Suppose $AB$ is a straight line. Then the locus (in a plane) of points $O$ such that $AO^2 + BO^2 = AB^2$ is a circle with diameter $AB$. This is closely related to Pythagoras's Theorem, using the fact that an angle in a semicircle is a right angle.

Question If $AB$ is a straight line, is there a name for the curve defined as the locus of points $O$ such that $AO^3 + BO^3 = AB^3$, and can any points on the curve (other than $A$ and $B$) be constructed by ruler and compass?


We may consider the curve as part of a family defined by


Then $k=1$ gives just the line segment and $k=2$ gives the ciurcle with diameter $AB$. As k approaches infinity the locus approaches the union of arcs $CAD$ and $CBD$, where $C$ and $D$ are the distinct points in the plane for which $AB$, $AC$, $AD$, $BC$ and $BD$ are all congruent (think of how you would construct an equilateral triangle).

The locus for $k=3$ then lies between the $k=2$ circle and the arcs for infinite $k$. It looks something like an ellipse with $AB$ as a minor axis, but it's not actually an ellipse and I do not know a specific name. The curve in the garage-shelf construction, referred to in the comments, does have a name, the "four-cusped hypocycloid". But that is actually a different curve.

Infinitely many points can be constructed on the $k=3$ curve. To see how, let $AB=1$ and set up the equation

$(1+a)^{3}+(1-a)^{3} =2+6a^{2}=s^{3}$

We select any constructible value for $s$ between $2^{1/3}$ and $2$ itself, and then there will be a pair of constructible additive-inverse roots between $-1$ and $+1$ or $a$. We then construct points $O$ such that $AO$ is $(1+a)/s$, allowing either sign for $a$, and $BO$ is $(1-a)/s$. Similarly the identity

$(1+a)^{5}+(1-a)^{5}= 2+20a^{2}+10a^{4} (=s^{5})$

leads to constructible points on the $k=5$ curve.

  • $\begingroup$ Thank you, so as expected the point equidistant from A and B corresponding to $s=2^{1/3}$ is not constructible, but infinitely many other points can be constructed, the simplest having $s=3/2$ giving $a=(11/48)^{1/2}$. $\endgroup$ – Adam Bailey Mar 6 '16 at 23:08

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