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Math Help - Polynomial, area of a polygon

  1. #1
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    Polynomial, area of a polygon

    Hi!

    Could you help me solve this problem?

    Integers (x,y) satisfying the equation
    xy^2 - y^3 - xy + x^2 + 5 =0

    are vertexes of a certain polygon. What is its area?

    Thank you.
    Last edited by gollum; October 12th 2011 at 09:06 AM.
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  2. #2
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    Re: Polynomial, area of a polygon

    Hello, gollum!

    I used some Olympic-level gymnastics . . .


    Integers (x,y) satisfying the equation: xy^2 - y^3 - xy + x^2 + 5 \:=\:0

    are vertexes of a certain polygon. .What is its area?

    We have a quadratic in x:\;\;x^2 - (y-y^2)x - (y^3-5) \:=\:0

    Quadratic Formula: . x\;=\;\frac{(y-y^2) \pm \sqrt{(y-y^2)^2 + 4(y^3-5)}}{2}

    . . x \;=\;\frac{(y-y^2) \pm\sqrt{y^4 + 2y^3 + y^2 - 20}}{2} \;=\;\frac{(y-y^2) \pm \sqrt{y^2(y+1)^2 - 20}}{2}


    Since x is an integer, the radicand must be a square.
    . . . . . . y^2(y+1)^2 - 20 \:=\:k^2

    Hence, we have: . \big[y(y+1)\big]^2 - k^2 \:=\:20

    The difference of two squares is 20.
    . . The only such squares are: . (\pm6)^2\text{ and }(\pm4)^2


    This gives us: . y = 2,\;\begin{Bmatrix}x = 1 \\ x=\text{-}3\end{Bmatrix}

    . . . . . . and: . y = \text{-}3,\;\begin{Bmatrix}x = \text{-}4 \\ x = \text{-}8\end{Bmatrix}


    We have four vertices:

    . . \begin{array}{ccccccc} && * &&& * \\ && (\text{-}3,2) &&& (1,2) \\ \\ \\ * &&& * \\ (\text{-}8,\text{-}3) &&& (\text{-}4,\text{-}3) \end{array}


    We have a parallelogram with base 4 and height 5 . . .

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