Here's a puzzle I've come across that I don't know how to solve. Any guidance is appreciated.?

Question

There is a large circle with radius R. Inside this circle is placed another circle with radius r (< R) so that the two circles are concentric. Now, in the region between the circumferences of the two circles, place N circles (of diameter R-r) such that each small circle has one and only one point of contact with each large circle and with two adjacent small circles.

Basically, imagine that this is like a bearing system.

Is this possible? If so, what is the radius of the smaller of the two large circles (r) in terms of the radius of the largest circle (R). Also, how many "small" circles can fit in the space between the two "large" circles?

I hope this is clear. I thought about this puzzle earlier today and can't figure out how to solve it. I guess that's what 15 years of no formal math can do to a person... *sigh*

It should be clear but I'll state for formality that N is a whole number.

Answer 1

Consider two of the smaller circles tangential to each other.
Now form a triangle between the centers of these two circles and the center of the larger circles.
The sides of this triangle are (R+r)/2 twice and R-r.
Let θ be the angle between the two equal sides.

You can use the cosine rule to obtain:
sin(θ/2) = (R-r)/(R+r) = (1-r/R)/(1+ r/R)
θ = 2*arcsin [(1-r/R)/(1+ r/R)]
N = no. of small circles that can fit
N = 2π/θ
= π / arcsin [(1-r/R)/(1+ r/R)]

eg. if r/R = 0.07180, exactly 3 smaller circles will fit
if r/R = 0.52786, exactly 10 smaller circles will fit.
if r/R = 0.9005, exactly 60 smaller circles will fit.
As this ratio tends to 0, N tends to 2.
As this ratio tends to 1, N tends to infinity.
So there is no upper bound on the number of circles that can fit.

Answer 2

Yes, this is very possible. But there is no single answer. It is dependent on the number of bearings that you have or the specific values of R and r.

In the simplest case, image that you had exactly 3 ball bearings. This would make the inner r rather small compared to the outside R. The 3 bearings would touch and their centers would form a triangle. There would be an inscribed circle r and circumscribed circle R.

Now image four bearings (centers forming a square). Again you can have an inscribed circle and circumscribed circle.

You can repeat with 5 forming a pentagon with their centers.

At the point when you get to 6 you will find that the radius of the bearings is equivalent to the radius of the inner circle r.

You can continue with smaller bearings and a larger r pretty much indefinitely.

I disagree with those trying to use the circumference of the circle between, or the area between because neither of those account for the space between the ball bearings. In the case of the middle circle circumference, the centers actually cross through the tangent points in a polygon, not a circle.

Similarly, when you consider the area, there is a lot of space in between that you can't fit more bearings into.

I'm still trying to figure out the relationship, but I think it might be easier to imagine a fixed value for R, and a specific value of n. Then you could figure out r = f(n) for each value of n from 3 upwards.

That would probably lead you to a generalized formula that you could then solve for n = g(r).

Answer 3

You have reasonably well clarified your problem!

People usually follow problem solving culture! You too have done it.

Another manner is setting problems in a manner solution of these become easy. It is problem setting!

You can move backward from 'need' to a point where you began to imagine ( accommodating small circles inside the space of two concentric circles)!

1) fix diameter of small circle that fit inside the space of two concentric circles!
2) decide number of said circles
3) Apply polygon relations relating said number of circles!
4) Find radius of circle where all small circle centers lie! It is most probable that said radius is not whole number!
5) Now you can fix small circle and big circle as you had imagined which is not a need!

Your problem has been solved in a different manner!

Remember that forward as well as backward computing is possible and in some occasions backward computing (as shown ) will work far better!

Regards.

