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I know how to make a pentagon (there are youtube videos), and I kinda thought I knew how to do a hexagon using 5 circles:

one in the center, shift-drag 2 guide blocks.

  • 2 below it, each shift-drag 2 guide blocks, one started one block to the left, the other one block to the right, and both 1 block below the center one.
  • 2 above it, each shift-drag 2 guide blocks, one started one block to the left, the other one block to the right, and both 1 block above the center one.

The interior intersections delineating the hexagon. Kinda like this:

regular hexagon plan

But I was wrong, the two "side" corners don't intersect (there's no three-way intersections on the sides -- I thought that was just an abhorration of my drawing by hand) and it is not very regular.

The rules for drawing roads in SimCity are:

  • Straight roads can either be free rotation, or 45º, or 90º.
  • There is an underlying grid of "blocks" so you can use Cartesian coordinates
  • The road tool has modes: Straight line, Free draw, Arc, Rectangle (drawn at once), Oval/Circle

There are some actions you can't do in SimCity that are used for typical regular constructions:

  • There is no setting of a compass to a given span (can't take a measure of N).
  • There is no function to find the midpoint of anything.
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Perhaps you could implement this. –  David Mitra Jun 17 '13 at 11:21
    
In a circle with center $O$ and diameter $AB$, let $M$ and $N$ be the midpoints of $OA$ and $OB$. Build perpendiculars to the diameter at $M$ and $N$. The points where the perpendiculars meet the circle, along with points $A$ and $B$, are vertices of a regular hexagon. If you make your circle $4n$ horizontal "blocks" wide, then the grid should make the rest pretty clear. (I think. I'm not entirely clear on the SimCity drawing rules.) –  Blue Jun 17 '13 at 11:41
    
Blue post your comment as an answer, converting midpoints to Cartesian measures. I can see how that could work for circles of specific diameters. –  roberto tomás Jun 17 '13 at 12:46

1 Answer 1

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In a circle with center $O$ and diameter $AB$, let $M$ and $N$ be the midpoints of $OA$ and $OB$. Build perpendiculars to $AB$ at $M$ and $N$. The points where the perpendiculars meet the circle, along with the points $A$ and $B$, are vertices of a regular hexagon.

(If the perpendicular through $M$ meets the circle at $P$, then by construction, $|OM|/|OP| = |OM|/|OA| = 1/2 = \cos 60^\circ$, and we can conclude that $\triangle OAP$ is equilateral. Likewise with the other points of intersection.)

If you make your circle $4n$ "blocks" wide, then the grid should make the rest pretty clear.

More specifically, put the center at $(0,0)$, and let the diameter of length $4n$ (for any whole number $n$ you like) extend from $A(-2n,0)$ to $B(2n,0)$; then the midpoints are $M(-n,0)$ and $N(n,0)$. (For instance, if the diameter is $400$, then $M$ and $N$ are at $(\pm 100,0)$.) Follow the grid-lines up and down from $M$ and $N$ until they hit the circle to get the final four vertices of the hexagon.

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