Re: Some theorems about polyhedra


On Dec 19, 5:45 am, Charles Delorme <c...@xxxxxx> wrote:
Dr Tim wrote:
1. If five faces of a distorted cube are cyclic quadrilaterals, then
the sixth is too.
2. If five faces of a distorted cube are quadrilaterals-of-tangents
(QOTs), then the sixth is too,
3. If all-but-one of the faces of a trihedral polyhedron are cyclic,
then the last is too.
4. If all-but-one of the faces of a polyhedron-with-quadrilateral-
faces is a QOT, then the last is too.

Does anyone recognise these theorems?
If not, you saw them here first!

Tim Robinson

Dear colleague,
the first property is the same as a Miquel theorem (in the plane)
if each of the five 4-uplets of points ABCD , ABA'B' , BCB'C' , CDC'D'
and ADA'D' is cocylic, then the sixth one A'B'C'D' is also cocyclic.

An inversion sending the sphere with your eight points to a plane
relates the properties.

Sincerely

For number (1), first you can prove if you take any two adjacent
facets which are cyclic quadrilateral then their 6 vertices lie on a
sphere, and then consequently it pops out that all the vertices of
this distorted cube must lie on this sphere if the other three facets
have to be cyclic quadlateral. And since the sixth facets lies on the
intersection of the plane containing that facets and this sphere
[ which would be a circle that has all the sixth facets' vertices on
its perimeter ], it must be also a cyclic quadlateral.

I think by the same argument and just a little bit of work on the some
different configuration of the polyhedron, number (3) would be solved.
Like if you assume that the polyhedron is convex, also you don't need
to take it to be trihedral, and it has that property that every facets
except maybe one is a cyclic quadrilateral, then by the same argument
I used for in (1) you can prove that all the vertices of this
polyhedron lie on a sphere and the intersection of this sphere with
the plane containing that special facet is the cyclic circle of that
facet.

Number (2) is easy, it only requires one simple lemma. a quadrilateral
ABCD is tangential quadrilateral iff AB+CD=BC+CA.
now label your distorted cube by ABCDD"A"B"C" such that facets ABCD,
CDD"C", ADD"A", BCC"B", and A"B"C"D" are tangential quadrilateral, and
we want to prove that the last facet ABB"A" is also tangential
quadrilateral.

by lemma we have

AB + B"A" = (BC+DA-CD) + (B"C"+D"A"-C"D")
BB" + AA" = (BC+B"C"-CC") + (AD+A"D"-DD")
CC" + DD" = CD + C"D"
------------------------------------
thus AB+B"A"=BB"+AA". So the facet ABB"A" is also tangential
quadrilateral.

number (4) probably would get solved by using this lemma many times.
.

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