A simple trigonometry problem: Put eight coins around a central coin

[Charles Link]

TL;DR Summary

A simple trigonometry problem: Put eight coins around a central coin. Compute the required ratio of the radii of the inner coin to the outer coin(s).

I have a simple trigonometry problem. I thought of making this as one of the math challenge problems, but it is almost too easy for that. $\\$ It is well known that 6 coins (circles) of equal size can be put around a center coin of the same radius, with the outer coins each touching two other outer coins plus the center coin. The problem is to compute with outer coin of radius "a" and inner coin of radius "b", what is the ratio of $\frac{b}{a}$, if there are 8 coins of radius $a$ around the center coin of radius $b$? $\\$

Once you have solved for the answer, you might find it of interest that 8 U.S. pennies fit almost (but not quite) perfectly around a JFK 50 cent piece. If you google the sizes, you will see why.

$\\$ Just for the fun of it, I welcome any high school level students to post an answer. $\\$ My post here is of a different nature than what I normally post= I'm hoping the section I picked is suitable for this post= @fresh_42 Hopefully this one is ok.

 

[Charles Link]

For coin sizes, see https://en.wikipedia.org/wiki/United_States_Mint_coin_sizes
$\\$ And a hint to the above problem: It is a simple application of the law of cosines. $\\$ I will post the solution in a day or two if no one else posts it.$\\$
Edit: 30.6 mm/19.05 mm=1.606. As we see below, (post 3), the answer is 1.613 (approximately) so the JFK half dollar is just ever so slightly smaller than necessary to put the 8 pennies around it.

 

[Charles Link]

And here is the solution: Let $a$ be the radius of the smaller outer coins, and $b$ the radius of the coin in the center. By the law of cosines $C^2=A^2+B^2-2AB \cos{\theta}$.
Here $\theta=45^{\circ}$ because the circle gets divided in eight parts.
$C=2a$ , and $A=B=a+b$.
That makes $4a^2=2(a+b)^2-2(a+b)^2 \frac{\sqrt{2}}{2}$.
This gives $\sqrt{\frac{4}{2-\sqrt{2}}}-1=\frac{b}{a} \approx 1.613$.

 

[Charles Link]

Here's the 8 pennies around the JFK half dollar. The last penny pushes the others out ever so slightly when it is put into place.

 

[Charles Link]

a follow-on to this: Today I tried similar to the above, but doing it with twelve smaller circles instead of 8. The result I got once it is reduced to its simplest form is $b/a=\sqrt{3}$ .

Edit: This last result I determined to be incorrect. I made an arithmetic/algebraic error. The correct answer for this is instead $b/a=\sqrt{6}+\sqrt{2}-1$.

@BRUCE A RATCLIFFE You might find this thread of interest.

 

[Charles Link]

The algebra for the above is kind of simple: $b/a+1=(4/(2-\sqrt{3}))^{1/2}=2(2+\sqrt{3})^{1/2}=\sqrt{2}(4+2 \sqrt{3})^{1/2}$.
Now $(4+2\sqrt{3})^{1/2}=\sqrt{3}+1$, so that the expression simplifies somewhat. My error above (in post 5) was to have the leading 2 in the top line inside the parenthesis. In any case, I'm glad I spotted my mistake, per the post above.

 

[OmCheeto]

It is a simple application of the law of cosines.

Since I'm not familiar with the law of cosines I just looked at the problem and solved it the best I could. It appears I've generated a universal solution for the ratio of the radii of the outer circles to the radius of the inner circle.

r = R sin ( π / n ) / ( 1 - sin ( π / n ) )

where
r is the radii of the outer circles
R is the radius of the inner circle and
n is the number of outer circles

The only assistance I got was how to calculate the length of the chord of a circle: chord length = 2 ( R + r ) sin ( theta / 2 )

where theta is the angle made by the centers of two adjacent outer circles and the center of the inner circle.

 

[Charles Link]

@OmCheeto The chord length $a=2r$, so that $(R+r) \sin(\theta/2)=r$, which then gives

$\frac{R}{r}+ 1=\frac{1}{\sin(\theta/2) }$ which is basically the same thing that the law of cosines gives if you convert the half angle formula to a full angle.
Note: $\cos(2 \theta)=\cos^2(\theta)-\sin^2(\theta)=1- 2 \sin^2(\theta)$,
so that $\sin(\theta/2)=(\frac{1-\cos(\theta)}{2})^{1/2}$.

