# Homework Help: Solve for unknown radius without trig

Tags:
1. Dec 6, 2016

### caters

1. The problem statement, all variables and given/known data

R = radius of outer circle = 5
r1 = radius of largest inner circle = 3
r2 = radius of second largest inner circle = 1
2. Relevant equations
C = 2πr
A = πr2
d = 2r
3. The attempt at a solution
Make a triangle with 2 sides of length r2+a and 1 side of length d(d for diameter)

Set the perimeter equal to x:

$$(r_2+a)+(r_2+a)+d = x$$

Plug in known values:

$$(1+a)+(1+a)+2 = x$$
$$5+2a = x$$
$$2a=x-5$$
$$a = \frac{1}{2}x-\frac{5}{2}$$

Substitute into original equation:

$$(r_2+\frac{1}{2}x-\frac{5}{2})+(r_2+\frac{1}{2}x-\frac{5}{2})+d=x$$

Simplify:

$$-3+2 = 0$$
$$-1=0$$

Okay something's wrong. I should be able to solve for x, not get something as ridiculous as -1 = 0 so that I can then solve the perimeter equation for a to find the unknown radius.

So what am I doing wrong here? do I need another linear equation in order to solve this?

2. Dec 6, 2016

### QuantumQuest

Why are you doing this? Is this the way to create the next outwards circle? If it is, make some drawing for this to be clear.

You are utilizing the first equation, substitute some values, find $\alpha$ and then you return to the same equation and substitute again. Does this make sense? Or you need another equation as well?

3. Dec 6, 2016

4. Dec 6, 2016

### caters

Here is what I am trying to compute:

That part of the isosceles triangle that is inside that smaller circle, that is the unknown radius a. I know it is a rough drawing but that longer side is d which in this case is 2 and that smaller side that makes it isosceles is r+a which in this case is 1+a

So I am setting the perimeter of that triangle as x.

But when I try to solve for x in order to find that unknown radius by first solving for the unknown radius in terms of x and then substituting, I get -1=0 which is obviously wrong since for 1 thing, a circle can never have a negative radius and another is that the statement itself isn't true. It is like saying that you can divide by 0 in normal mathematics. Obviously false.

The only other thing I can think of though would be either the altitude of the triangle or the area for finding that unknown radius and I don't think either one is going to be of much help.

5. Dec 6, 2016

### haruspex

Your diagram looks wrong to me. Shouldn't the circle of radius a be tangent to the outermost circle?

6. Dec 6, 2016

### caters

No because that is not the case for this kind of Apollonian gasket. It is actually the next layer that is tangent to the outermost circle

Anyway, why does that make a difference? I would still be using the perimeter of the triangle to solve for the unknown radius. So basically in this kind of Apollonian gasket, if multiple circles are touching the outside of 1 circle and going outwards from it like in the layer right next to the center circle, those are tangent circles. If it is the other way around that the same circle is touching multiple circles a layer before it like in the case of the circle with the unknown radius, that is not a tangent circle.

So while yes, the circle with the unknown radius is touching tangent circles, it itself is not a tangent circle.

And what I mean by a layer before it is the closest circles of a larger radius not taking into consideration that outer circle.

7. Dec 6, 2016

### haruspex

Ok, I think I see it.
Suppose we try to generate the diagram thus:
Draw the innermost circle.
Draw eight circles around it of the right radius to fit.
Draw the outermost circle tangent to those.
Draw the circles radius a, where a is any value less than that to make them tangent to the outermost circle.
Proceed to fill in with circles of the max possible radius everywhere.

There is a place where this procedure breaks down, and this is the key to determining a. Where is that?

8. Dec 6, 2016

There is a number of circles near the perimeter that for some reason don't touch the perimeter that has 3 circles between it and the perimeter. Perhaps there is a reason for this, but I lack an explanation for it. IMO, these circles should touch the perimeter to go along with the pattern. Otherwise, it is at the whim of where you want to include extra circles.

9. Dec 6, 2016

### haruspex

Not sure which ones you refer to. Each circle radius 'a' is tangent to 8 circles that are also tangent to the perimeter. If 'a' were made slightly smaller or larger, what would go wrong?

10. Dec 6, 2016

There are 8 such circles including the one at the center of the apex of the isoceles triangle of OP's diagram in post #4. Why aren't these circles made larger to touch the perimeter? Also, if you look more closely at the drawing, there are places where circles should touch in 3 places, but only make contact at two places with other bigger circles.(i.e. the circles that are almost adjacent to the innermost circle of the diagram. If they don't make complete contact, the diagram is very arbitrary.)

11. Dec 6, 2016

### haruspex

Because they are following some other recursive rule, which I have not fully divined.
No, not arbitrary. I made that mistake too at first. Try to answer my question in post #9.

12. Dec 6, 2016

Thank you. I'm going to need to look more closely at this one. I might be a little slow=I haven't spotted the pattern yet. :-)

13. Dec 7, 2016

### caters

Because if their radius was any larger, they would intersect with the tangent circles and if it was any smaller, they wouldn't be touching those tangent circles. That is why there are 8 non-tangent circles in both the outward direction towards the perimeter and in the inward direction towards the center circle that themselves are touching 16 more tangent circles.

