9 Main Theorems

Theorem 63

✓

There does not in fact exist a noperthedron Rupert solution with

\begin{align*} \theta _1,\theta _2& \in [0,2\pi /15] \subset [0,0.42], \\ \varphi _1& \in [0,\pi ] \subset [0,3.15],\\ \varphi _2& \in [0,\pi /2] \subset [0,1.58],\\ \alpha & \in [-\pi /2,\pi /2] \subset [-1.58,1.58]. \end{align*}

Proof ▶

By 58, there is a valid solution table containing a valid row whose pose interval is a superset of the 5-d interval above. By 62, this means there is no Rupert solution in that interval.

Theorem 64

✓

There is no 5-parameter pose that makes the noperthedron have the Rupert property.

Proof ▶

Theorem 63 says there is no tight pose that makes the noperthedron Rupert. Corollary 8 says that this suffices for the general case.

Theorem 65

✓

There is no purely rotational pose that makes the noperthedron have the Rupert property.

Proof ▶

Suppose there were a purely rotational pose. Then convert that to an equivalent 5-parameter pose with Theorem 19 and appeal to Theorem 64.

Theorem 66

✓

There is no pose that makes the noperthedron have the Rupert property.

Proof ▶

By Theorem 20, we need only show that the noperthedron is pointsymmetric to see that if it is Rupert, then it must be Rupert via a purely rotational pose. But Lemma 6 shows exactly this. And yet we know via Theorem 65 that the noperthedron is not rotationally Rupert, so we have a contradiction, hence the noperthedron has no pose that makes it Rupert.

Theorem 67

✓

The noperthedron is not a Rupert set.

Proof ▶

By Theorem 66, there is no pose that makes the noperthedron a Rupert set.

Theorem 68

✓

The noperthedron is not a Rupert polyhedron.

Proof ▶

By Theorem 18 it suffices to show that the convex hull of the noperthedron vertices is not a Rupert set. But this is exactly what Theorem 67 shows.