Math

A Trisectrix from a Carpenter’s Square

Posted by Dave Richeson on

UPDATE: The article is now published. Read it in Mathematics Magazine.

Yesterday I posted an article to the arXiv, “A Trisectrix from a Carpenter’s Square.”

Abstract: In 1928 Henry Scudder described how to use a carpenter’s square to trisect an angle. We use the ideas behind Scudder’s technique to define a trisectrix—a curve that can be used to trisect an angle. We also describe a compass that could be used to draw the curve.

I also made a GeoGebra applet to accompany the article. Give it a try.

Screen Shot 2016-03-11 at 2.38.49 PM

2 Comments

  1. Jan Van lent says:
    March 18, 2016 at 3:34 pm

    I played around a bit with your Carpenter’s Trisectrix. The curve has the following paremetrizations:
    (t - 4 t/(1+t^2), 3 - 4 t/(1+t^2)) for -\infty < t < \infty and
    ((cos(2 a) - cos(a))/sin(a), 1 - 2 cos(a)) for -\pi < a < \pi.
    The "drop" part of the curve is given by |t| < \sqrt{3} or |a| < 2 \pi / 3.
    The drop has area 3\sqrt{3} \approx 5.20. The perimeter can be written using Elliptic functions (approximately 4.12).
    If the drop of the curve is considered to be the cross-section of a solid of revolution, its volume is (8 \ln(2) - 3) \pi \approx 8.00 and its surface are is 3 (9 - 4 \sqrt{3}) \pi \approx 19.53.

    1. Dave Richeson says:
      March 18, 2016 at 3:43 pm

      Cool! Thanks. My first approach was to parametrize the curve. I parametrized it in terms of the slope of one of the legs of the T (actually I parametrized it both ways) and used that to find the non-parametrized version. Then I came up with this other approach that didn’t require going through the parametrization. I didn’t go the trig route. Thanks!

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