Spending the weekend folding polyhedra and finding convex internal angles

So I’ve managed to resolve the circle construction question from a few posts back and determined that any number n can be constructed by a fairly small number of operations; more precisely the 2-logarithm the number but I’ve put off putting together a good description of the proof as I’ve been preoccupied with school, work, and another new mini-project: Polyhedra.

I originally got the idea that I should look back into Polyhedra after I overheard some students mentioning the concept of the Euler characteristic of a graph (V + F – E) and recognizing that I remembered very little of that particular theory I set to refreshing on the theory and making lots of paper polyhedra, drawing Hamiltonian paths, and colorings and so on as I stumbled over them.


As I reviewed the E + F – E = 2 proofs I also had to recognize that I had some gaps in my knowledge regarding the formal existence of triangulation of polygons and continued on the merry chain of ‘better check that’ untill I at the bottom of the chain ended up with

Every polygon has a strictly convex angle. (An internal angle less than \pi)

A statement for which I designed the following proof (which I unfortunately realized had some issues as I was typing it)

Let  P be the polygon A_1 A_2 \cdots A_{n-1}A_{n}spanned by the n points A_1, A_2, ..., A_n in the plane. Consider the line L formed by extending the segment A_1 A_2 and let A_k be a vertex on $P$ which is of maximal distance from $L$. Or more precisely has the following property

  • For any other vertex A on P the distance from A to the line L is shorter or equal to the distance between B and $L$.

Now (using the Paralell postulate) let L' be a line through B which is parallel to L. By construction all vertices on P either lie on L or on the same side of L' as L (A_1 and A_2). Let us for convenience say that P lies on the ‘south side’ of L'.

At most one of Bs neighbors in the polygon on P lie on L' (for otherwise it would be a redundant vertex) and so the angle formed by these three points contained on the south side of $L’$ is strictly convex and it is an internal angle of P. (It is internal for a ray from due north into B intersects the polygon 0 times (an even number). We have therefore found a strictly convex internal angle!

I will have to admit that I realized an issue with this proof as I finished it as I had somewhat embarrasingly not properly considered how to formulate (prove) that the ‘south’ angle at B was indeed the internal one and it sort of exposes that I haven’t collected enough formal devices to differentiate between angles of the form CDE and $latex EDC without coordinates or images and the proof isn’t as satisfactory as I’d like and I will see if I can fix it.

The image I had in my mind for why this was true was after all pretty straight forward.


Using higher maths the existence of a strictly convex internal angle can be straightforwardly deduced from the existence of the convex hull as the convex angles of the hull put upper bounds on at least 3 internal angles of any polygon. But it of course has the issue of just kicking the ball further down the corridor as you’d have to prove the existence of a convex hull.


Constructing more integers using compass. 7 and 15.

Tired an a little bit hungover today I decided to play some more with the concept of constructing a multiple lengths with a compass and as few operations as possible; the problem I sketched yesterday in: https://seriouscephalopod.wordpress.com/2015/04/03/o/

Nothing interesting really except firstly I realized I had missed that there was a 3-circle construction of ‘7’ while I had originally displayed a 4-circle construction in the original post and I suppose I should correct myself plus I thought it was kind of cute.


Besides for that i suppose I have tried to gather some more empirical data regardning these constructions. I tried to write a program to runt through all simpler construction as if in a tree diagram with a new node added in every step but not having figured out a good cutoff condition I got stuck with too many ‘stupid constructions’ since I counted them all and the data set grew to quickly in relation to the data I got so I decided to scrap the algorithmic approach before I’ve figured the system better.

So I just tried to find constructions by hand and though my ‘minimas’ are only upper limits than unquestionable minimas as a result but here are the first minimal construction numbers which feel somewhat certain at least.

n m_n
2 1
3 2
4 2
5 3
6 3
7 3
8 3
9 4
10 4
11 4
12 4
13 4
14 4
15 4
16 4
17 5

At least in the beginning it has turned out that the powers of 2: 4,8,16 have formed good proto-constructions from which I can add just one circle to get to a nearby number so you can see that the number of circles necessary increases as you pass 4, 8, and 16. This might just be a quirk of the fact that these numbers are still pretty close to eachother or it is something fundamental. Nevertheless I can as the following question for further inquiry:

Question 2:  For some n is there a number k such that 2^n < k such that the minimal number of circles necessary to construct k is less than n (m_k \leq n)?

I suspect there is but we’ll see… I was however able to find a counterexample to Question 1 in the previous post.

Question 1:  Is the inequality

m_n \leq a_1 m_{p_1} + ... + a_{k}m_{p_k} \qquad (n = p_1^{a_1}\cdots p_k^{a_k})

in fact an equality or what is the smallest n such that it is a strict inequality?

