Fun geometry stuff

[Kyriakos]

I may have more to post in the future, but here is a first:

Already in the ancient era (as Proclus notes in his comments on Euclid) there were a few people who thought construction in geometry as being bypassed when the premise seems to allow you to do things in a single move.
A good example of this would be Proposition II in the first book of Euclid (pic provided - I actually modeled this in Blender ), where Proclus (5th century byzantine from Constantinople) says that some people just used the compass to secure a radius of BC, then created a circle from A.
But it seems that Euclid's reason for presenting this so early on was exactly to highlight that such a thing wouldn't be a valid construction, and in logic it's known as "begging the question" (petitio principii). This becomes important, since likewise in formal systems (computers and other stuff) you can't just go outside the system and derive a new formula, even if it is obvious outside the system; you have to mention either specific theorems or alter the previous statement by one of the allowed rules. It's the same here: you have to construct BC from something else (despite using a radius BC to get there), and just using the compass to move length BC around would be analogous to examining a statement from outside the system.



(the circles from A and B, with radius AB, are how to construct the equilateral triangle; the other vertex is where those circles intersect- Proposition I)

(I also provided an alternative construction, without use of Proposition I but still using radius BC only from a center in known points of BC)

 

[Kyriakos]

Ok, to finish off with Proposition II (all unique cases)

Here is the construction in the method of Euclid (that is, with the use of an equilateral triangle), in the case that AB>BC and AB=BC.




The only other different cases are symmetrical (A is to the other side of BC, or is below/above them).
But If A is actually in the segment BC (which likely one can infer isn't the case, from the proposition's wording) then you can just use the same idea as in the alt method I posted, and which you can see in the lower pair of constructions (do notice that in them you never construct the circle with radius equal to BC; I provided a construction with equilateral triangle and one without).

 

[Kyriakos]

Ok... Here is the first part (one case) of Proposition III

It should be stressed again that Euclid proceeds always from the previously proven constructions, in a type of procession of theorems or true sentences.
So in Proposition II you use Proposition I (how to construct an equilateral triangle given only a line segment and compass), and in III you use II (how to move a segment so that it now starts from another given point).



(ps: you might notice that I deliberately made it so CD = 1/2 of AB I like it, since now it looks like something Paul Klee would have drawn)

 

[Kyriakos]

Proposition 4 is perhaps notable due to the (rare for Euclid) use of superposition to show something.
He argues that if two triangles have two sides equal and the angle they contain also equal, they will be identical, since if you have ΑΒ=ΔΕ, ΑΓ=ΔΖ, angle A=angle Δ, and you superpose Α on Δ, B will be superposed on E, and since angle A= angle Δ it follows that ΑZ will be identical (premise was it is equal to) to ΑΓ. From this follows that EZ=BΓ.

Moving on to Proposition 5, which is:
To show that in an isosceles triangle, the base angles are equal and if the equal sides are extended below the base they will form also equal angles.

Now there are many proofs for this, already in the ancient era. First let's show the one Euclid provides (it is the more laborious proof of the group)

AΔ and ΑΕ are the extensions of AB, AΓ.
There we draw points Z and H, with BZ=ΓΗ.
From this follows that triangles AΓΖ and ΑΒΗ are equal, since they have two sides (AΓ, ΑΒ and ΑΖ, ΑΗ) be equal and also the contained angle A be equal.
From that follows that the other two angles in those triangles are also equal (AZΓ, ΑΗΒ and ΑΓΖ, ABH) as well as that BH=ZΓ.
Finally, we can show that also triangles BZΓ abd ΒΓΗ are equal (two sides and one angle being equal), therefore angles ABΓ = ABH-GBH = AΓΖ-ΒΓΖ= ΑΓΒ.

Aristotle provided a few proofs, very different to the above. For example:

The problem here is only that he makes use of various previously proven stuff (such as about angles to arcs).

