<s>[INST] Number of right isosceles triangles that can be formed with points lying on the curve $8x^3+y^3+6xy=1$ Number of right isosceles triangles that can be formed with points lying on the curve $$8x^3+y^3+6xy=1$$
MY ATTEMPT :
We have,
$$8x^3+y^3+6xy=1$$
adding both the sides  $$6xy^2+12x^2$$ and simplifying we get ,
$$y^2+y(1-2x)+4x^2+2x+1=0$$
after this I got struck pls help me out with this question
Answer given is 3
 [/INST] Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we see that our equation it's
$$(2x+y-1)(4x^2+y^2+1-2xy+2x+y)=0,$$ which gives $$2x+y-1=0$$ or
$$(2x-y)^2+(2x+1)^2+(y+1)^2=0.$$
The second gives a point $A\left(-\frac{1}{2},-1\right)$ and two other points are placed on the line $2x+y-1=0$.
Thus, the vertex of the right angle of the triangle may be in three these points.
Easy to see that all these possibilities  occur.
Indeed, let $AD$ be a perpendicular to the line $l:2x+y-1=0$ and $\{B,C\}\subset l$ such that $D$ be a mid-point of $BC$ and $AD=DB=DC.$
Thus, $\Delta ABC$, $\Delta ADB$ and $\Delta ADC$ they are right-angled  isosceles triangles.
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