isosceles triangle

**disclaimer** · December 6th 2008, 12:35 PM

Hi;

The coordinates of the vertices of an isosceles triangle are integers. Show that the square of the length of its base is an even number.

Any hint would be helphul. Thanks in advance.

**Moo** · December 6th 2008, 12:54 PM

Hello,

Originally Posted by disclaimer

Hi;

The coordinates of the vertices of an isosceles triangle are integers. Show that the square of the length of its base is an even number.

Any hint would be helphul. Thanks in advance.

Let $(x_1,y_1)$ be the coordinates of the summit : $A_1$
Let $(x_2,y_2)$ and $(x_3,y_3)$ be the coordinates of the two other points : $A_2,A_3$

We know that $A_1A_2=A_1A_3$ . Hence $A_1A_2^2=A_1A_3^2$

So this can be written :

$\boxed{(x_2-x_1)^2+(y_2-y_1)^2=(x_3-x_1)^2+(y_3-y_1)^2}$

Now you want to prove that $A_2A_3^2$ is even.

$A_2A_3^2=(x_3-x_2)^2+(y_3-y_2)^2$

Now write $x_3-x_2=(x_3-x_1)+(x_1-x_2)$ and $y_3-y_2=(y_3-y_1)+(y_1-y_2)$

and expand the two squares following this model : $\left((x_3-x_1)+(x_1-x_2)\right)^2=(x_3-x_1)^2+(x_1-x_2)^2+2(x_1-x_2)(x_3-x_1)$

Then simplify and conclude

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