Orthogonal parabolas

Algebra Level 5

The two orthogonal parabolas $\begin{cases} y=(x+1)^2 \\ (y-2.5)^2=x+4 \end{cases}$ intersect at four distinct points, $A=(x_1,y_1), B=(x_2, y_2)$ , $C=(x_3, y_3)$ and $D=(x_4, y_4)$ . Let $S=x_1+x_2+x_3+x_4$ and $R=y_1+y_2+y_3+y_4$ . Find the value of $\lfloor{1000(R+S)}\rfloor$ .

Remark : If the orthogonal parabolas intersect at four distinct points, then these four points are always on a same circle.

This problem is part of Curves... cut or touch? .

The answer is 6000.

1 solution

Chan Lye Lee
Nov 16, 2015

It is given $\displaystyle \begin{cases} y=(x+1)^2 \\ (y-2.5)^2=x+4 \end{cases}$ .

Eliminating the $y$ , we have $\displaystyle \left((x+1)^2-2.5\right)^2=x+4$ , which yields $\displaystyle x^4+4x^3+\ldots =0$ . This means that $S=-4$ , by Vieta's Formula.

On the other hand, Eliminating the $x$ gives $\displaystyle y=\left((y-2.5)^2-4+1\right)^2$ , which yields $\displaystyle y^4-10y^3+\ldots =0$ . This means that $R=10$ , by Vieta's Formula.

Now $\lfloor{1000(R+S)}\rfloor = 6000$ .

The floor function had me thinking R+S was not an integer and that the problem was harder than it actually was.

Matt O - 5 years, 5 months ago

Orthogonal parabolas

This problem is part of Curves... cut or touch? .

The answer is 6000.

1 solution

0 pending reports