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@mrflip
Last active March 2, 2026 10:47
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Ancient Greek Geometry walkthrough / answers / cheats

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

Most solutions taken from the about thread. See the comments below for more additions since my last check-in.

Polygons

Circle Packs

Circumscribed Polygons

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):

@ILoveMath62
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ILoveMath62 commented Feb 24, 2026

@ILoveMath62
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Since you have a heptadecagon in-origin 38 moves, maybe 36 moves could possibly exist? I'm not entirely sure...

@Eddy119
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Eddy119 commented Feb 25, 2026

I don't think so, because the x=-1/2 vertical line doesn't exist in the non-in-origin version. Drawing that line there will cost one move. I've been thinking and trying.
I suggest you think and try if you can reduce the number of moves especially during the initial setup, since the scale doesn't matter at all for non-in-origin, you can just make it any size (halving or finding perpendicular costs 3 moves); or see if you can get rid of the stepping off the starting side circle for the final star trick (prob not possible), or maybe different {17/k} stars happen to be simpler (prob not either?). I still don't fully understand the requirements for the star trick.
Another idea I had was scaling the Carlyle circles' parameters to find twice or half the roots, and/or making themselves twice or half as big, but I think the current ones are already very simple, maybe the simplest.

One concept to simplify is basically seeing if you can reuse existing points/lines/circles, as maybe some things you want is already constructed, or you can use existing ones to do a task. Extending lines don't cost a move.

Feel free to also try to reduce the number of moves for the in-origin version, maybe it's possible...

Maybe at this point even if it's possible, we'll have to basically recalculate the method, but this method might already be the simplest, as the paper title suggests. We might have simplified so much already, you can only simplify so much.

@ILoveMath62
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ILoveMath62 commented Feb 26, 2026

Heptagon by approximating the side length as sqrt(36.898)/7. Relative error of the first side: 6.8456*10⁻⁹

@ILoveMath62
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Good job.

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