MHB Can an Equilateral Triangle be Formed with Rational Vertices on the X-Axis?

[kaliprasad]

in a plane point is said to be rational if both x and y co-ordinates are rational. Show that if in a $\triangle$ ABC all the vericles A , B, c are rational then $\triangle$ cannot be equilateral

 

[Opalg]

[sp]Suppose that $ABC$ is an equilateral triangle whose vertices are all rational. By translating the axes through rational distances, we may assume that $C$ is at the origin. Let $A$ be the point $(x,y)$ and $B$ be the point $(u,v)$, where $x,y,u,v$ are all rational. The linear transformation of rotation through $\pi/3$ takes $A$ to $B$. But this transformation is given by the matrix $\begin{bmatrix} \cos(\pi/3) & -\sin(\pi/3) \\ \sin(\pi/3) & \cos(\pi/3) \end{bmatrix} = \begin{bmatrix} \frac12 & -\frac{\sqrt3}2 \\ \frac{\sqrt3}2 & \frac12 \end{bmatrix}.$ It follows that $\begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} \frac12 & -\frac{\sqrt3}2 \\ \frac{\sqrt3}2 & \frac12 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac12 x -\frac{\sqrt3}2y \\ \frac{\sqrt3}2 x + \frac12y \end{bmatrix},$ so that $u = \frac12 x -\frac{\sqrt3}2y$. But then $\sqrt3 = \frac{x-2u}y$ – a contradiction since the right side is rational and the left side is not. Therefore no such triangle can exist.[/sp]

 

[kaliprasad]

great answer by Opalg. This gives me another insight of the solution

my solution

Spoiler

Without loss of generality we can chose the co-ordinates ( we can make this by proper shift of co-ordinates) as A= (0,0), B= (x,y), C=(a,b) with x y
a and b being rationalNow slope of AB = $\frac{y}{x}$ is rational

Slope of AC = $\frac{b}{a}$ is rationalSo $\tan\angle (BAC) = \frac{\frac{y}{x} –\frac{b}{a}} {1 + \frac{by}{ax}}$ which is rational as both numerator and denominator are rational

As $\tan \, 60^0 = \sqrt(3)$ is irrational so $\angle (BAC)$ cannot be $60^0$ as tan of the angle is rational.

so the triangle cannot be equilateral

Hence proved

 

[Albert1]

in a plane point is said to be rational if both x and y co-ordinates are rational. Show that if in a $\triangle$ ABC all the vericles A , B, c are rational then $\triangle$ cannot be equilateral

Spoiler

if $\triangle ABC$ is equilateral with side length a
let x and y coordinates of points being :$A(0,0),B(a,0),C(\frac{a}{2},b)$
(here $a,b$ are all rational)
the area of $\triangle ABC=\dfrac {\sqrt 3a^2}{4}=\dfrac {ab}{2}$
$\therefore b=\dfrac {\sqrt 3a}{2}$ is not rational
and we conclude that $\triangle ABC$ cannot be equilateral

 

[kaliprasad]

Spoiler

if $\triangle ABC$ is equilateral with side length a
let x and y coordinates of points being :$A(0,0),B(a,0),C(\frac{a}{2},b)$
(here $a,b$ are all rational)
the area of $\triangle ABC=\dfrac {\sqrt 3a^2}{4}=\dfrac {ab}{2}$
$\therefore b=\dfrac {\sqrt 3a}{2}$ is not rational
and we conclude that $\triangle ABC$ cannot be equilateral

Spoiler

B(a,0) does not cover all cases of B for example B=(1,1) by rotation cannot be a rational point in x-axis

 

[Albert1]

we fix point A and B on the x axis,and point C(a/2 ,b)
(here a,b rational) and show that b cannot be rational
if the triangle is equilateral
so we don't even have to rotate