Straight Lines and Curves

Question

If the line $y=x\sqrt3$, intersects the curve $x^3+y^3+3xy+5x^2+3y^2+4x+5y-1=0$, in three points $A,B,C$. If $O$ is the origin, then what's the product $OA\cdot OB\cdot OC$?

Answer 1

Hint: Let $A\equiv(A_x,A_y),B\equiv(B_x,B_y),C\equiv(C_x,C_y)$: the question is to determine the product $$\sqrt{A_x^2+A_y^2}\sqrt{B_x^2+B_y^2}\sqrt{C_x^2+C_y^2}=\\\sqrt{A_x^2+3A_X^2}\sqrt{B_x^2+3B_x^2}\sqrt{C_x^2+3C_x^2}=\\2A_x\cdot 2B_x\cdot 2C_x=\\8A_xB_xC_x$$ But the product $A_xB_xC_x$ can be easily found from the equation $x^3+y^3+3xy+5x^2+3y^2+4x+5y-1=0$ obtained by substituting $y=x\sqrt3$. In fact for a cubic equation $Ax^3 + Bx^2 + Cx + D = 0$, the product of the roots is $-\frac{D}{A}$. Thus the requested product is $$8\cdot-1\cdot \frac{-1}{1+3\sqrt3} =\frac{8}{1+3\sqrt3}\frac{1-3\sqrt3}{1-3\sqrt3}=\frac{8(1-3\sqrt3)}{1-27}=\frac{8(3\sqrt3-1)}{26}\approx 1.2911$$