Brain Teaser 2 - Unusual challenge

[Pekis]

Hello Franz,

Well done !

Your solution is a fast & straight one ... and mine was rather unusual:

I used the graphic intersection of the so-called parallel curves.
Except in the case of a line or a circle, a parallel curve is a rather heavy polynomial => the parallel curve of a parabola is not a parabola !
But if the curves are expressed in parametric form, the world becomes simple:
For a curve defined as (f(t),g(t)), the parallel curve at a given distance d (-d for the other side) is (f(t)+d*g'(t)/sqrt(f'(t)^2+g'(t)^2),g(t)-d*f'(t)/sqrt(f'(t)^2+g'(t)^2))

I had then to find the parametric form of the parallel curves of y=x^2+1, y=x-2, x^2+y^2=2 and find the value of d (with same sign for all curves) which makes them intersect simultaneously.

y=x^2+1:
Original curve: (t/2,t^2/4+1)
Parallel curve: (t/2+d*t/sqrt(1+t^2),t^2/4+1-d/sqrt(1+t^2))
(d>0 towards the intersection)

y=x-2:
Original curve: (2*t,2*t-2)
Parallel curve: (2*t-d/sqrt(2),2*t-2+d/sqrt(2))
(d>0 towards the intersection, but had to change the sign of d)

x^2+y^2=2:
Original curve: (sqrt(2)*cos(t),sqrt(2)*sin(t))
Parallel curve: ((sqrt(2)+d)*cos(t),(sqrt(2)+d)*sin(t))
(d>0 towards the intersection)

And here is the result on the impressive free online grapher desmos, with my graph

I used the slider to change d and refined it with the appropriate zoom level, and voilà !



Still using the slider, one can easily find another solution around x=-12 (distance around 10.6). Franz, how would you transform your system to also find it ?

I had much fun on my way to the solution, like Steve McQueen on the way to San Mateo . You too ?

Thanks to Carl & Franz