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After constructing the two rays at an angle of approximately 60 degrees, the author goes about constructing an actual (approximate) equilateral triangle by draw
by tomsmeding 8y ago
After constructing the two rays at an angle of approximately 60 degrees, the author goes about constructing an actual (approximate) equilateral triangle by drawing two perpendiculars. That might be useful if one needs a triangle of precisely that size, but given a pair of compasses (or a straightedge with two markings), one gets any ≈equilateral triangle by just marking the two intersection points of any circle around the origin with the two rays. I believe this method doesn't give anything less precise than the method outlined in the post.
Given that the idea was to make the construction simple and small, I'm surprised the compasses method wasn't chosen.
- Timothee 8y agoI was confused as well, but I think the idea is that he's using only graph paper and a straightedge without measurement. (or markings) So he has to place points with integer coordinates on the graph paper, and draw lines between them.
- rocqua 8y agoThat's an interesting number set. The set of all points that can be produced from a straight-edge and the grid of integers. I'm pretty sure this is just the grid of rational numbers. Interestingly, this is a subset of the 'constructible' numbers. That is, the numbers that can be produced from a compass and straight-edge (without even needing a grid).
- joppy 8y agoIt’s not so surprising, since once you have a compass, straightedge, and a line segment you call “unit length”, you can construct all of the integer valued points in the plane.
- deleted 8y ago[deleted]
- svat 8y agoThe post is titled “60-degree angles on a lattice”, and links to “how to draw a good approximation to an equilateral triangle on a piece of graph paper” — that is, the problem being considered is one where all coordinates have to be integers. If you're willing to use compasses and get arbitrary coordinates, you don't need any of this, not even the first part — it's an entirely different (and much simpler) problem.
- tomsmeding 8y agoRight, of course; thanks for noting.