Parabola's Golden Bridge

We have,

$y = m x + c = x^2$

which has two solutions $x_1, x_2$ with $x_1 > 0$ and $x_2 < 0$ , the corresponding y-coordinates are,

$y_1 = m x_1 + c$

$y_2 = m x_2 + c$

Hence, we have,

$AP = \sqrt{ {x_1}^2 + (y_1 - c)^2 } = x_1 \sqrt{1 + m^2 }$

$BP = \sqrt{ {x_2}^2 + (y_2 - c)^2 } = -x_2 \sqrt{1 + m^2 }$

from which,

$| AP - BP | = (x_1 + x_2) \sqrt{1 + m^2}$

but from Vieta's formula applied to $x^2 - m x - c = 0$ , we know that $x_1 + x_2 = m$

Hence, $m \sqrt{1 + m^2} = 1$

Squaring, $m^2 (1 + m^2) = 1$

Solving for $m^2$ using the Quadratic formula, we obtain

$m^2 = \dfrac{ -1 + \sqrt{5} }{2}$

Hence, $p = 5, q = -1, r = 2$ , making the answer $5 - 1 + 2 = 6$ .

Since the solution apparently must be independent of $c$ , we'll just look for that point on the parabola $x^2$ that is at a distance of $1$ from the origin. We then have the following to solve:

$\sqrt{x^2+x^4}=1$

which has as a solution

$x = \sqrt { \dfrac{\sqrt{5}-1} {2} }$

leading us to the slope of the line from the origin to that point, which is the same as $x^2/x=x$

If we do an affine transform of this parabola, plotting the following equation, we'll end up with the plot below, where the independence of c becomes obvious.

$y={ \left( kx \right) }^{ 2 }-kx\sqrt { \frac { 1 }{ 2 } \left( \sqrt { 5 } -1 \right) }$

where

$k= \sqrt { \frac { 1 }{ 2 } \left( \sqrt { 5 } -1 \right) }$

Let $A(a,a^2)$ and $B(b,b^2)$ .

Point $B$ is reflected about point $P$ giving point $B_{1}$ .

$m=\dfrac{a^2-b^2}{a-b}=a+b=B_{1} E$

$m=\dfrac{AE}{B_{1} E} \Rightarrow AE=m^2 \Rightarrow AB_{1} =AP-BP=|m| \sqrt{m^2+1}$

This means that this difference in lengths is constant for a given slope, regardless of the value of $c$ .

$|m| \sqrt{m^2+1}=1 \Rightarrow m^2 =\dfrac{\sqrt{5} - 1}{2} \Rightarrow p+q+r=5-1+2=\boxed{6}$

Note: Inspired by Is there an elegant method?