This curved surface constructed inside a rectangular prism

analytic geometrygeometric-constructiongeometrysurfaces

Given is a rectangular prism with one side $\square ABCD$ and its opposite side $\square EFGH$. Point $J$ lies on $\square ABCD$ some distance $y$ between diagonal $AC$ and edge $AB$, and point $K$ lies on $\square EFGH$ $y$ between diagonal $EG$ and edge $GH$. The midpoint of line segment $JK$ is the exact center of the prism (i.e., it is also the midpoint of diagonal $AG$), but $JK$ need not be perpendicular to any side of the prism.

Draw curve $AJC$ on side $\square ABCD$ and curve $EKG$ on side $\square EFGH$, so that the curves are geometrically similar. Now you have a curved surface, $AEGC$, with opposite straight sides $AE$ and $GC$, and opposite curved sides $AJC$ and $EKG$. Stipulate also that all points on $JK$ lie on this surface, as do all points on the line segment between the midpoint of $AE$ and the midpoint of $GC$.

What is this three-dimensional, curved surface? Does it have a name and general equation? Possibly an application or analog in real life? If not, how can one find the equation?

Edit: These are images of the surface.

The surface

Also the surface

Best Answer

This is my interpretation of the construction, confirmed by OP in a comment to the question:

As point $J$ traces a curve from $A$ to $C$ in plane $ABCD$, the point $K$ (defined as the reflection of $J$ in the center of the prism) traces an identical, but reoriented, curve in plane $EFGH$. The target surface appears by joining corresponding points on these curves in reverse tracing order; ie, parameterizing $J=J(t)$ for $0\leq t\leq 1$, with $K=K(t)$ the reflection of $J(t)$, the surface is the union of segments $\overline{J(t)K(1−t)}$.

Without reverse-tracing, the surface is, of course, a cone; I don't know a name for the reversed version ... "anti-cone"? :)

Now, for a prism with its center at $M$ (so that $\frac12(J(t)+K(t)) = M$), we can parameterize a point $P=P(s,t)$ on the surface by

$$P(s,t) = (1-s) J(t) + s K(1-t) = (1-s) J(t) + s(2M-J(1-t)) \tag{1}$$

So $P(0,t)$ traces the defining curve of $J$, while $P(1,t)$ traces the reflected curve. Note that there's no particular reason here to restrict the defining curve to a plane.

With $M$ conveniently at the origin, $(1)$ reduces to

$$P(s,t) = (1-s) J(t) - s J(1-t) \tag{1'}$$

Some examples ...

  • If the "curve" is merely a segment joining, say, $J_0$ to $J_1$ (OP's $A$ and $C$), then $$J(t) = (1-t) J_0 + t J_1 \tag2$$ and $(1')$ becomes $$\begin{align} P(s,t) &= (1-s) ((1-t) J_0 + t J_1) - s (t J_0 + (1-t) J_1) \\[4pt] &= J_0 - s(J_0+J_1)-t(J_0-J_1) \end{align} \tag{3}$$ giving the plane through $J_0$, $J_1$, and the origin.

  • If $J_0 = A=(p,p,r)$ and $J_1 = C=(-p,-p,r)$ (giving our prism a square face) and the curve is a semicircle in the plane $z=r$, then $$J(t) = (0,0,r) + (p,p,0) \cos \pi t + (p,-p,0)\sin\pi t \tag4$$ and we have the cylinder $$P(s,t) = (1-2s)\;\left(\;p (\cos\pi t + \sin \pi t), p (\cos\pi t - \sin\pi t), r \;\right) \quad\to\quad x^2 + y^2 = 2 p^2 \tag{5}$$

  • More generally, we can replace $\cos\pi t$ in $(4)$ with any $f(t)$ such that $f(0) = -f(1) = 1$, and $\sin\pi t$ with any $g(t)$ such that $g(0)=g(1)=0$. How far we can get in eliminating the parameters depends upon the exact nature of $f$ and $g$.

    For, say, $$f(t) = 1-2t \qquad g(t) = 2t(1-t) \tag{6}$$ we obtain this cubic surface $$z(x+y)^2 + 4 p ( r x - r y - p z ) = 0\tag{7}$$ Here's a rendering with $p=r=1$:

enter image description here

In response to a comment: One can solve $(7)$ for, say, $x$ using the Quadratic Formula to get $$x = \frac{-2 p r - y z \pm 2 \sqrt{p^2 r^2 + 2 p r y z + p^2 z^2}}{z} \tag{7'}$$ For points restricted to the prism, we take the "$\pm$" to be "$+$".

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