[Math] Can area be irrational

analytic geometrycoordinate systemstriangles

I'm stuck in a question of my book which says:

If in an equilateral triangle the coordinates of two vertices are integral then what can we say about the coordinates of the third?

The answer is that at least one of the coordinate of third vertex is irrational.

Now I don't know how to prove this although there is an explanation also in my book that I may not agree to. But first here is my attempt:

I have assumed an equilateral triangle with coordinates of 2 vertices as $(0,0)$ and $(2,2)$.

Now, let the 3rd vertex be $(x,y)$.

Using the fact that all sides are equal in equilateral triangle and using distance formula I get

$x^2+y^2=(x-2)^2+(y-2)^2$

Then we get $x+y=2$ so $x$ and $y$ are not irrational for sure !!!!!!

P. S= i know that the above line is wrong now but its better that is there so as to know how i attempted it…

Coming to what my book says…
It says that lets assume the coordinates of 3rd vertex to be rational so when we use the area of a triangle formula if 3 vertices are given we get a rational area. But we know area of equilateral triangle is $\frac{\sqrt{3}}{4}side^2$ so the area is irrational that's not possible so one coordinate has to be irrational.

Now why can't area be irrational? Isn't it the case with a circle also? And why am I proving the wrong thing?

Please help.

Best Answer

If $ABC$ is an equilateral triangle and $A=(0,0),B=(x,y)$, then $C$ lies in: $$ C=\left(\frac{x\mp \sqrt{3}y}{2},\frac{y\pm \sqrt{3}x}{2}\right),$$ so assuming that $B$ is a point with rational coordinates, $C$ is not.

Moreover, the area of an equilateral triangle is just $\frac{\sqrt{3}}{4}l^2$ where $l$ is the length of a side.

In the same way, if the side length is a rational number, the area is not.


With an alternative approach, if all the three vertices of an equilateral triangle have rational coordinates then the area is a rational number by the shoelace formula. But in such a case the squared length of the side is also a rational number, hence the area is at the same time a rational number and $\sqrt{3}$ times a rational number, contradiction.