Question

Prove that a triangle which has one of the angle as 30 degree, can not have all vertices with integral coordinates.

Answer 1

Let $△ABC$ is triangle with one angle $={30}^o$

All vertices are not Integral coordinates

Let $x,y,a,b$ & $z$

slope of line $AB=\frac{y}{x}=m_1$

slope of line $BC=\frac{b}{a}=m_2$

Angle between lines $AB\times BC=\tan {30}^o=\frac{m_1-m_2}{1+m_1{m}_2}$

i.e, $\tan {30}^o=\frac{\frac{y}{x}-\frac{b}{a}}{1+\frac{y}{x}\frac{b}{a}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{ay-bx}{ax+by}=\frac{1}{\sqrt{3}}$ Here $LHS$ is rational & $RHS$ is irrational

we cannot find any integers $a,b,x,y$ that

So, all vertices are not Integral coordinates.