14. Dec 7, 2016

I googled Apollonian gasket, and the designs vary somewhat, but for the case at hand, I can't see anything that determines the size of the circles in question. Perhaps there is a very subtle reason for the choice that they make, but so far, I haven't spotted it.

15. Dec 7, 2016

### haruspex

That does not quite do it, since ordinarily you would simply adjust those.
The key is that there are exactly six smaller circles tangent to it and to the outermost circle. That fixes the size of a, relative to the larger circles.
Note that the same happens between the innermost circle and the eight surrounding it, but how that turns into a general algorithm I have not figured out.

It occurs to me that the same principle applies to the innermost and outermost circle, and the eight tangent circles they share. The relative radii of these ten can be computed from that, but these do not match to the given ratios. They involve cosec(π/8). Where did you get the radii 1, 3 and 5 from?

Last edited: Dec 7, 2016
16. Dec 7, 2016

@haruspex Good observation that the 1,3, and 5 given in the definition of the original problem were incorrect. I computed the ratios (using the law of cosines, etc.), and got for the second and 3rd largest radii that $\frac{r_2}{r_3}=\sqrt{2(2+\sqrt{2})}-1$ with the largest radius $r_1=r_2+2 r_3$. If we let $r_1=5$, this would make $r_2=2.22$ and $r_3=1.38$ (approximately).

Last edited: Dec 7, 2016
17. Dec 8, 2016

### caters

The radius of the outer circle and inner centered circle were given and then all I had to do was subtract 3 from 5 giving me a diameter of 2 and thus a radius of 1. I knew that if I subtract the radius of a centered circle from the radius of a bigger circle, I get the diameter of the largest tangent circle and that the sum of all the different radii(no repeats so if same radius appears more than once in a given layer it is considered as 1 radius) converges to the radius of the outer circle just as the area sum converges to the area of the outer circle.

But I still don't see how all this talk about patterns within circles is going to help solve for a, the unknown radius. Really, this should be as simple as solving for a in terms of x, substitution, solving for x, substitution, and then solving for a once more with x having been solved for. That all is just algebra. I don't see how algorithms to make the Apollonian gasket or your mention of $cosec(\frac{π}{8})$ is going to help find the perimeter of the triangle in my diagram(which is key to solving for the unknown radius with algebra alone.

Last edited: Dec 8, 2016
18. Dec 8, 2016

To put 6 circles of radius one around another circle will have all of the circles including the center one to have a radius equal to one. To put 8 circles of radius one around another circle will require that $r_2/r_3$ is as given above (post #16). That is actually a simple calculation involving the law of cosines to determine the ratio $r_2/r_3$. You can set $r_3=1$ and you find $r_2=1.61$ (approximately) and is not equal to 3. You can not put 8 circles of radius 1 around a circle of radius 3 and have them all touching. @haruspex was the first to notice this incorrect result in the original post. $\\$ The radius of the next circle $r_4$ is still undetermined, at least for me, because I don't see the pattern that is being applied. Normally these circles are all made tangent to the outer circle as is seen in my post #3 and also if you google Apollonian gasket. Somehow this pattern is a modified Apollonian gasket, and the packing to me still appears arbitrary for $r_4$.

Last edited: Dec 8, 2016
19. Dec 8, 2016

### haruspex

In the original numbering, that is r3. The outermost is just R.
Suppose we draw it in at some radius, smaller than needed to touch R.
We can now draw in two circles tangent to R, r2 and r3. These do not touch other in the diagram. Next we draw in circles each tangent to one of those and to R and r3. The diagram shows these just happen to touch other, no overlap, no gap. Clearly the value of r3 to achieve that is unique. Calculating it is another matter.

@caters, are you familiar with the equation that relates the radii of four mutually tangential circles?

20. Dec 8, 2016

### caters

I am pretty sure that for 2 given radii, 1 for the outer circle and 1 for an inner centered circle that the radius of the largest tangent circle(the one that is tangent to both the center circle and the outer circle) it is simply the difference between the 2 radii divided by 2 so I am pretty sure that for the largest tangent circle, the formula for the radius would be this:

$$\frac{R - r_1}{2}$$

21. Dec 8, 2016

### haruspex

You are missing the point. The given diagram cannot be drawn with the radii specified. Try drawing the centre circle and its eight largest neighbours with those radii.

22. Dec 9, 2016

@caters It may interest you, if you try the result with some coins, you can take 6 pennies and make a ring around the center penny. If you want to use standard coins and make a ring of 8, I found the penny is almost the right size for a 50 cent piece in the center, but the pennies are perhaps about a thousandth of an inch too large to make perfect contact in the ring of 8. Calculations I did by googling the size of the coins support this experimental find. (penny=.750" diameter; 50 cent=1.205" diameter.) I need to actually refine my $r_2/r_3 =1.61$ calculation above (see posts #16 and #18) to get a more precise answer, but I don't have a calculator handy.