From a smooth 30-inch-diameter circular plywood disk, two smaller circular disks
of diameters 20" and 10" are cut. What is the largest circular disk that can be cut
from one (either) piece of the remaining plywood?
Descartes' Kissing Circles Formula says for curvatures a, b, c, d (= reciprocals of the radii), there is a "close relatonship"
. . (a^2 + b^2 + c^2 + d^2) = (1/2)(a + b + c + d)^2. .The curvature is negative if you're inside the circle.
Descartes' formula comes from 1638; Fredrick Soddy's poem about it came 293 years later
With this relation in mind, it's a simple enough matter to plug in the known and find the unknown:
Known radii : 5 , 10 , and -15 (neg sign for concave big circle), Unknown radius : r.
Curvatures : 1/5 , 1/10 , -1/15 , a = 1/r. Formula of Descartes:
[(1/5)^2 + (1/10)^2 + (-1/15)^2 + a^2] = (1/2)[(1/5 + 1/10 - 1/15 + a)^2]
(1/25 + 1/100 + 1/225 + a^2) = (1/2)(1/5 + 1/10 - 1/15 + a)^2 . . mult both sides by 900 = 30^2:
36 + 9 + 4 + 900 a^2 = (1/2)(6 + 3 - 2 + 30a)^2 ==> 2(49 + 900 a^2) = (7 + 30a)^2 ==>
98 + 1800 a^2 = 49 + 420 a + 900 a^2 ==> 900 a^2 - 420 a + 49 = 0 ==> (30 a - 7)^2 = 0
So a = 7/30 giving r = 30/7 ; the diameter of the circular disk is d = 60/7 = 8 4/7 inches.
The double root means the two circles are the same size, normally there's a 'big' and 'small' solution
León-Sotelo
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De Ignacio Larrosa Cañestro viene esta generalización:
Las curvaturas ki (i = 1 .. n+2) de n + 2 esferas n-dimendionales son ki (i
= 1 .. n+2), mutuamente tangentes estan relacionadas por
n*Sum((ki)^2, i, 1, n+2) = (Sum(ki, i , 1, n+2))^2
Donde las curvaturas, inversos del radio, se toman negativas para la
circunferencia en que la tangencia es interior.
En el caso de 4 circunferencias en el plano
2(k1^2 + k2^2 + k3^3 + k4^2) = (k1 + k2 + k3 + k4)^2
Es decir en eun espacio de dos dimensiones(el plano) tenemos n+2=4 circulos tangentes que nos conducen a la fórmula de Descartes.
En un espacio de 3 dimensiones tendremos 3+2=5 esferas tridimensionales besandose,en el espacio de cuatro dimensiones tendriamos 4+2=6 hiperesferas besandose etc...
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