New answers tagged

6

It is not possible with 7 (i.e., with a triangle $T$ and a quadrilateral $Q$). I write a rough proof. First, any quadrilateral $Q$ lying in a plane $\pi$ can be partitioned in two triangles $Q_1$ and $Q_2$, whose common edge is a diagonal $d$ of $Q$. Now the intersection of the triangle $T$ with $\pi$ consists of two points (otherwise they are coplanar and ...


4

Here is another example with 8 vertices: a short fat Star Trek symbol and a square in orthogonal planes. Since the distance between the base points of the red figure is greater than its height, one cannot rotate the square to take it out. Addendum: As Saúl Rodríguez Martín mentions, this example may not work. If we assume, however, that the links are ...


14

As Sam Hopkins commented, 8 vertices are enough. Let $Q$ be the pentagon from the picture and let $\pi$ be the plane containing it. Now we can define the triangle $P$ as a triangle of less diameter than the black segment and intersecting $\pi$ at two points: one point $a_0$ in the open blue region $B$ and one point $b_0$ in the open green region $G$. $P$ and ...


3

I think that 8 might be possible, by interlocking two Star Trek symbols as shown below. Adendum: This candidate may not work, as quarague points out, but I leave it as a potential "how not to" example. There are other ways to cross the two figures, while they remain unlinked, and one of these variations could be more promising. Also I think it is ...


1

Here is a proof using analytic tools, complex numbers, and formulas for the involved points in terms of them. Points in the plane will be denoted by capital letters like $A,B,C,D;N;P,Q,R,S$ and the corresponding affixes will be their lower cousins, respectively $a,b,c,d;n;p,q,r,s\in\Bbb C$, and decorations (sub- or upper indices) will be kept. First let us ...


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