Answer 4

So you want an integral number of ball bearings of diameter
R-r in the races of the two rollers. The point of contact between the ball bearings then-- it's a circle in between the
inner and outer races. But does it fall on the center of the
ball bearings? If it does, 4*R/R-r equal line segments form a
regular polygon connecting the points of lateral contact and center pointsLets look at some examples. Say r=0, then2 ball bearings of diameter R, fit in race R. They touch nowhere laterally and a circle of radius R/2 connects their centers. What about 3 packed ball bearings. What size circle would they enclose? First of all You can see that the
line of contact is an equilateral triangle each leg of which is
a chord of the ball bearings not a diameter. So in general,
the circle of the centers will lie outside the circle of the con-
tact points. To the problem. Use the cosine law to figure
out the radius of the inner and outer races given the ball size in the three size x ball case:
(2x)^2=(2r)^2-(2r)^2*cos(120)
(r)^2(1-cos120)=(2x)^2; (r)^2=(x)^2/(1-cos120)
r=x/sqrt(1-cos120). then x/sqrt(1-cos120)-x=radius of inner
race. x/sqrt(1-cos120)+x = radius of outer race.
For 4 size x ball bearings x/sqrt(1-cos360/4)+/-x and so on.

Answer 5

Well, with no formal math experience either, here's my (un)educated guess:
- Each of the new circles should have a diameter equal to the difference of the 2 radii (R-r)
- So, imagine a 3rd large circle that sits half-way between the two original large circles - so its radius should be r + (R-r) / 2
- Calculate the circumference of this imaginary circle.
- Now, divide that circumference by the diameter of the new circles.
- Drop off any remainder and you'll have a whole number equaling the maximum number of small circles that can fit.

I guess I could reduce that to a formula, but my brain hurts!

Answer 6

Its' a tough question. Here's the answer :

N = Floor [ 4(R^2 - r^2) / (R-r)^2 ]


So how did I come up with the magical answer....

Area of Large circle = (Pi)R^2
Area of small circle = (Pi)r^2

Difference of area = (Pi)(R^2 - r^2)
Now the situation you described (of many small circles between the 2 circles is physically possible), requires the difference in area of the 2 concentric circles to be some multiple of the area of the small circles between the 2. Since the sum of the areas of these circles won't exactly equal this difference, you have to use the floor function.

You know the area of one of the small circles is =
(Pi)((R-r)/2)^2, so then...

N * (Pi)((R-r)/2)^2 = (Pi) (R^2-r^2)
solving for N gives you

N = Floor [ 4(R^2 - r^2) / (R-r)^2 ]

----

Now if you want to solve for r ... you will have to get the above equation into a format in which you can solve for r quadratically:

this will simplify to :
NR^2 - 2NRr + Nr^2 + r^2 - 4R^2 = 0

or more specifically:

r^2(N+1) - r (2NR) + (NR^2 - 4R^2) = 0

so solve the quadratic equation with parameters
a = N + 1
b = -2NR
c = NR^2 - 4R^2

to get your value of r given that you have R and N

Answer 7

Condition for"filled tangency" : Central angle determined by
two adjacent filler circle centers and big circle center is
1/N of a circle (360). Call this angle a. Then doing the trig
gives a = arccos {1 - (R - r)^2 / 2r^2 } = 360/N for integer N.
Thus R and r have to be just right.

Answer 8

The centres of the small circles lie on a circle of radius r + ½ (R - r) = ½ (R + r)

the circumference is pi (R + r) = n (R - r)

Answer 9

i know the answer!

r=R{[1-cos (pi/2+pi/N)] / [1+cos (pi/2+pi/N)]}
and
N=180*/ (90* - cos^-1 [(R-r)/(R+r)] )

from your Q, it's obvious the small circles build N-gon with sides (R-r) when their centers connected.

if you take out 1 slice of N-gon(like a pizza's) and cut it into 2(so from an isosceles we have a right tri), the hypotenus is [(R-r)/2 + r] and the side that was connecting small circles is [(R-r)/2].


edit
but maybe using sine is much easier

[(R-r)/2] / [(R-r)/2 + r]
=sin (360/2N)
=sin (pi/N)

R-r=(R+r) sin (pi/N)

r=R[(1-sin (pi/N)) / (1+sin (pi/N))]

N=180 / sin^-1 [(R-r)/(R+r)]