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 $\frac{a}{b}=3+2 \sqrt{3} \approx 6.464$.

 

[Charles Link]

I looked at this problem again the other day and one thing that might be worth mentioning is that it can be solved by using the law of sines instead of the law of cosines. For the law of sines, you can do it in two different ways: you can use the angles of 90 degrees and 22.5 degrees, with opposite sides of $a+b$ and $a$, or you can use angles of 45 degrees and 67.5 degrees, with opposite sides of $2a$ and $a+b$.

One solution above basically used the 90 degrees and 22.5 degrees, where $\sin(90^{\circ})=1$.

For completeness, I thought it might be worth mentioning that the law of sines works almost as well as the law of cosines for this problem, but it involves computing the sine of the half angle of 45 degrees, or the cosine of the half angle of 45 degrees (i.e. $\sin(67.5^{\circ})=\cos(22.5^{\circ})$).

For the case of using 90 and 22.5 degrees, it is basically the definition of sine, that $\sin(22.5^{\circ})=a/(a+b)$, but this same formula is the law of sines with $\sin(90^{\circ})= 1$ simply not having an active role in the division of $( a+b)/(\sin(90^{\circ})$.

 

[Charles Link]

Today I had someone ask me if I could solve this one in 3 dimensions by putting spheres around a center sphere. I think I have it correct=I had to google this part=that there are 5 regular polyhedrons of 4,6,8, 12, and 20 faces. This should make it so we only have a choice of these 5 number of spheres around the center sphere. I proceeded to work out the case for 12 outer spheres, each with a radius of $a$, when the inner sphere has a radius of $1$.

The solid angle formed by the cone with apex at the center of the center sphere and encompassing one outer sphere must be $\pi /3$ steradians. This gives upon using $2 \pi \int \sin{\theta} \, d \theta$ that the cosine of the half angle $\theta_o$ of this cone is 5/6.

Next $a/(a+1 ) =\sin(\theta_o)$. This gives the result that $a \approx 1.25$.

I tried testing the result by putting a dozen baseballs around a slightly smaller center ball, and the result does look to be correct. (I also need to check my arithmetic above=I didn't use a calculator, so it may need a correction in the second decimal place).

Note: Checking the above, $a=(11+6 \sqrt{11})/25$ is my exact result for $a$, which computes to 1.236 to 3 decimals.

 

[Charles Link]

With some additional thought, I realize that my solution of the 3-D case in post 12 is in error. The cones encompassing the spheres do not completely fill the $4 \pi$ steradian space, but instead, I need to work out the solid angle of the tetrahedron where the base is an equilateral triangle with sides of $2 a$, and where the other sides are length $1+ a$ and assign solid angle of $\pi/3$ to that tetrahedron. Not the simplest calculation, but I hope to update this in a day or so. Otherwise, others may also want to try it. I welcome anyone else who solves it first. $1.236$ is close, but now seen to be in error.

 

[Charles Link]

For this 3-D problem, the solid angle made by a tetrahedron is necessary, but so far I haven't succeeded in simplifying a 2-D integral over the face of a triangle that seems to lack symmetry. @PeroK Please see posts 12 and 13 above. Perhaps there is an easy way to solve this, but so far I don't see a simple way. Note: My first method of post 12 is in error. Thanks :)

Edit: See https://math.stackexchange.com/ques...ngle-subtended-by-a-tetrahedron-at-its-vertex

I was able to compute the solid angle of the tetrahedron from this, but when I assumed there were 12 of these building blocks when there are 12 balls around a central ball, I think this may not be the case. I did a sketch of the configuration, and if I counted them correctly, there are 15 and not 12. This comes as a surprise=I need to check this more carefully, but the value for $a$ gets a much more consistent result when I use 15. When I used 12, $a$ computed to be around 1.55, but is somewhere near 1.2 when 15 is used. I'm still working on it, but hopefully these results are indeed correct. Edit: I'm still at the drawing board=there are a dozen outer balls, and each one has 5 triangles it interacts with, giving a count of 60, but each one will get counted 3 times, so there are apparently 20 distinct triangles, and not 15.

Edit/update: Things are much more consistent now, using 20. I get that $a \approx 1.15$, and this is now consistent with what I got incorrectly in post 12 where I overestimated the solid angle subtended by the spheres: It is something less than $4 \pi /12$ because the cones do not fill the $4 \pi$ space. Thereby the 1.236 is too large, and the number for $a$ needed to be less than that. If time permits, I may post the details of this solution in the next day or two. :)

 

[Charles Link]

and a follow-on to the above: The 20 triangles that result from using a dozen outer balls around the central ball is apparently why there is a 20 sided regular polyhedron. If you place a ball at the center of each of these triangles, you do not get the closed packed arrangement that you get with a dozen outer balls if my assessment is correct, so that I tend to believe that one dozen outer balls may be the largest number you can have with a closed-pack arrangement. I welcome any and all feedback. :)

Edit: and it is very easy to miscount them, but there clearly are 20 triangles. This can be seen from my baseballs in a bucket, where I have one at the south pole and wrapped a little plastic around and below it. Then on the next layer there are 5 balls in a ring just above and to the outside of it. This makes for 5 triangles of 3 balls each. We can then do the same thing at the north pole, at the top of the bucket. The two sides fit together, but rotated 36 degrees, and ten triangles each with 3 balls result when the two halves are merged. This is even easy to see in a sketch in two dimensions.

