Which quadrilaterals are possible as the intersection of a plane and a rectangular pyramid?

Question

Suppose I have a pyramid with a rectangular base, and consider a plane that does not intersect the base or the apex (it only intersects the four faces incident to the apex).

When the plane is parallel to the base, the intersection is similar to the base.

When that plane is rotated only along an axis perpendicular to one of the sides of the base, the resulting shape is a trapezoid.

But if the plane is rotated in another direction, some other quadrilateral is formed.

I suspect that that the intersection is always a convex quadrilateral, though I'm unsure how to prove it.

I would be surprised if that condition is also sufficient for a quadrilateral to be an intersection of this type, but I have yet to find a counterexample.

Answer 1

Let's parameterize the vertices of a "pyramidic section" as follows: Put the vertex of the pyramid at the origin, and let its edges pass through the points $$(\pm\cos s\cos t, \pm\cos s\sin t, \sin s ) \tag1$$ Here, $t$ is half of an angle between diagonals of the pyramid's base (in any plane parallel to the $xy$-plane); that is, $\tan t$ gives the "aspect ratio" of the rectangle. The edges of the pyramid make angle $s$ with the $xy$-plane.

Let the cutting plane have normal vector $$(-\sin p\cos q, -\sin p\sin q, \cos p) \tag2$$ Here, $p$ is the angle the plane makes with the $xy$-plane, and $q$ is an angle of rotation about the $z$-axis. Let the plane pass through point $H=(0,0,h)$. Let the plane meet the edge-lines in points $A$, $B$, $C$, $D$ corresponding to respective sign choices in $(1)$ of $(+++)$, $(-++)$, $(--+)$, $(+-+)$. We (and by "we", I mean Mathematica) find that, for instance, $$\begin{align} A &= \frac{h}{\tan s-\tan p\cos(q-t)}(\phantom{-}\cos t,\phantom{-}\sin t,\tan s) \\[6pt] B &= \frac{h}{\tan s+\tan p\cos(q+t)}(-\cos t,\phantom{-}\sin t,\tan s) \\[6pt] C &= \frac{h}{\tan s+\tan p\cos(q-t)}(-\cos t,-\sin t,\tan s) \\[6pt] D &= \frac{h}{\tan s-\tan p\cos(q+t)}(\phantom{-}\cos t,-\sin t,\tan s) \end{align} \tag3$$ and we can calculate the following distances: $$\begin{align} a:=|HA| = \frac{h\,\sqrt{1+\tan^2p\cos^2(q-t)}}{\tan s-\tan p\cos(q-t)} \\[6pt] b:=|HB| = \frac{h\,\sqrt{1+\tan^2p\cos^2(q+t)}}{\tan s+\tan p\cos(q+t)} \\[6pt] c:=|HC| = \frac{h\,\sqrt{1+\tan^2p\cos^2(q-t)}}{\tan s+\tan p\cos(q-t)} \\[6pt] d:=|HD| = \frac{h\,\sqrt{1+\tan^2p\cos^2(q+t)}}{\tan s-\tan p\cos(q+t)} \end{align} \tag4$$ Also, with $\theta := \angle AHB$, $$\cos\theta = -\frac{h^2 (\cos2t+\tan^2p \cos(q-t)\cos(q+t))}{ ab( \tan s-\tan p \cos(q-t)) (\tan s+\tan p \cos(q+t))} \tag5$$

We can formally recapture the parameters from these metrics. Defining $a':=1/a$, $b':=1/b$, $c':=1/c$, $d':=1/d$, we have (barring transcription errors) $$\begin{align} h^2 &= \frac{(a' - b' + c' - d') (a' + b' + c' + d')}{(a'b'-c'd')(b'c'-d'a')} \\[6pt] \tan^2 s &= \frac{(a'-b'+c'-d')(a'+b'+c'+d')}{(a'-b'-c'+d')(-a'-b'+c'+d')}\\[6pt] \tan^2t &= \frac{ a'b'-c'd'}{u_{ab}-u_{cd}}\cdot\frac{u_{bc}-u_{da}}{ b'c'-d'a'} \\[6pt] \tan^2p &= \frac{(a'-b'+c'-d')(a'+b'+c'+d')\left(\begin{array}{c} u_{ab}+u_{bc}+u_{cd}+u_{da}\\ -(a'+c')^2-(b'+d')^2 \end{array}\right)}{(u_{ab}-u_{cd})(u_{bc}-u_{da}) } \\[6pt] \tan^2q &= \frac{(a'b'-c'd')(u_{ab}-u_{cd})}{(b'c'-d'a')(u_{bc}-u_{da})} \end{align} \tag6$$ where $$u_{ab}:=a'^2 + b'^2 - 2a'b'\cos\theta = \frac{|AB|^2}{a^2b^2} \qquad u_{bc}:=\frac{|BC|^2}{b^2c^2} \qquad \text{etc}$$

