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Thread: Tricky equation?

  1. #1
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    Tricky equation?

    a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w

    Given a, u and v, w = ?
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  2. #2
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    Re: Tricky equation?

    This is basically Heron's formula. So, if you know the area of the triangle and the lengths of two of its sides, then $w$ is the length of one of the sides.

    So, we have $a = \dfrac{1}{2}uv\sin \gamma$ where $\gamma$ is the angle between $u$ and $v$.

    We also have (by the law of Cosines): $w^2 = u^2+v^2-2uv\cos \gamma$ (again, where $\gamma$ is the angle between $u$ and $v$).

    So, we have $\sin \gamma = \dfrac{2a}{uv} = \dfrac{\text{opp}}{\text{hyp}}$, so $\cos \gamma = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{\sqrt{u^2v^2-4a^2}}{uv}$.

    This gives $w = \sqrt{u^2+v^2-2\sqrt{u^2v^2-4a^2}}$
    Last edited by SlipEternal; Jun 14th 2018 at 10:53 AM.
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    Re: Tricky equation?

    Quote Originally Posted by SlipEternal View Post
    This gives $w = \sqrt{u^2+v^2-2\sqrt{u^2v^2-4a^2}}$
    Thanks Slip. Your 1st - sign should be a +, right?

    Seems that works only if w = longest side.
    Tried it with u,v,w = 4,13,15 : area = 24
    If u=4, v=15, then w<>13.
    Do I need another coffee?!
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  4. #4
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    Re: Tricky equation?

    Quote Originally Posted by DenisB View Post
    Thanks Slip. Your 1st - sign should be a +, right?

    Seems that works only if w = longest side.
    Tried it with u,v,w = 4,13,15 : area = 24
    If u=4, v=15, then w<>13.
    Do I need another coffee?!
    I think the first - sign should be a $\pm$. One of those will give the correct answer. We just need to figure out which. We will have to use guessing and checking.

    With the - sign as is, it worked with u=4,v=15 and with u=13,v=15. It did not work with u=4,v=13.
    With the - sign changed to a +, it worked with u=4,v=13, but not with the other two.

    I think this is because $\cos \gamma$ could be negative (if $\gamma > \dfrac{\pi}{2}$), but $\dfrac{\sqrt{u^2v^2-4a^2}}{uv}>0$ always (so long as it is a valid triangle).

    Edit: Yes, it is definitely a $\pm$. It could be either. Example: $u=4,v=13,w=\sqrt{145},a=24$ works just as well as $u=4,v=13,w=15,a=24$.

    This gives:

    $$w = \sqrt{u^2+v^2 \pm 2\sqrt{u^2v^2-4a^2}}$$
    Last edited by SlipEternal; Jun 14th 2018 at 01:32 PM.
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  5. #5
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    Re: Tricky equation?

    Ahhhh................thanks loads.......
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  6. #6
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    Re: Tricky equation?

    Two addendums:
    It is possible for $w $ to be negative. Example: $u=4,v=13,w=-15,a=24$

    So there is another $\pm $ on the entire formula. This gives four solutions for $w $ which makes sense given that the original equation works out to a 4th degree polynomial in $w $.

    Second issue: the creation of the formula relied on $uv $ in the denominator. We should consider either value being zero as a degenerate case and solve that separately.
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    Re: Tricky equation?

    Quote Originally Posted by SlipEternal View Post
    Second issue: the creation of the formula relied on uv in the denominator. We should consider either value being zero as a degenerate case and solve that separately.
    Can we not simply add "......where u>0 and v>0"?
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  8. #8
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    Re: Tricky equation?

    Quote Originally Posted by DenisB View Post
    Can we not simply add "......where u>0 and v>0"?
    Yes, that should work.
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  9. #9
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    Re: Tricky equation?

    Quote Originally Posted by SlipEternal View Post
    Yes, that should work.
    Lol, if $u=0$ or $v=0$, then the area is going to be zero, as well. So, my formula will still work. Turns out there are no degenerate cases. I asked on math.stackexchange.com if I caught all degenerate cases, and it turns out there were none.
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