Dynamic Geometry: P12

The two parts of $y=\dfrac 1x$ curve is symmetrical about the straight line $y=-x$ . Therefore the center of the circle $Q$ is along $y=-x$ . Let the tangent of the positive $y=\dfrac 1x$ and the circle be $P \left(u, \dfrac 1u\right)$ . Then the tangent at $P$ has a gradient of $\dfrac {dy}{dx} = - \dfrac 1{x^2} \ \bigg|_{x=u} = - \dfrac 1{u^2}$ . The gradient of the normal at $P$ is $u^2$ and the equation of the normal is $\dfrac {y-\frac 1u}{x-u} = u^2$ , $\implies y = u^2x -u^3 + \dfrac 1u$ . Then the point $Q (x_q, y_q)$ is on this line where $y=-x$ . Then we have

$- x_q = u^2x_q - u^3 + \dfrac 1u \implies x_q = \dfrac {u^3 - \frac 1u}{u^2+1} = \dfrac {u^4-1}{u(u^2+1)} = \dfrac {u^2-1}u = u - \dfrac 1u \implies y_q = \dfrac 1u-u$

The radius of the circle is:

$r = \sqrt{(u-x_q)^2+\left(\frac 1u-y_q\right)^2} = \sqrt{\left(u- u+\frac 1u \right)^2+\left(\frac 1u- \frac 1u +u\right)^2} = \sqrt{u^2+\frac 1{u^2}}$

The width of the triangle $w=PP'$ , where $P' = \left(-\dfrac 1u, - u \right)$ . Then:

$w = \sqrt{\left(u+\frac 1u\right)^2+\left(\frac 1u+u\right)^2} = \sqrt 2 \left(u+\frac 1u\right)$

The height of the triangle $h=MQ$ , where $M$ on the line $y=-x$ is the midpoint of $PP'$ . Hence $M = \left(\dfrac {u-\frac 1u}2, \dfrac {\frac 1u-u}2\right)$ , Note that $M$ is also the midpoint of $(OQ)$ , where $O(0,0)$ is the origin. Then $h= MQ = \dfrac {OQ}2 = \sqrt 2 \cdot \dfrac {u-\frac 1u}2 = \dfrac {u-\frac 1u}{\sqrt 2}$ .

Then the area of the triangle,

$\begin{aligned} \frac {wh}2 & = \frac 7{24} \\ \frac 12 \cdot \sqrt 2 \left(u+\frac 1u\right) \cdot \frac {u-\frac 1u}{\sqrt 2} & = \frac 7{24} \\ u^2 - \frac 1{u^2} & = \frac 7{12} \\ 12 u^4 - 7u^2 - 12 & = 0 \\ (3u^2 - 4)(4u^2 + 3) & = 0 & \small \blue{\text{Since }u > 0} \\ \implies u^2 & = \frac 43 \\ \implies r & = \sqrt{u^2+\frac 1{u^2}} = \sqrt{\frac 43 + \frac 34} = \frac 5{2\sqrt 3} = \frac {5\sqrt 3}6 \end{aligned}$

Therefore $a+b+c = 5 + 3 + 6 = \boxed{14}$ .