So, whenever these parameters "make sense", a given quadrilateral is (probably) a pyramidic section.

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What could possibly go wrong?

Well, all the formulas are ostensibly squares, so negative values are a problem. This is the case with the kite in @intelligenti pauca's answer. Writing @ip's $a$ parameter as $k$, we have $$(a,b,c,d)=(1,k,2,k) \;\to\; (a',b',c',d') = (1,1/k,1/2,1/k) \;\to\; h^2=(4 - 3 k) (4 + 3 k)$$ This is negative for $k>4/3$.

@ip's answer also gives the counterexample of a non-rectangular parallelogram. In this answer's notation, $a=c$ and $b=d$, but $a\neq b$; hence, $a'=c'$ and $b'=d'$, but $a'\neq b'$. This is problematic in that $h^2$ gets a zero denominator. (Reconciling the rectangular case $a'=b'=c'=d'$, which should be valid, provided the rectangle matches the aspect ratio of the pyramidal base (so that we can take a "horizontal" cutting plane), is left as an exercise to the reader. (I may revisit this case in a future edit.))

A known aspect ratio for the pyramidal base constrains resulting quadrilaterals. For instance, a square base has $s=\pi/4$, so that $\tan s = 1$; substituting into the corresponding formula in $(6)$ and simplifying gives $$a'^2+c'^2 = b'^2 + d'^2$$

Answer 2

A counterexample is any parallelogram which is not a rectangle, i.e. with different diagonals.

Let $ABCD$ be a parallelogram, with unequal diagonals $AC$ and $BD$ meeting at their common midpoint $O$. If $ABCD$ is a section of a right pyramid of vertex $P$ with rectangular base, then $PO$ is the bisector of both $\angle APC$ and $\angle BPD$. Line $PO$ must then be perpendicular to $AC$ and $DC$, hence $PO$ is perpendicular to plane $ABCD$.

But we should also have $\angle APC=\angle BPD$ and that is impossible because $$ \tan{\angle APC\over2}={AO\over PO}\ne{BO\over PO}=\tan{\angle BPD\over2}. $$ When $ABCD$ is not a parallelogram, the locus of $P$ turns out to be a circle. But finding whether or not $\angle APC=\angle BPD$ for some $P$ on the circle is not easy. Experimenting with GeoGebra I found that, even in that case, this condition is sometimes impossible to meet.

EDIT.

Let's see a more interesting example, a kite with vertices $$ A=(-1,0,0),\quad B=(0,a,0),\quad C=(2,0,0),\quad D=(0,-a,0), $$ where $a>0$ (in figure below $a=1$). The locus of points $P$ such that $\angle BPO=\angle DPO$ is plane $y=0$. The locus of points $P$ such that $\angle APO=\angle CPO$ is also the locus of $P$ such that $PC/PA=2$, i.e. an Apollonian sphere with diameter $OE$, where $E=(-4,0,0)$.

The vertex $P$ of the pyramid lies then on the intersection of the above loci, i.e. on a circle of diameter $OE$, perpendicular to plane $z=0$.

Let then $P=(t-2,0,\sqrt{4-t^2})$ a generic point on the upper half of that circle, where $-2\le t\le2$. We can easily compute: $$ PA^2=5-2t,\quad PC=2PA,\quad PB^2=PD^2=8+a^2-4t. $$ The cosine law gives then: $$ \cos\angle BPD={8-4t-a^2\over8-4t+a^2}, \quad \cos\angle APC={8-5t\over10-4t}. $$ The pyramid has a rectangular base only if $\angle APC=\angle BPD$, which gives: $$ t={9\over4}a^2-2. $$ Hence we have a solution in this case, but only if $a<4/3$: for $a>4/3$ we would get $t>2$, which is impossible.