That is as there better constructions than just reusing the constructions for it’s factors. The counterexample I found was 15 = 3 * 5 which using first the 3-step construction for 5 and then multiplying that by 3 using 2-circles takes a total of 5 circles as seen below


Red circles first construct 5 and blue circles then construct 15 from multiplying 5 by 3.

However using first the 3-step construction of 8 and then adding one more circle you can get to 15. Creating15Using4Circles

So the inequality was not an equality. That’s nice because on one hand it would have been a scarily powerful result had it been true but this way (it being not) I don’t have to think about it anymore.

How many circles does it take to make an integer?

A few days ago I made a post about me beginning to play a little more with the concept of ‘minimality’ when it comes to compass and straight edge constructions. That is to try to complete a geometric construction in as few operations as possible at least for now considering the act of drawing circles and lines equally arduous and extending existing lines to infinity to be free.

Today I’ve begun to play a little with perhaps the simplest of the construction problems; that of creating a length which is a multiple of another length both starting at the same point.

It is fairly straightforward to realize that constructing an n-length will take at most n-1 operations as you can work your way over to n by means of iterated drawings of circles of radius 1 using the extremities as new centers. See the picture below for the example of constructing the length 5.

Draw5by5It is however evident that this is not the way which requires the fewest number of operations as if you instead make of use of some greater circles you can cut down the 5-construction to at least 3 operations as seen below.

I suspect that some larger n-lengths might have their minimal construction by means of moving ‘off’ the line and  making use of some triangles or other shapes but if I for the time being limit myself to only making these kind of ‘circles along a line’ constructions I suppose the minimal number of circles necessary to make the numbers 2 through 8 should be visualized by


One thing you notice quite quickly when building these constructions is that they have something to do with prime numbers. Not surprising perhaps given that I was playing with multiples but I just wasn’t expecting it. You see this in the examples of 6 and 8 which are the composite numbers that the 6:construction involves first the construction of 3 and then doubling that length by the 2 construction and for the construction of 8 = 2*2*2 we resure the construction of 2 three times.

Thus if we call m_{n} the minimal number of circles necessary for constructing a length n we may put an upper limit on m_{n} for composite numbers. If n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}

m_n \leq a_1 m_{p_1} + a_2 m_{p_2} + ... + a_k m_{p_k} (1)

This is about as far I’ve come today with regards to interesting results playing with this problem but I am posing the following question for later

Question 1:  Is the inequality of (1) in fact an equality or what is the smallest n such that it is a strict inequality?

I’ve yet to make much progress with regards to some systematized way of investigating this construction problem except for the fact that you can reduce the construction to a symbolic scheme letting {0,1} be the set of the two numbers you start with and {0,1,2} the set you get when you put a circle about 1 with radius 1 and {0,1,2,4} what you get when you put a circle about 2 with radius 2 and so forth.

The idea is that to any set of numbers you can add any number which is the difference of existing numbers To {0,1,2,4} you may add 7 because 4 – 1 = 3 and you add this to 4 to get 7. I don’t have the time to think more on this right now but since today is easter and I will be tied up in family stuff for the weekend I want to make sure I remember to get back to it afterwards.

Working with a compass and straight edge takes time…

I am somewhat embarrassed to admit today was the first time I actually played with a physical compass despite being well into my 20s,having studied university maths for years and worked with the euclidean constructions for a year. WP_003271

Having run though most of the simple geometry questions I’ve had recently I thought I should return to just playing a bit more with the actual construction parts instead of the proof parts and the puzzle like questions of trying to figure out construction which consume the smallest number of operations. I was originally mesmerized by this sort of question when I looked up some of the constructions for how to trisect a line and finding it kind of surprising how few operations (circles and lines) you can reduce it to. When sketching constructions by hand it’s easy to loose track of how many steps you need to perform an action in practice as you bisect and angle there or draw a perpendicular here while the number of circles and lines being implicitely necessary can be quite staggering. On top of that most of the constructions in the Elements employ the operation of ‘moving a line segment to a point’ which if you’re actually to do it with circles and lines makes you go “HELL NO!. because (as far as I’ve counted) takes 4 circles and 2 lines for that simple thing…

Playing more by hand should probably add the necessary frustration to push me to find or look up some new methods.A for my weekly TikZ practice here is the fastest way to draw a parallel line through a point I’ve come up with so far:

TodrawaparallellLinewhich relies on the similarity of inscribed triangles.I would be genuinely surprised (or embarrassed) and not a little bit impressed if there was a faster one but I suppose eventually I’ll bother looking up whatever theory there is concerning minimality problems.