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# mrflip/Ancient Greek Geometry.md

Last active Jul 24, 2022
Ancient Greek Geometry walkthrough / answers / cheats

# Solutions for Ancient Greek Geometry (https://sciencevsmagic.net/geo)

## Non-Constructible Figures

Abuse of floating-point math can make the widget approve non-constructible polygons (polygons with edge count 7, 9, 11, 13, 14, 18, 19, 21, 22, 23, 25, 26, 27, 28, 29, 31, 33, 35, ..., which cannot be precisely constructed using straightedge and compass):

### Eddy119 commented May 4, 2020 • edited

If someone can trick it into accepting a 13-gon and 19-gon, please tell me... I wonder how small the error has to be for it to be tricked

### Eddy119 commented May 4, 2020

Also if someone can find the approximate or exact length or the angle of the side of the 11-gon approximation then please post it here too

### mrflip commented May 5, 2020

@Eddy119 — thanks for the links, I've incorporated them. I think https://commons.wikimedia.org/wiki/File:01-Elfeck-3.svg gives the sizes and angles for the pseudo-11-gon, is that not what you need?

### Eddy119 commented May 6, 2020

Hmm... is my construction of the 17-gon the fastest (least amount of moves) for drawing it inside the origin circle?

### mrflip commented May 6, 2020

Yes it still is! Will update. You can save a lot of moves by doing the star trick -- instead of drawing all the scalloped circles and then the lines, get two edges to intersect at the points of the 17-sided star, and then walk them around the shape.

### Eddy119 commented May 6, 2020

I guess you're not going to add any more approximations? Is there an easy way to make approximations for an n-gon?

### Eddy119 commented May 7, 2020 • edited

If you check the Wikipedia article on the 17-gon, it says that another more recent construction is given by Callagy, and the source is here: https://www.jstor.org/stable/3617271 (if you need to go over the paywall I have a copy from my university account)
It finds the central angle of the 17-gon directly, but I'm not sure if it's faster...
Maybe this is faster?

### Eddy119 commented May 8, 2020

I guess the next challenge is the 51-gon!
I gave up on the 13-gon... Still looking for a nice simple construction that tricks the code... Please send here if anyone finds one
Also, anyone know how small the error has to be to trick it?

### Eddy119 commented May 8, 2020

Well, to construct a 13-gon, you have to trisect a certain angle... check http://web.archive.org/web/20151219180208/http://apollonius.math.nthu.edu.tw/d1/ne01/jyt/linkjstor/regular/7.pdf out for 3θ (3 times the desired angle) or https://commons.wikimedia.org/wiki/File:01-Dreizehneck-N%C3%A4herung.svg for a very good approximation... I tried to do both but I just couldn't...

### Eddy119 commented May 8, 2020 • edited

Once we know how to approximately trisect an angle well enough to trick the website, we can pretty much make all of the polygons up to 40 except 23, 25, 29, 31

### Eddy119 commented May 8, 2020 • edited

41, 43, 47, 53, 59, 67, 71... basically all the prime numbers that are neither fermat primes nor pierpoint primes

### Eddy119 commented May 8, 2020 • edited

51-gon should be constructible as it is 17*3 and (2^n)*, 3* or 5* (7, 9, 11, 13, 19 , 37) should be approximately constructible as 7, 9, 11 is already done and 13, 19, 37 are pierpoint primes

### Eddy119 commented May 8, 2020

Should be recognised by website except Bold:
2 (...)
3 (fermat no.)
4 (2^2)
5 (fermat no.)
6 (3*2)
7 (approx. constructible, pierpoint prime)
8 (2^3)
9 (3^2, approx. const.)
10 (5*2)
11 (approx. constructible as shown by me)
12 (3*2^2)
13 (should be approx. const. , please share, pierpoint)
14 (7*2, approx. const.)
15 (7*5)
16 (2^4)
17 (fermat prime, next fermat is 257)
18 (9*2, approx. )
19 (pierpoint, should be approx. const., please share)
20 (5*2^2)
21 (7*3, should be approx. please share)
22 (11*2)
23 (not pierpoint, if someone can approximate this, congrats, please share)
24 (3*2^3)
25 (5*5, not pierpoint, if someone can approximate this, congrats, please share)
27 (9*3, should be approx. const., please share)
28 (7*2^2)
32 (2^5)
34 (17*2)
35 (7*5, approx const because one is pierpoint and other is fermat)
36 (2*2*3*3, approx)
37 (pierpoint, approx)
39(13*3)
40 (5*2^3) ...
Italic involves trisecting an angle and Bold can't be made either by bisecting nor trisecting

### mrflip commented May 10, 2020

Thanks Eddy! I'll batch these up at some point. I found a new (for me) trick when stellating a polygon to fill out the edges: a diameter of the circle goes through the points of the star, so if your construction can naturally have two diameters with enough of an angle spread you can avoid drawing a circle to reflect the vertices off. You can see that in the 20-gon solution I just put up, where it saves one move.

### Eddy119 commented May 10, 2020

I'm not sure how floating point math works but I guess it results in some less approximate constructions being recognised and some more approximate constructions not being recognised.