A Trisectrix from a Carpenter’s Square

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.

2 Comments

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 $3 (9 - 4 \sqrt{3}) \pi \approx 19.53$ .

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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