Inside a particular isosceles triangle there are four congruent unit circles (radius=1) tangent along its base. The circles on the end are also tangent to the lateral sides of the triangle.
Then a circle of radius R is tangent to the two middle unit circles and the two lateral sides of the triangle.
Finally one more unit circle is placed atop the larger circle tangent to it and the two lateral sides.
What is the radius of the large circle?
Can you generalize for N circles along the base?
For comparison, the classic problem
Four Tangent Circles is a version of this one with only two unit circles at the base.
R = (1/9) (2+sqrt(7))^2 = 2.3981116938...
Call OB = P
Note, CE = CF = R
Consider triangle COG
OC^2 + OG^2 = CG^2
OC^2 + 1 = (1+R)^2
OC = sqrt(R^2+2R)
P = 1 + R + OC
P = 1 + R + sqrt(R^2+2R) eqn. 1
Now, find a second relation between P and R using three colinear tangents:
H=(3,0)
The line BH is y = -P/3 x + P
or, using the form a x +b y + c = 0:
-P/3 x -y + P =0
The distance d between a point (x1,y1) and a line is:
d = |a x1 + b y1 + c| / sqrt (a^2+b^2)
Assign d the length of CD, where C = (x1,y1) = (0, sqrt(R^2+2R) )
and the line is BH. We use R = 1 + d
R = 1 + [P-sqrt(R^2+2R)] / sqrt(P^2/9 + 1) eqn. 2
Wolfram Alpha gives the intersection of these two eqn.s at
R = (1/9) (2+sqrt(7))^2 = 2.3981116938064848...
P = 4 + sqrt(7)
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For any N in general:
We note the geometry is different for odd and even cases of N, since the large circle rests on top of a single small circle for odd N, and between two circles for even N. For N = 2,4,..., the only change above is that the denominator of the P^2 term in eqn. 2 becomes (N-1)^2. So, denom = 1,9,25,49,... respectively.
I plugged-in 25 for the N=6 (on the bottom) case. Wolfram Alpha soved the intersection without a closed form this time.
R=3.16702 (P=8.21226). It worked... See the plot here.
Looking back at Four Tangent Circles solutions, I see that using trig functions, BS determined R as the solution to a cubic. Since adding incremental baseline here is what changes with larger N, I imagine all general N solutions are cubic solutions, where the cubic coefficients have changed.
Edited on January 14, 2024, 3:30 pm
soln
