Locus of Points $O$ such that $AO^3+BO^3=AB^3$ 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?
 A: We may consider the curve as part of a family defined by
$OA^{k}+OB^{k}=AB^{k}$
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.
