Equilateral Triangles.

• Jan 5th 2011, 11:02 AM
Also sprach Zarathustra
Equilateral Triangles.
Prove the condition that a triangle $XYZ$ should be equilateral is that:

$X^2+Y^2+Z^2-XY-XZ-YZ=0$
• Jan 5th 2011, 11:10 AM
snowtea
$X^2+Y^2+Z^2-XY-XZ-YZ = \frac{1}{2}((X - Y)^2 + (Y - Z)^2 + (X - Z)^2)$
• Jan 5th 2011, 11:45 AM
Quote:

Originally Posted by Also sprach Zarathustra
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...

$2X^2+2Y^2+2Z^2-XY-XZ-YZ=X^2+Y^2+Z^2$
• Jan 5th 2011, 04:12 PM
Soroban
Hello, Also sprach Zarathustra!

Quote:

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

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

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

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

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

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: . $X \;=\;\dfrac{(Y+Y) \pm 0}{2} \:=\:Y$

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

• Jan 6th 2011, 05:48 AM
Also sprach Zarathustra
Quote:

Originally Posted by Also sprach Zarathustra
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:

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