreciate its neatness/clumsiness... It is pretty simple anyway.

    The idea is to use barycentric coordinates: for any point P in the plane, there is a unique triplet (\lambda,\mu,\nu) such that \lambda+\mu+\nu=1 and P=\lambda A+\mu B+\nu C.
    Given these coordinates, P lies in the triangle ABC if, and only if the three numbers \lambda, \mu, \nu are positive (or zero, corresponding to bounderies).

    It is easy to express a,b,c in terms of \lambda,\mu,\nu. We have \overrightarrow{AP}=\mu\overrightarrow{AB}+\nu\ove  rrightarrow{AC}, hence a^2=AP=\|\mu\overrightarrow{AB}+\nu\overrightarrow  {AC}\|^2=\mu^2 AB^2+2\mu\nu\overrightarrow{AB}\cdot\overrightarro  w{AC}+\nu^2 AC^2.
    Because ABC is equilateral with unit sides, we conclude a^2=\mu^2+\mu\nu+\nu^2.
    By circular permutation of letters, we get similar expressions for b^2 and c^2. Thus, b^2=\nu^2+\nu\lambda+\lambda^2.

    What we need in fine is expressions for \lambda,\mu,\nu in terms of a,b,c. This can be laborious, but I found a soft way to write it. We have a^2-b^2=\mu^2-\lambda^2+\mu\nu-\lambda\nu=(\mu-\lambda)(\mu+\lambda+\nu), hence a^2-b^2=\mu-\lambda.
    Again by circular permutation of the letters, we have c^2-a^2=\lambda-\nu. We deduce (a^2-b^2)-(c^2-a^2)=\mu+\nu-2\lambda=1-3\lambda.

    Finally, we have \lambda=\frac{1}{3}(1-2a^2+b^2+c^2). And, similarly, \mu=\frac{1}{3}(1-2b^2+c^2+a^2) and \nu=\frac{1}{3}(1-2c^2+a^2+b^2).

    Remembering what I said first about barycentric coordinates, the conclusion is then straightforward: P lies inside the triangle if, and only if 2a^2-b^2-c^2\leq 1, 2b^2-c^2-a^2\leq 1 and 2c^2-a^2-b^2\leq 1.

    If the triangle had side r, it would suffice to replace 1 by r^2 in the conditions, making them more "homogeneous".
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    (minor variant of a problem due to Roy Barbara)

    Let a,\ b,\ c be three positive real numbers.

    Find necessary and sufficient conditions on a,\ b,\ c for there to exist an interior point P in the equilateral triangle ABC with unit side, such that |PA|=a,\ |PB|=b,\ |PC|=c.

    I have a solution to this, but as I have not looked it up so I cannot tell if it is the originator's solution but as it is not as neat as I would like it is probably clumsy compared to the best solution, so lets see what we can do

    CB

    Is the solution to this problem not simply a = b = c??

    Cheers,

    David
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by David24 View Post
    Is the solution to this problem not simply a = b = c??

    Cheers,

    David
    No, see Laurents answer

    CB
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  6. #6
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    Well, I would draw three circles with centres at A,B,C, where P is on each of the circles, and then continue from there.
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