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.
Hello,
Let $\displaystyle (x_1,y_1)$ be the coordinates of the summit : $\displaystyle A_1$
Let $\displaystyle (x_2,y_2)$ and $\displaystyle (x_3,y_3)$ be the coordinates of the two other points : $\displaystyle A_2,A_3$
We know that $\displaystyle A_1A_2=A_1A_3$. Hence $\displaystyle A_1A_2^2=A_1A_3^2$
So this can be written :
$\displaystyle \boxed{(x_2-x_1)^2+(y_2-y_1)^2=(x_3-x_1)^2+(y_3-y_1)^2}$
Now you want to prove that $\displaystyle A_2A_3^2$ is even.
$\displaystyle A_2A_3^2=(x_3-x_2)^2+(y_3-y_2)^2$
Now write $\displaystyle x_3-x_2=(x_3-x_1)+(x_1-x_2)$ and $\displaystyle y_3-y_2=(y_3-y_1)+(y_1-y_2)$
and expand the two squares following this model : $\displaystyle \left((x_3-x_1)+(x_1-x_2)\right)^2=(x_3-x_1)^2+(x_1-x_2)^2+2(x_1-x_2)(x_3-x_1)$
Then simplify and conclude