Equilateral triangle problem

[DariusBotusanu]

Good evening, gentlemen, I have a rather tricky problem for which I would kindly require your assistance.
The problem was on the subject of the 2014 University of Babes-Bolyai contest for the Faculty of Mathematics and Informatics.
It states the following:
-we have an equilateral triangle OAB
-in the cartesian system xOy, the points have the following coordinates: O(0,0); A(m,n) with m,n being non-negative natural numbers; B(x,y), with x,y being non-negative real numbers
-PROVE that x and y cannot be natural numbers simultaneously

As my part of the solution, by constructing 2 circles centered at O and at A, both with radius r=[MATH]sqrt(m^(2)+n^(2))[/MATH], I showed that the coordinates of B verify the equations:

 

[tkhunny]

Do we lose anything by assuming n < y?

What's stopping you from rotating your triangle so that:

n_rotated = 0 and x_rotated = m_rotated/2

?

And, of course, does that get us anywhere?

 

[DariusBotusanu]

The triangle cannot be rotated as both A and B have non-negative coordinates.
Furthermore, the assumption y<n is not helpful as it does not lead to any fruitful answer.
Also, non-negative means: m>0,n>0,x>0,y>0

 

[Farzin]

Good evening, gentlemen, I have a rather tricky problem for which I would kindly require your assistance.
The problem was on the subject of the 2014 University of Babes-Bolyai contest for the Faculty of Mathematics and Informatics.
It states the following:
-we have an equilateral triangle OAB
-in the cartesian system xOy, the points have the following coordinates: O(0,0); A(m,n) with m,n being non-negative natural numbers; B(x,y), with x,y being non-negative real numbers
-PROVE that x and y cannot be natural numbers simultaneously

As my part of the solution, by constructing 2 circles centered at O and at A, both with radius r=[MATH]sqrt(m^(2)+n^(2))[/MATH], I showed that the coordinates of B verify the equations:
View attachment 16935
View attachment 16936

If I understood your question correctly, an equilateral triangle can NOT have integer vertices on the Cartesian system.
The reason is that the area of an equilateral triangle is [MATH]\dfrac{\sqrt{3}}{4}a^2[/MATH] which [MATH]a[/MATH] is the side length of the triangle. Now a triangle with three rational vertices must have a rational area but we know that the area of an equilateral triangle is always irrational number.

 

[tkhunny]

The triangle cannot be rotated as both A and B have non-negative coordinates.
Furthermore, the assumption y<n is not helpful as it does not lead to any fruitful answer.
Also, non-negative means: m>0,n>0,x>0,y>0

Nonresponsive. Build your triangle where ever you like, with all the non-negative constraints, and THEN rotate it so that there are two vertices on the x-axis, just to simplify your equations. We can rotate it back, later.

n < y suggests only that we rotate (n,m) the the x-axis, rather than (x,y). It is a restriction that loses no generality.

Anyway, just a thought. I hadn't decided if it would lead anywhere.

 

[DariusBotusanu]

If I understood your question correctly, an equilateral triangle can NOT have integer vertices on the Cartesian system.
The reason is that the area of an equilateral triangle is [MATH]\dfrac{\sqrt{3}}{4}a^2[/MATH] which [MATH]a[/MATH] is the side length of the triangle. Now a triangle with three rational vertices must have a rational area but we know that the area of an equilateral triangle is always irrational number.

Thank you, sir. That indeed seems like the most reasonable answer.
I appreciate your help and I wish you good luck and fortune!

 

[Harry_the_cat]

Good evening, gentlemen,

Just wanted to make you aware that there are actually some female helpers on this site too!

 

[Farzin]

Thank you, sir. That indeed seems like the most reasonable answer.
I appreciate your help and I wish you good luck and fortune!

You are welcome.