Pappos wrote a rather elegant and very short proof (Hofstadter mentions it at some point in his GEB, claiming that once a computer "came up" with something very similar, but of course the computer didn't actually think to achieve it).
It's the following:

Proof by Pappos of Alexandria:
Triangle ABΓ is equal to triangle AΓΒ (two common sides and a common angle). Therefore angle γ is equal to angle β.

Other proofs had also been given in the ancient era, such as the one where you bring the distance from A to the base (since this is an isosceles, it divides the base to two equal parts, and you then compare the equal triangles' angles).

 

[Kyriakos]

A recap of V and also VI, which is its inverse (to show that in this case if the statement is true then so is its inverse and vice versa).

 

[Kyriakos]

In a while (in the end? ) we will get to this:

The spiral of Theodoros of Cyrene. It is (as happens with Plato) part of the subject of one of the dialogues, where the important geometer Theaetetos appears as a character.
Notice that each new hypothenuse is linked to a side 1 and a side of the previous hypothenuse.
The spiral was used to calculate incommesurable numbers - which is what irrational numbers were known as in ancient Greece.

 

[Kyriakos]

See how easy it is to arrive from tic-tac-toe to Gary Oldman's Dracula?

 

[Kyriakos]

Is this the Academy, and people don't want to enter if they are not into geometry?

Anyway, here are propositions VII and VIII
Interestingly I first misunderstood the scope of proposition VIII, thinking that the sides just had to be above the halfplane.
But in reality it was about sides formed with the same edge point in the base being equal with end points C,D outside the base.
No harm done, however, since I noted the generalization and special case; you can't have different triangles with equal sides formed from the same base vertex and two points outside the base, unless you don't have a halfplane defined by the base as a restriction.
You can have two different triangles with the equal sides, as long as those are linked to different endpoints of the base.
The axis of symmetry just shifts to the perpendicular to the base's midpoint, instead of the base itself.

By the way, a similar construction by Ptolemy (following figure) can be used to provide another proof of the law of cosines. Obviously in Prolemy's circumscribed triangles with similar angle of symmetry, all the vertices are presented as points in the periphery of a (different to mine) circle.



Anyway, here is Ptolemy's circle (highlighted and superimposed on my construction), which has as center the point (O) where the larger equal sides intersect when rotated to reach the smaller sides. (going from the pic above, which is from wiki, I infer that it is the other way around re properties of smaller and larger side if DC is over a certain length, which I suspect is tied to the distance of the center of the circle to the base).



More curiously, it seems to have the same distance from base AB that its final point on the vertical axis has to another circle in the construction (will look into that)

 

[Kyriakos]

This is part of the famous geometrical calculator of Hippocrates of Chios. You can calculate any multiplication (or division) with it, as long as you mark even segments and are able to multiply without other help one of the numbers with 1 ).

Picked up again by Descartes, it is purely the Thales intersection theorem - above you see the calculation of 2 times 2: first you form line a which is from horizontal axis 1 to point 2 in any other axis, and this stands for 2 times 1, then form parallel lines to it starting from horizontal axis point 2; those lines arrive at 4 and are twice as large as their respective first.
It becomes obvious that the ratio of any first parallel to another of its group, is equal to the ratio of the analogous for any other group.

From a glance, it seems that this is pretty much a massive (potentially infinite) generalization of Pascal's Triangle as well (I think Conway once did one such thing; see, he isn't that special ^_^ ).

 

[Ferocitus]

Nature is working on a similar problem that puzzled Kyriakos.

 

[Kyriakos]

If you wondering what the above awesome graphic is, no it's not a boa constrictor which has fully digested the elephant, but my lazy presentation of 1/4+1/16+1/64+... = 1/3.
Basically each time you divide by four, you immediately lose half of the operating area you had, with the first fourth being automatically filled and the remaining second fourth being now the new operating area, which means that ultimately you lose twice as much as the area you get to work on for the next step, so end up with 1/3 of all.
Of course this isn't a special case; it is analogous for all such geometric progressions. In the simplest case, that of 1/2+1/4+1/8+..., you never lose ANY area, so naturally end up filling the entire provided space (=1).