Results 1 to 5 of 5

Thread: Equilateral Triangles.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Equilateral Triangles.

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


    $\displaystyle X^2+Y^2+Z^2-XY-XZ-YZ=0$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2010
    Posts
    470
    $\displaystyle X^2+Y^2+Z^2-XY-XZ-YZ = \frac{1}{2}((X - Y)^2 + (Y - Z)^2 + (X - Z)^2)$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Quote Originally Posted by Also sprach Zarathustra View Post
    Prove the condition that a triangle $\displaystyle XYZ$ should be equilateral is that:


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


    By the Law of Cosines...

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

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

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

    Summing these...

    $\displaystyle 2X^2+2Y^2+2Z^2-XY-XZ-YZ=X^2+Y^2+Z^2$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, Also sprach Zarathustra!

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

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

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


    We have a quadratic equation:

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


    Apply the Quadratic Formula:

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


    This simplifies to:

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


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

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


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

    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by Also sprach Zarathustra View Post
    Prove the condition that a triangle $\displaystyle XYZ$ should be equilateral is that:


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

    Here is another solution using the complex numbers:

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

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

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

    So:

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

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


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

    $\displaystyle X-Z=(Y-Z)z_1$

    Now the negative direction:

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


    Hence again, algebraically:

    $\displaystyle X-Z=(Y-Z)z_2$
    ================================================== =========

    Now we simplify $\displaystyle X-Z=(Y-Z)z_1$


    $\displaystyle X+Yz_1-Z(z_1+1)=0$

    or:

    $\displaystyle X+Yz_1+Zz_2=0$ [1]


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


    $\displaystyle X+Yz_2+Zz_1=0$ [2]


    Now we multiply 1 and 2:

    $\displaystyle (X+Yz_1+Zz_2)(X+Yz_2+Zz_1)=X^2+Y^2+Z^2-XY-XZ-YZ=0$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Are equilateral triangles isosceles?
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Apr 11th 2010, 04:12 AM
  2. Equilateral triangles
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Jan 18th 2010, 12:52 PM
  3. equilateral triangles
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Apr 21st 2009, 02:03 AM
  4. Replies: 5
    Last Post: Dec 22nd 2008, 10:06 AM
  5. Circles insribed in equilateral triangles.
    Posted in the Geometry Forum
    Replies: 2
    Last Post: Mar 25th 2008, 02:48 PM

Search Tags


/mathhelpforum @mathhelpforum