How could I make this polygon in Inkscape?

May 12, 2019 5:25 AM

See this triangle? See how one of the corners is removed? Imagine the other two are removed as well. You're left with a hexagon with three sides of length x, and three sides of length 2x. What's the easiest way to make this in Inkscape so I can rotate / resize / tessellate?
posted by some little punk in a rocket to Technology (6 answers total)

Response by poster: (I guess it doesn't have to be Inkscape. If you know how to do it in something else, I can try to apply that process to what I have - thanks!)
posted by some little punk in a rocket at 5:29 AM on May 12, 2019

Best answer:

1. Create equilateral triangle using "Polygons & Stars" tool.
2. Select the triangle. Under the "Path" menu, convert object to path. You now have a polygon with six control nodes, three at the corners and three on the sides.
3. Select the "Edit Paths by Nodes" feature. Select all of the nodes. Click the "Insert Nodes into Selected Segments" tool. You now have a polygon with twelve control nodes, evenly spaced along the sides.
4. Delete the nodes at the corners of the triangle. This will give you control nodes in the correct locations, but the ones in the new "corners" are rounded.
5. Select the nodes in the new "corners" and click the "Make Selected Nodes Corner" button.

posted by Johnny Assay at 5:34 AM on May 12, 2019 [3 favorites]

Some more ways (not saying they're better than the above, it's just nice to see different approaches):

* Make one little equilateral triangle with Polygons & Stars
* Make 12 copies so you have 13 little triangles
* Flip 6 of them upside down
* Drag and drop your 13 triangles into the shape of a hexagon, using snap settings to get them to snap together perfectly
* Merge them into a single hexagon with Path > Union

Or

* Make an equilateral triangle
* Duplicate it, and shrink the duplicate to 25% scale
* Make 2 more copies of the little triangle
* Drag and drop the 3 little triangles to the corners of the big triangle, using snap to get them exactly right
* One at a time, subtract the little triangles from the big one using Path > Difference
posted by moonmilk at 7:26 AM on May 12, 2019 [1 favorite]

It's the hull of thee equilateral triangles, flipped and translated so their tips intersect, then rotated one to each of the other points of the original triangle. In OpenSCAD:

module hexa_gone(d) {
    hull() {
        for (i=[0:2]) {
            rotate(i*120)translate([0,d])rotate(150)circle(d=d, $fn=3);
        }
    }
}

hexa_gone(20);

posted by scruss at 6:40 PM on May 12, 2019

Response by poster: Thanks everybody! Johnny Assay - worked first time for this rank amateur, so that says something about the clarity of your instructions. moonmilk - that's how I was trying to make it, but the whole Path > Union / Path > Difference thing was eluding me. Thanks for showing me how those two tools work. scruss - my 3D printer thanks you!
posted by some little punk in a rocket at 12:38 AM on May 13, 2019

JohnnyAssay - I had to "Insert Nodes into Selected Segments" twice to get 12 nodes. My triangle started with 3 nodes.
posted by scruss at 4:19 PM on May 13, 2019