(8) - Incoming and Outcoming

If we consider the two perpendicular lines as axes, the family of circles that are tangent to both have equations $(x-r)^2+(y-r)^²=r^2$ (the parameter $r$ corresponds to the circles' radii). Which of these pass through the point $(p,p)$ (the centre of the blue circle)?

Plugging in, we have $(p-r)^2+(p-r)^²=r^2$ , or $r^2-4pr+2p^2=0$ .

This is a quadratic in $r$ . The sum of the two solutions is $4p$ (by Vieta).

But clearly the two solutions are $m$ and $n$ . So $m+n=4p$ , and the answer is $\boxed4$

Consider a standard unit circle with its circumference tangent to the $x$ -axis and $y$ -axis. Let the shortest distance from the origin $O$ to the circumference be $a$ , then we note that $a = \sqrt 2 -1$ .

• For the orange circle with radius $m$ , the corresponding distance $a_m = (\sqrt 2-1)m$ $\implies m = \dfrac {a_m}{\sqrt 2-1}$ .
• And we note that $a_m = (2+\sqrt 2 -1)n = (\sqrt 2+1) n$ $\implies n = \dfrac {a_m}{\sqrt 2+1}$ .
• Similarly, $a_m = (1+\sqrt 2-1)p = \sqrt 2 p$ $\implies p = \dfrac {a_m}{\sqrt 2}$ .

Therefore, $\dfrac {m+n}p = \dfrac {\frac {a_m}{\sqrt 2-1}+\frac {a_m}{\sqrt 2+1}}{\frac {a_m}{\sqrt 2}} = \dfrac {\sqrt 2 +1 + \sqrt 2-1}{\frac 1{\sqrt 2}} = 2\sqrt 2 \times \sqrt 2 = \boxed 4$ .