Question

Consider three distinct lattice points $A(a,a{)}^2,B(b,b{)}^2,C(c,c^2)$ on $y=x^2$ where $a,b,c\in$ then

Answer 1

Area of $△ABC$ = $\frac{1}{2}|bc(c-b)-ac(c-a)+ab(b-a)|$

Now, considering that $a,b,c\in N$, they can be either all odd, all even, 2 odd 1 even or 1 odd 2 even. In all the above cases, the expression gives an integral answer on solution.

Length of any side is given as $\sqrt{(a-b{)}^2+(a^2-b^2{)}^2}=|a-b|\sqrt{(a+b{)}^2+1}$ which is clearly irrational as there exists no perfect square root for $\sqrt{x^2+1}\forall x\in N$.

Let length of altitude be $h$.

$\therefore area\left(△ABC\right)=\frac{1}{2}\times h\times |a-b|\sqrt{(a+b{)}^2+1}$

$⟹h=\frac{|bc(c-b)-ac(c-a)+ab(b-a)|}{|a-b|\sqrt{(a+b{)}^2+1}}$

which clearly is an irrational number.

Now, equation of tangent is $y-t^2=2t(x-t)$ at $(t,t^2)$

Intersection point of tangents at $A$ and $B$ is $(\frac{a+b}{2},ab)$

similarly, other points are $(\frac{b+c}{2},bc)$ and $(\frac{a+c}{2},ac)$

$\therefore \text{area of triangle of tangents}=\frac{1}{4}|(b+c)ac-bc(a+c)-(a+b)ca+ab(a+c)+(a+b)bc-ab(b+c)|$

(using determinant form for area of triangle)

$=\frac{1}{2}\times \text{area}\left(△ABC\right)$ which may be an integer or a rational number.