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Math Help - Maximum Area Problem in a Triangle - Not Solved Yet

  1. #1
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    Maximum Area Problem in a Triangle - Not Solved Yet

    The following extremum problem in a triangle has not been solved. It looks simple, and intuitively "P = mid point of BC" seems to be the answer, but noone was able to prove it yet. The closest solution brings a set of 4 equations which was not able to be solved neither by WolframAlpha nor by Maple.

    Thanks.

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  2. #2
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    Quote Originally Posted by oswaldo View Post
    The following extremum problem in a triangle has not been solved. It looks simple, and intuitively "P = mid point of BC" seems to be the answer, but noone was able to prove it yet. The closest solution brings a set of 4 equations which was not able to be solved neither by WolframAlpha nor by Maple.

    Thanks.
    This isn't the complete solution, only a first step:

    1. \angle(QPR) = 90^\circ because:

    \angle(p_1)=\dfrac{\gamma + \alpha_2}2


    \angle(p_2)=\dfrac{\beta+\alpha_1}2

    \angle(PQR)=\angle(p_1) + \angle(p_2) = \dfrac{\gamma + \alpha_2}2 + \dfrac{\beta+\alpha_1}2 = \frac12 \underbrace{((\alpha_1+\alpha_2)+\beta+\gamma)}_{\  text{sum up to 180}}

    2. Therefore the area of the triangle PQR is calculated by:

    A(\Delta(PQR))=\frac12 \cdot |\overline{PR}| \cdot |\overline{PQ}|

    3. Since you "know" () the angles of the triangles \Delta(BPR) and \Delta(PCQ) and at least one side of these triangles it must be possible to find the optimal place of P on \overline{BC}.
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    Observing m(RPQ)=90 degrees, is a good start, but then it gets complicated:
    Let BP=x, as a variable in the final extremum function, say A(RPQ)=f(x).
    sine rule in triangles:
    PR / sin(beta) = x / sin(180-beta-p1) => PR = x [sin(beta) / sin(beta+p1)]
    Similarlay, PQ = (BC - x) [sin(gamma) / sin(gamma+p2)]

    A(RPQ) = PR*PQ/2 = x*(BC-x)*[(sin(beta)*sin(gamma)) / (2*sin(beta+p1)*sin(gamma+p2))] = f(x)

    Here is the problem, angles, p1 and p2 also depend on x.

    What should be the next step? Any ideas?
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  4. #4
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    Quote Originally Posted by oswaldo View Post
    Observing m(RPQ)=90 degrees, is a good start, but then it gets complicated:
    ...
    What should be the next step? Any ideas?
    Quote Originally Posted by oswaldo View Post
    The following extremum problem in a triangle has not been solved. It looks simple, and intuitively "P = mid point of BC" seems to be the answer, but noone was able to prove it yet. ...
    1. I've drawn several examples of the triangle in question if P runs from B to C.

    2. Perpendicular to BC over P I have drawn a point whose distance from BC correspond to the value of the area of triangle PQR. How I got the area's value is shown in the box at the upper left part of the image.

    3. All these points lie on a curve which pretty much looks like a "simple" parabola. But unfortunately I can't find the equation which belongs to this parabola.
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  5. #5
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    Thank you earboth for the simulation. Indeed red looks like a parabola. This means that f(x)= K * x(BC-x), and K is a constant. In other words,
    (sin(beta)*sin(gamma)) / (2*sin(beta+p1)*sin(gamma+p2)) must be a CONSTANT. The rest is just trigonometry. I worked on it a little, but could not succeed.
    We also know that p1+p2=90.
    I will try again later..
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