If you do a little algebra on your universal formula, you get $\frac{R}{r}=\frac{1}{\sin(\pi/n)}-1$ where $\frac{\pi}{n}=\frac{\theta}{2}$,
so I agree with it. Very good. :)

I do recommend you read up on the law of cosines=it can be very handy at times.

Note: Your $R$ is my $b$ above, and your $r$ is my $a$. You used an $a=2r$, different from my $a$.

 

[SSequence]

Here is how I thought about this problem, which I suppose is just a minor variation on the usual methods (or basically just a slightly different way of looking at the same thing).
Edit: I noticed that this post is very similar to solutions posted in some of the posts above.Let's first consider the problem of having all coins with same radius $r$. We have a circle-1 of radius $r$ with centre coordinates $(-r,0)$. We have a circle-2 of radius $r$ with centre coordinates $(r,0)$. Now we consider a circle-3 of radius $r$ with centre coordinates $(0,h)$. Here $h$ is some positive real number. Now what we want is to find $h$ in terms of $r$ such that circle-3 intersects with circle-1 and circle-2 at only one point.

In particular, consider circle-1 and consider a line segment (say $AB$) joining centre of circle-1 to centre of circle-3. That is, the end points of the line segment are $(-r,0)$ and $(0,h)$, which we denote as $A$ and $B$ respectively. Now if we also include the origin $O$ then we have a right angled triangle AOB. The main thing we can note is that if circle-1 and circle-3 intersect at only one point then we must have length of the hypotenuse as $2r$. The length of the other two sides will be $r$ and $h$. So we get the equation:
$(r)^2+(h)^2=(2r)^2$
Solving for $h$ gives:
$h=\sqrt{3}r$
I think if we work out the angle $OAB$ it should turn out to be 60 degrees for this working to be correct. Further the angle $OBA$ should be 30 degrees.Now let's consider the case where the radius of circle-3 is $R$. In this case we will have coordinates of points $A$ and $B$ as $(-r,0)$ and $(0,h)$ again. So we should have the following equation now:
$(r)^2+(h)^2=(R+r)^2$
Further if we denote the angle OAB as $x$, then it can be solved by the equation:
$2x+45^\circ=180^\circ$
Solving it gives:
$x=67.5^\circ$

Also, let's write $R/r=k$. Basically we now have the equation:
$\cos(67.5^\circ)=r/(r+R)$
Once we substitute $R=kr$ we get:
$\cos(67.5^\circ)=1/(1+k)$
Solving for $k$ seemed to give me $1.613$, which seems to agree with approximate value written in post#3.Edit2:
Few additional comments. It seems that one thing that we would probably want to check is that whether there exists a line [other than $y=0$] that satisfies the following two conditions [apparently, it seems to me that it might be necessary to check]:
(a) It passes through the centre of circle-3.
(b) It intersects with circle-1 at exactly only one point.The point here being that by reflecting circle-3 along the above line we get another one of our required circles [just as we would get circle-2 from circle-1 by reflecting it along the line $y=0$]. We would then want to make sure we can iterate this process.Edit3:
Briefly, here is something that I found reasonably instructive while thinking about this. Why is it that for the case of $n=3,4,5$ outer circles the radius of outer circles is bigger than the inner circle but for all $n \geq 6$ we get the outer circles having equal or smaller radius.

It seems to me that one qualitative way of looking at this is to consider two circles (of unequal radii) intersecting at only one point. Now consider the two lines that are:
(a) tangent to the bigger circle
(b) pass through the center of smaller circle.

Then the angle between the lines mentioned above can't be made equal to or lower than 60 degrees. That seems to explain why for all $n \geq 6$ we can't have the outer circles with bigger radius than the inner circle.

 

[Charles Link]

The other day I just worked one more simple case of this problem=the case of 3 larger circles around a (small) center circle. This one has the result that the ratio of the radii $\frac{a}{b}=3+2 \sqrt{3} \approx 6.464$.