{{ message }}

Instantly share code, notes, and snippets.

# mrflip/Ancient Greek Geometry.md

Last active Oct 5, 2020
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):

### bikerusl commented Sep 11, 2017 • edited

 Here is circle 7 in origin in 14 moves: https://sciencevsmagic.net/geo/#0A1.1A0.0L1.0L4.3L1.1L6.6L4.1L2.9A2.1A9.1A14.19A0.21A3.1L10.1L5.24A10.20A5.22A6 I was nowhere near solving this one but after I tried your solution (THANKS! 👍 ) was able to optimise it by removing one of the origin size circles. I find it easier to optimise than to come up with original solutions. Sorry I'm not quite familiar with git enough to fork and put it in there myself. Thanks for putting this up. Damn that 8 move square in origin is slick! I love it. I've solved all the main puzzles now. Now going for easter eggs :-)

### 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 7, 2020

 Here it is. https://i.imgur.com/g7yLseA.png

### mrflip commented May 7, 2020

 If you look at my versions you’ll see what I mean: after doing the two three-node hops using the tangent line to N (total 9 hops), I mark a circle all the way across, giving two edges that converge at the tightest angle (largest star), and draw the giant circle at the star’s diameter. Since the 8 hop is relatively prime to 17, you can then use symmetry to walk the edges around without marking anything else off. Most of the other shapes in the catalog end up with some shape in the construction giving you a reflection for “free”, like the pentagon one does, but I haven’t found one here. Also, using G’ adds a step; constructing N on G gives you a tangent annoyingly close to the perp thru M but works and saves a step. Actually both tangents to N clip hit at vertices, Might be worth seeing if that can save any steps. One note: when you make the giant star the script can get super confused and crashy and evening sometimes adds three steps for extending a line for free. it seems to not like super long line segments, which is why you’ll see in some of those that I extended them in parts. Once you have the star set up it should only be 15 steps to the end. … On Thu, May 7, 2020 at 2:27 AM Eddy119 ***@***.***> wrote: *@Eddy119* commented on this gist. ------------------------------ https://sciencevsmagic.net/geo/#0A1.1A0.0L1.1L5.0L4.2L3.4A1.1A4.12L13.6A0.0A6.24L23.28A0.28L18.52A0.0A52.69L70.79A28.70L93.93A92.92A93.141L142.155A28.17A174.189A17.228L229.243A17.108A0.0A108.335L334.352A0.0A277.108A420.N.452A0.13L490.490A0.0A490.547L548.570A18.571A18.597A571.584A570.640A584.656A640.672A656.0A1.686L672.672L656.656L640.640L584.584L570.570L18.18L571.571L597.597L616.616A597.616L938.938A616.938L992.992A938.992L1046.1046A992.1046L1100.1100A1046.1100L1158.686A672.1219L686.1219A686.1219L1266.1266L1158 here's that version: I tried to star it but I'm not sure how to do it so — You are receiving this because you authored the thread. Reply to this email directly, view it on GitHub , or unsubscribe .

### 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) 26 (13*2, please share 13) 27 (9*3, should be approx. const., please share) 28 (7*2^2) 29 (not pierpoint, please share) 30 (5*2*3, please share) 31 (not pierpoint, please share) 32 (2^5) 33 (11*3, approx, please share) 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) 38 (19*2, please share) 39(13*3) 40 (5*2^3) ... 41 (not pierpoint, please share) 42 (7*2*3, please share) 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 think the problem with the 25-gon is that it places the point to the right of the bottommost point of the 25-gon somewhere else to where it should be... if someone wants to research this problem here's how I found out https://sciencevsmagic.net/geo/#0A1.1L0.0L2.1A0.3A0.1L3.3L9.9L4.9L0.0A10.10A0.26L25.30A2.30A1.25L51.10L45.73A10.45L92.92A10.10A92.115L116.N.14A0.141L142.171A0.0A171.218L217.217L246.246A0.246L288.288A287.287A288.332L331.N.0A1.234L233.233L353.381A233.233A381.464L0.14A493.493A14.538L493.493L14.14L518.518A14.518L659.538A493.538L720.720A538.772L720.772A720.830L772.659A518.659L885.885A659.885L949.949A885.949L1016.1016A949.1084L1016.1084A1016.1152L1084.1152A1084.1212L1152.1212A1152.1212L1264.830A772.830L1317.1317A830.1317L1369.1369A1317.1369L1409.1409A1369.1409L1453.1453A1409.1499A1453.1499L1453.1499L15.15A1499.1264A1212.15L494.494L1264

### 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.