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Math Help - Equilateral Triangles.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Equilateral Triangles.

    Prove the condition that a triangle XYZ should be equilateral is that:


    X^2+Y^2+Z^2-XY-XZ-YZ=0
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  2. #2
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    X^2+Y^2+Z^2-XY-XZ-YZ = \frac{1}{2}((X - Y)^2 + (Y - Z)^2 + (X - Z)^2)
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Prove the condition that a triangle XYZ should be equilateral is that:


    X^2+Y^2+Z^2-XY-XZ-YZ=0
    \displaystyle\ cos\frac{\pi}{3}=\frac{1}{2}


    By the Law of Cosines...

    \displaystyle\ X^2=Y^2+Z^2-2YZ\frac{1}{2}

    \displaystyle\ Y^2=X^2+Z^2-2XZ\frac{1}{2}

    \displaystyle\ Z^2=X^2+Y^2-2XY\frac{1}{2}

    Summing these...

    2X^2+2Y^2+2Z^2-XY-XZ-YZ=X^2+Y^2+Z^2
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    Hello, Also sprach Zarathustra!

    \text{Prove the condition that triangle }XYZ\text{ should be equilateral is that:}

    . . X^2+Y^2+Z^2-XY-XZ-YZ\:=\:0

    We assume that X,Y,Z are lengths of the sides (positive real numbers).


    We have a quadratic equation:

    . . X^2 - (Y + Z)X + (Y^2 - YZ + Z^2) \:=\:0


    Apply the Quadratic Formula:

    . . X \;=\;\dfrac{(Y+Z) \pm \sqrt{(Y+Z)^2 - 4(Y^2-YZ + Z^2)}}{2}


    This simplifies to:

    . . X \;=\;\dfrac{(Y+Z) \pm\sqrt{-3(Y- Z)^2}}{2} .[1]


    Since \,X is a real number, Y-Z \,=\,0 \quad\Rightarrow\quad Y \,=\,Z

    Then [1] becomes: . X \;=\;\dfrac{(Y+Y) \pm 0}{2} \:=\:Y


    \text{Therefore: }\:X \,=\,Y\,=\,Z \;\hdots\;\text{ the triangle is equilateral.}

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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Prove the condition that a triangle XYZ should be equilateral is that:


    X^2+Y^2+Z^2-XY-XZ-YZ=0

    Here is another solution using the complex numbers:

    The "corner" XY is actually the "corner" YZ turned through with an angel of \frac{2\pi}{3} in the positive(or negative) direction!

    Now, let us observe the complex number: cis{\frac{2\pi}{3}}.

    This complex number is a solution of the equation: z^3=1.

    So:

    z_1=cis{\frac{2\pi}{3}}

    z_2=cis{\frac{4\pi}{3}}


    By turning XY with positive direction \frac{2\pi}{3} we can describe algebraically:

    X-Z=(Y-Z)z_1

    Now the negative direction:

    cis({-\frac{2\pi}{3}}) = \frac{1}{z_1}=z_2


    Hence again, algebraically:

    X-Z=(Y-Z)z_2
    ================================================== =========

    Now we simplify X-Z=(Y-Z)z_1


    X+Yz_1-Z(z_1+1)=0

    or:

    X+Yz_1+Zz_2=0 [1]


    And same simplification on X-Z=(Y-Z)z_2 :


    X+Yz_2+Zz_1=0 [2]


    Now we multiply 1 and 2:

    (X+Yz_1+Zz_2)(X+Yz_2+Zz_1)=X^2+Y^2+Z^2-XY-XZ-YZ=0
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