Monday, December 6, 2010

Fox 322

This has not been solved yet. What'd you think?

12 comments:

  1. This problem leads to very long equations. Even Maple refuses give a solution

    César Lozada

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  2. César, you might have already seen that m(RPQ)=90 degrees. Does that help to simplify the set of equation?

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  3. Cesar is right
    it is a very long equation and one has to solve the terms like xsin(y-x)=0

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  4. Does unsolved mean that you foxes do not possess a solution either?

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  5. "Unsolved" means that we don't know a solution. It does not necessarily mean that the problem is very very hard.

    "Not confirmed" means that a problem might be solved only once, and no one has confirmed it yet.

    For this case (Fox 322), we were expecting to find a solution in the following format:
    Max A(PQR) = f(|BC|, angles A, B, C), but those who tried could not find yet. (We don't know either). It may very well be possible (for this Fox), that the solution can only be identifed numerically for a given ABC triangle.

    Hope this helps.

    -8foxes

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    Replies
    1. Use the theorem of Stewart. Let PC = t*c for some t in (0,1), then

      p[t]:=|AP|=Sqrt[t*c^2 + (1 - t) b^2 - t*(1 - t)*a^2].

      Furthermore, one can calculate via Stewart's Thorem the squares of the length of the bisectors |PR|^2 and |PQ|^2; multiplying these squares gives

      (2F[t}])^2 = t*a*p[t]*(1 - (b/(p[t] + t*a))^2)*(1 - t)*a*
      p[t]*(1 - (c/(p[t] + (1 - t)*a))^2),

      where F[t] denotes the area of PQR. And no: t=1/2 is i. g. *not* the maximum value. Mathematica refuses to solve F'[t]=0.

      Also, F[t] can be written as

      F[t]=2*c*sin(beta)*(1-t)*t*a^2*p[t]/((1-t)*t*a^2+a*p[t]+p[t]^2),

      but Mathematica doesn't find a solution for F'[t]=0 either.

      Michael

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    2. Ups, typo: Let PC = t*a, of course.

      Michael

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  6. Maximum A(PQR) = (|BC|/2)^2
    This when P is midpoint of BC and A is infinitely distant on the perpendicular at P. Angles B and C both approach 90 degrees. PQR is 45-45-90 with edge 0.5*|BC|*sqrt(2).

    Consider B at (0,0), C at (1,0), A at (x,y) and P at (p,0). We can find points Q and R without sines or cosines:
    R.X = A.X * P.X / (|AP| + P.X)
    R.Y = A.Y * P.X / (|AP| + P.X)
    Q.X = (|AP| + A.X * (1 - P.X)) / (|AP| + 1 - P.X)
    Q.Y = A.Y * (1 - P.X) / (|AP| + 1 - P.X)
    With this, we can find the area of PQR as a function of x, y and p:
    A(PQR) = f(x, y, p) = |PR| * |PQ| / 2
    Expanding |PR| and |PQ| should be fun :o)

    This will solve A(PQR) for any triangle ABC, with A(PQR) = f(|BC|, angle B, angle C). Anyone care to write that function?

    -DM

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  7. P ıs the mıd-poınt of BC ıs the usual suspect.
    And Maximum A(PQR) = (|BC|/2)^2 looks belıevable.
    but we have to follow the rest of the steps for a complete proof.
    thank you DM.

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  8. (|BC|/2)^2 is not actually the maximum. It is the limit of A(PQR) as |AP| goes to infinity, at which point AB, AP and AC are essentially parallel. Solving that case gives maximum A(PQR) = (|BC|/2)^2.

    P as the midpoint of BC doesn't give maximum A(PQR) in most cases. It may be that A(PQR) is maximized when RQ is parallel to BC.

    Another observation: A(PQR) <= 0.25 * A(ABC)

    I'll try to get an image up tomorrow to show some of my work for the function.

    -DM

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  9. http://i11.photobucket.com/albums/a190/DMathias/Fox322.png

    D on AP, |PD| = |BP|
    E midpoint of BD

    F on AP, |PF| = |PC|
    G midpoint of FC

    Finding point R:

    D.X = P.X + (A.X - P.X) * P.X / |AP|
    = (|AP| + A.X - P.X) * P.X / |AP|
    D.Y = A.Y * P.X / |AP|

    E.X = D.X / 2
    E.Y = D.Y / 2

    R.Y / R.X = A.Y / A.X
    R.Y / (P.X - R.X) = E.Y / (P.X - E.X)

    R.X * A.Y / A.X = (P.X - R.X) * E.Y / (P.X - E.X)

    Solve for R.X
    R.Y = R.X * A.Y / A.X

    Similar solution for finding point Q

    A(PQR) = f(x, y, p) = |PR| * |PQ| / 2

    g(|BC|, angle B, angle C, p) = |BC|^2 * f(tan C / (tan B + tan C), tan B * tan C / (tan B + tan C), p)

    -DM

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  10. The area A(PQR) is maximum when RQ // BC

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