I found a general equation for a circle tangent to the other two (geometrically, using Pythagoras) and then used a discriminant argument to see when this touched the parabola. Which is messy, but gives a very nice radius in the end! It makes me think there must be a gentler way...
Building on the previous and going back to basics, an incircle stack in the parabola y=x^2 has successive radii which differ by 1. Also an incircle radius has a simple relationship to the circle centre and also the tangent points which are 1/2 beneath the y center ... www.desmos.com/calculator/n...
Here there are blue and green incircles of radii 3 and 4 with centers at 9 1/4 and 16 1/4 and a common horizontal tangent at 12 1/4. The third (black) circle has a tangent point at 12 (previous relationship) so it generates an incircle with centre at 12 1/2 and radius 3 1/2, an "average". ....
The radius of the black circle is 1/4 of the radius of the orange circle, so the black circle has a diameter which passes through the tangent point, the center and the horizontal tangent and the diameter is 1/2 the 7/2 radius. It explains why the numbers are relatively simple, once you have X.
With a bit of scribbing for a gentler way, expressing the green circle centre/radius locus in terms of y (other post) and then finding the distance from the parabola (X,X^2-1) and equating distance and derivative gives the result but computation heavy. Subbing the derivative into the distance. ...
essentially gives (Dy)^2 sin^2 = r^2 which is just simple trig at the tangent point on parabola. Its a quadratic in y so looking at the y^2 coeff, its zero when 4X^2 - 4n^2 + 1 = 0 so a single solution/tangent point at X = sqrt(n^2-1/4) The discriminant of the quadratic (function of x) is >= 0 ..
and = 0 at X=n which isnt a valid. So the "interesting" thing about this solution is that with a fair bit of work it comes out as a quadratic (in y), and the single solution corresponds to the y^2 coefficient being zero which is reasonably efficient, rather than arguing about the discriminant.
Went through the incircle stack in the parabola in y=x^2. The incircles have radii r = (2n-1)/2 = 1/2, 3/2, 5/2, ... and their horizontal tangents are n^2 = 1, 4, 9, .... The green circles have radii r = n/4 and centres (x,y) = (3/8 sqrt(4n^2-1), n^2-1/8) and the stuff in the other posts
Final(?) thing is the two circles surrounding point tangent are 1/4, 2 1/4, 6, 1/4, 12 1/4 height from directrix / distance from focus. So subracting 1/4 and averaging the two surrounding points gives the 1, 4, 9, ... tangent point(s) Next post ...
Obv very spedulative, but thinking it of a sequence of a 3 circle problems on a parabola base rather than the classic 3 circles on a straight line common tangent base might give something. math.stackexchange.com/questions/39...
Another thing that seems to hold is that the "hard" tangent point(s), so the one between the green circle(s) and the parabola are a distance 1, 4, 9, ... from the focus / vertical height from directrix. Id guess it would be easier to verify these relationships than derive them (this question).
I seem to remember the odd ukmt? question like this, but if you stack up the touching circles in the parabola you get something like.... So generalising from a rough simulation with 3 points, the "third circle radius" forms a linear sequence 1/4, 1/2, 3/4, ... www.desmos.com/calculator/v...
Where did you get your equation that you could discriminant? (I'm sure that's a verb). I did this but then had to use a repeated root/equating coefficients path to get to the value because I had a quartic with an x term.
Sorry! Think I had misremembered .. Looking at my messy notes it was a comparing coefficients job... but the final answer was nice!
No problem, just wanted to be sure I wasn't missing something 😃