Recall that a quadrisecant of a knot is a line that passes thru four points on it. If the points appear on the line in the order $a$, $b$, $c$, $d$ and on the knot in the order $a$, $c$, $b$, $d$, then the quadrisecant is called alternated.
It is well known that any nontrivial knot has an (alternated) quadrisecant. The statement looks very elementary, but all proofs I saw are quite involved, and it surprises me.
Is there an elementary proof that every knot has an (alternated) quadrisecant?
Here is how I like to think of these kinds of geometric problems. Rather than working in $\Bbb R^3$, let's work in $S^3$.
Given a knot $K$ in $S^3$, there is the configuration space of points in $K$. Let's consider $5$ points in $K$, i.e. $C_5 K$. This is a 5-manifold, diffeomorphic to a trivial 4-ball bundle over $K$.
In $C_5 S^3$ there is the subspace where all five points sit on a round circle. By "round circle" I mean the intersection of a 2-dimensional hyperplane in $\Bbb R^4$ with $S^3$. Let's call this the Round Circle Subspace, $RSS \subset C_5 S^3$.
$RSS$ has dimension $11$, or co-dimension $4$.
$RSS$ has a subspace that I call the pentacle subspace.
The idea is that the five points have a natural cyclic ordering $1 \to 2 \to 3 \to 4 \to 5 \to 1$. But five points on a circle also have a cyclic ordering. So we take the path-component of $RSS$ where adjacent points in the natural (or "label") ordering are not adjacent in the circle ordering. i.e. if you draw lines from $p_i$ to $p_{i+1}$ for all $i$ (index mod $5$) then you get a pentacle, provided you also draw the circle they are on.
Given a knot, generically the pentacles in $C_5 K$ are a $1$-dimensional submanifold. If you stereographically project at a point of $S^3$ that is a point of a pentacle, that pentacle is converted into an alternating quadrisecant for the projected knot -- which will now be a long knot, i.e. running off to infinity.
In my opinion this setup makes a lot of formal sense. If you can prove non-trivial long knots have alternating quadrisecants, it follows for closed knots by a closure operation.
A long knot you could view as a smooth embedding $\mathbb R \to \mathbb R^3$ that has a fixed axis as an asymptote. Define the collinear triple subspace of $\mathbb R^3$ as the space of triples that sit on a common straight line, as a subspace of $C_3 \mathbb R^3$. This space has three path components, usually denoted $Col_i$ where $p_i$ is in the middle of the linear ordering of the line.
Given a long knot $f : \mathbb R \to \mathbb R^3$ one takes the induced map $f_* : C_3 \mathbb R \to C_3 \mathbb R^3$ and the preimages $f_*^{-1}(Col_1)$ and $f_*^{-1}(Col_3)$. $C_3 \mathbb R$ is basically a copy of $\mathbb R^3$ and the alternating quadrisecants correspond to double points of $f_*^{-1}(Col_1)$ passing under $f_*^{-1}(Col_3)$, when forgetting the first coordinate of $C_3 \mathbb R$. You can also do it as a crossing count forgetting the third coordinate.
This does not answer your question, but it gives some context. If you don't have an alternating quadrisecant, it means $f_*^{-1}(Col_1)$ is purely "over top of" $f_*^{-1}(Col_3)$, from the point of view of projecting out the first coordinate of $C_3 \mathbb R$.
To me this is the primary geometric meaning associated to not having alternating quadrisecants.
edit: I should emphasize one other point. Non-trivial knots having alternating quadrisecants is equivalent to stating that non-trivial knots in $S^3$ have pentacles. The space of pentacles in $C_5 K$ is generically an oriented $1$-manifold. If you consider the forgetful map $C_5 K \to K$ that forgets all but one point, then the degree of this map, restricted to the pentacle subspace, is the type-$2$ Vassiliev invariant of the knot, i.e. the coefficient of $z^2$ in the Conway normalization of the Alexander polynomial.