 

[OmCheeto]

This is starting to remind me of @garrett Lisi's 'theory of everything' paper. I wonder if this is how he got started on the idea. Not that I understood a single word of his paper, nor how to begin to solve this problem.
Gad Zooks!

 

[Charles Link]

@OmCheeto I needed to use a formula for the solid angle for a tetrahedron from the link that is mentioned in post 14. (I could not have solved that on my own). I then set this equal to the $4 \pi /20$. This gave me, after a lot of algebra including trigonometric simplification of their formula, a 6th order polynomial equation for $a$. I already knew from my inexact estimate from post 12, the approximate value for $a$, (somewhere near 1.2 ). I think it is called Newton's rule for the first order Taylor series that allowed me to get a more precise solution, without really needing to solve the 6th order polynomial equation in its entirety. My final result is that $a \approx 1.13$.

Edit: One additional note: Using the formula from the Stack Exchange was kind of easy=the three angles at the vertex are all the same=call the angle $\theta$. The $\cos{\theta}$ is readily found from the law of cosines=just like for the case of the pennies above with side $C=2a$, and the other two sides have length $a+1$. That's how the $a$ gets into the polynomial expression.

Once I had their expression for the solid angle simplified as much as possible with the $a$ in it, I set that equal to $4 \pi /20$, and then took the cosine of both sides of the equation. The rest was all trigonometric identities and algebra. The result was a 6th order polynomial equation for $a$. (Basically repeating what I said above).

See also https://www.academia.edu/32720371/S..._vertices_Application_of_HCRs_cosine_formula_
This link comes from the Stack Exchange thread.

additional note: It is very easy to do an incorrect count on the number of triangles that you get from a dozen balls around a sphere=at first I thought it would also be a dozen, and then I later incorrectly counted 15. What I now am pretty sure is the correct answer is 20.

@PeroK I think you might find this 3-D problem in the last 5 or 6 posts of interest. One item of interest would also be how I think it was Plato or some other ancient Greek who first figured out the regular polyhedrons that are possible=simply 4, 6, 8, 12, and 20 sides. The link in post 14 from the Math Exchange is also interesting. I couldn't have worked this problem without it, (of calculating the radius $a$ of the outer spheres for a dozen balls).

See https://www.britannica.com/science/Platonic-solid

 

[Charles Link]

Just a general comment is that we must have rather limited viewing on Physics Forums these days, because I'm a little surprised that there aren't more viewers finding the 3-D problem of 12 spheres around a center and slightly smaller sphere of much interest. This latest part begins in post 12 above, and I found it to be a very interesting computation/calculation. The calculation is to find the radius $a$ of the outer spheres, assuming the center sphere has radius of $1$.

This problem beginning with post 12 is basically the 3-D counterpart of the 2-D calculation of the OP, to put 8 circles around a center circle.

Edit: and I really challenge the smarter ones out there=Can you solve for the radius $a$ of the outer spheres when there are a dozen balls around a slightly smaller inner sphere of radius $1$ ? I think very few could work this without at least a little help from a google or two.

 

[SSequence]

I think I would add one point here. It seems useful to describe it since it has not been mentioned directly in the thread yet.

Consider two circles $C_1$ and $C_2$ of radii $r_1$ and $r_2$ with their centers at points $(x_1,y_1)$ and $(x_2,y_2)$ respectively. Then the following statement is true:
"$C_1$ and $C_2$ intersect at only one point iff the distance between their centers is equal to $r_1+r_2$ or $|r_1-r_2|$."
A similar version of this statement is also true for spheres in 3D.

========

So suppose we are given a configuration of (finite) numbers spheres with their radii and coordinates of their centers specified to us. One way to test intersection is to use a 3D drawing software [which might be useful in general to build some intuition].

However, we can also verify intersections by checking for distances between the centers. For example, if we have 4 objects then there are 3+2+1=6 distances to check.

 

[Charles Link]

@SSequence Your suggestion is a good one=I found an alternative solution for $a$ that is a little simpler, using the distances between the centers, as well as a couple of the angles. I'll post the details in the next day or two, but I found $a$ satisfies $(2 \cos(\pi /5)+1) a^2-2 a- 1=0$, with the result that $a \approx 1.11$.

