# Can area be irrational?

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 improve the formatting of your question. It is an interesting question, but in such a format it is quite unappealing. – Jack D'Aurizio Mar 30 '15 at 15:55
• NB: That two numbers sum to an integer is not sufficient to conclude they are rational. – Ballunix Mar 30 '15 at 16:02
• Why can't the area be irrational? For example a circle?? – Jai Mahajan Mar 30 '15 at 17:04
• @Narasimham, certainly it can be, but it is not necessarily so. Consider, for example, a $(3, 4, 5)$ triangle, in which case $s = 6$ and the radicand is $6\cdot3\cdot2\cdot1 = 36$, whose square root is $A = 6$. – LSpice Mar 31 '15 at 0:11
• This 'Then we get $x+y=2$ so $x$ and $y$ are not irrational for sure !!!!!!' is completely wrong – verify $x=\sqrt 2$ and $y=2-\sqrt 2$. – CiaPan Mar 31 '15 at 9:08

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.

• @user166748: I do not get to which part of my proof you are referring to. I just proved that the it is impossible for the three vertices of an equilateral triangle to have all rational coordinates. – Jack D'Aurizio Mar 30 '15 at 17:08
• @user166748: have you really read the last lines? I said: by the shoelace formula, the area of a convex polygon having its vertices in points with rational coordinates is a rational number. By the pythagorean theorem, the area of an equilateral triangle having a rational squared length for its side is $\sqrt{3}$ times a rational number. The contradiction arises from the fact that a number (the area) cannot be at the same time a rational number and $\sqrt{3}$ times a rational number, since that would imply that $\sqrt{3}$ is a rational number. No circles involved. – Jack D'Aurizio Mar 30 '15 at 17:13
• Yes u r right...well proved... I thought u had proved it like my book....but please tell me if my book is wrong then in proving this question as it uses the fact that are cant be irrational to be a contradiction exactly like @user58220 did... But area can be irrational? – Jai Mahajan Mar 30 '15 at 17:16
• @user166748: you have to be clear about the hypothesis. The area of any convex polygon with rational vertices is rational. The area of any equilateral triangle with two or more vertices having rational coordinates is irrational. That's how the proof by contradiction works. – Jack D'Aurizio Mar 30 '15 at 17:20
• Ok now i think i have understood... We basically used 2 formulas and got different results from both so the assumption was wrong.. – Jai Mahajan Mar 30 '15 at 17:25

Your counterexample doesn't work, because $x+y=2$ doesn't imply that $x$ and $y$ are rational. In fact, we have also that $x^2+y^2=8$. By substituting $y=2-x$ into the latter equation, you will get a quadratic equation with two irrational roots.

You should re-read the book's proof...

Briefly,

1. Assume that the third point has rational co-ordinates
2. Then by the formula you cite, the area is rational
3. But the area is an irrational number times the length of a side squared
4. But by the original assumptions, two vertices have rational coordinates, so the side length squared is also rational
5. A rational times an irrational is irrational; therefore in this case, the area is irrational
6. Thus the assumption (Point 1) leads to a contradiction (Point 2 vs Point 5)

It's Points 4 and 5 that the book fails to state explicitly...

• Why can't the area be irrational? For example a circle??... That was my main question.. See the topic:) – Jai Mahajan Mar 30 '15 at 17:07
• you say that area is irrational so thats a contradiction... Why is it a contradiction? Cant the area be irrational like in the case of a circle? – Jai Mahajan Mar 30 '15 at 17:13
• @user166748 The area is irrational. The contradiction arises when we assume that the third point has rational coordinates, which means the area is rational, but this is not true. – Sawarnik Mar 30 '15 at 19:21
• @user166748 If the vertices of a polygon are all rational points (a rational point is one whose coords are rational), then the area has to be rational. A circle isn't a polygon, and even if you consider it to be an "infinite-sided polygon" where every point is a vertex, most of those vertices are not rational. (My italicized claim above is easier to prove for triangles.) – Akiva Weinberger Mar 31 '15 at 10:46

Area in general CAN be irrational. The Area of that particular triangle cannot.

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 3 √ 4 side 2 so the area is irrational that's not possible so one coordinate has to be irrational.

This quote mentions that a triangle with 3 rational vertices will have a rational area.

If it has a rational area, then it cannot also have an irrational area.

Thus, a triangle with 3 rational vertices cannot have an irrational area.

(Returning to your book's given answer, it argues that an equilateral triangle with two rational vertices must have an irrational area, therefore the third vertex cannot be completely rational since that could not produce an irrational area.)

Let the vertices of the triangle be $A=(0,0),B=(a,b),C=(c,d)$, and label the midpoint of $AB$ as $D=(a/2,b/2)$. Then the vectors $\overset{\to}{AD}=(a/2,b/2)$ and $\overset{\to}{DC}=(c-a/2,d-b/2)$ are perpendicular, being the altitude and (half-)base of the triangle, and $|DC|=\sqrt 3|AD|$. Rotating the vector $DC$ by $90^\circ$ to get $v=(b/2-d,c-a/2)$, we see that $v=\pm\sqrt 3\overset{\to}{AD}$, while both $v$ and $AD$ are rational vectors, so $\sqrt 3=\left|\dfrac{b/2-d}{a/2}\right|=\left|\dfrac{c-a/2}{b/2}\right|$ are two rational decompositions of $\sqrt 3$, either one of which is a contradiction. Similarly, the area of the triangle is $\rho=\frac{\sqrt 3}2|(a,b)|^2$, which yields the rational decomposition $\sqrt 3=\frac{2\rho}{|(a,b)|^2}$ under the assumption that $A$ is rational.

• What does it mean for two points to be perpendicular? – MJD Mar 30 '15 at 16:16
• @MJD Wow, got a lot of downvotes for that... Of course I meant that the vectors $(a/2,b/2)$ and $(c-a/2,d-b/2)$ are perpendicular, being the (half-)base and altitude of the triangle. – Mario Carneiro Mar 30 '15 at 20:14
• @MJD Edited to hopefully make the argument clearer. – Mario Carneiro Mar 30 '15 at 20:30