I think this is in agreement with my previous solution of $a \approx 1.13$, but this one is much easier to solve than the previous 6th order polynomial expression. I recomputed the solution to the 6th order polynomial, and instead of $1.13$, I'm getting very nearly $1.10$, so that the two solutions are showing to be very consistent. It might be worthwhile to plug the latest solution into the 6th order polynomial expression and show that it fits it precisely. [Edit: and I was now successful at this=the two solutions match identically]

This latest solution uses about 7 or 8 equations with 7 or 8 unknowns, but these were readily solvable. It uses the (x,y,z) coordinates of two of the balls near the equator, (one just above and the other just below=needed to solve for them=besides $a$, and one angle $\theta$, they were the unknowns), along with the coordinates of the two balls at the poles, whose coordinates are simple. The $\Delta \phi$ (azimuthal angle difference) between the two balls near the equator is known basically from symmetry to be $\pi/5$. (This was easy to see from the balls that I assembled in a bucket a week or two ago). Meanwhile $\theta$ is found from the law of cosines.

 

[fresh_42]

I have a different solution but I'm not sure whether I made mistakes, so I put it to the test here.

I assume that the radius of the larger coin at the center is $R$ and the radius of the $n$ smaller coins around it is $r.$ This led me to

where the black circle is defined by the centers of the smaller coins. $h$ is the height of the circle segment defined by the angle $\varphi =2\pi /n$ and $s$ the length of the chord. We only need the black circle to calculate them. I looked up the formulas on Wikipedia and got
\begin{align*}
s&=2(R+r)\sin\left(\dfrac{\varphi}{2}\right)\\
h&=(R+r)\left(1-\cos\left(\dfrac{\varphi}{2}\right)\right)
\end{align*}
What combines them with the red circle of a small coin is the equation $h^2+\dfrac{s^2}{4}=r^2$ and so
\begin{align*}
\dfrac{r^2}{(R+r)^2}&=\left(1-\cos\left(\dfrac{\varphi}{2}\right)\right)^2+\sin^2\left(\dfrac{\varphi}{2}\right)\\
&=2\left(1-\cos\left(\dfrac{\pi}{n}\right)\right)\\
&=\left(\dfrac{1}{1+\frac{R}{r}}\right)^2
\end{align*}
and finally
$$
\dfrac{r}{R}=\dfrac{\sqrt{1-\cos\left(\dfrac{\pi}{n}\right)}}{\dfrac{1}{\sqrt{2}}-\sqrt{1-\cos\left(\dfrac{\pi}{n}\right)}}.
$$
That means
\begin{align*}
\dfrac{r}{R}\approx 0.64 \, &, \,\dfrac{R}{r}\approx 1.563\text{ for }n=8\\[8pt]
\dfrac{r}{R}\approx 0.3533 \, &, \,\dfrac{R}{r}\approx 2.83\text{ for }n=12
\end{align*}
As already mentioned, I'm not sure whether I made a mistake somewhere.

I don't have other US coins than a silver dollar and no caliper so I couldn't test my figures.

 

[Charles Link]

Somewhat close @fresh_42 , but incorrect. The arc of radius $R+r$ does pass through the centers of all the smaller coins, but it intersects the circles of radius $r$ at a point slightly above where the circles make contact, which is found by drawing a straight line between their centers. You might want to look at the solutions earlier in the thread, where a couple of us solved it by the law of cosines as well as the law of sines.

For something even more challenging, try the similar problem in 3-D that starts around post 12.

 

[fresh_42]

The centers of all touching outer coins are at a distance $R+r$ from the center of the large coin per definition of touching. That draws my black circle. The question is whether two neighboring small coins need to touch at $P$ which I'm currently not sure about. If so, then $n\varphi = 2\pi$ defines the angle and the rest of the calculation.

 

[Ibix]

The centers of all touching outer coins are at a distance $R+r$ from the center of the large coin per definition of touching. That draws my black circle. The question is whether two neighboring small coins need to touch at $P$ which I'm currently not sure about.

I think the centers of the eight outer coins form a regular octagon. The contact point between two coins must lie on the straight line joining their centers, which would place the contact point between two outer coins on the sides of the octagon. I think your point $P$ is on the octagon's circumcircle, so would lie slightly outside the contact point of the circles.

 

[Charles Link]

@fresh_42 If you make $R$ very small, I think you will see how your diagram is misleading. In that case, the edges of the pie shape that make the angle $\phi$ will certainly pass inside the circle of radius $r$ to make contact at the point $P$.

 

[Ibix]

I think the centers of the eight outer coins form a regular octagon. The contact point between two coins must lie on the straight line joining their centers, which would place the contact point between two outer coins on the sides of the octagon.

The above observation leads directly to $\sin(22.5°)=\frac{r}{R+r}$, which solves to the ~1.613 ratio in #3.

 

[fresh_42]

I see. The location of $P$ is crucial and made it wrong. Now thinking about it, I have to agree with @OmCheeto's solution in post #7.

Here is a nice list of some explicit values:
https://de.wikipedia.org/wiki/Regelmäßiges_Polygon#Längen
(Chrome plus right-click and Translate to English works fine.)