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From: Thomas Nordhaus on 7 May 2010 05:54
Hi.
I'm trying to solve the following (self-posed ;-) ) problem:

"Let ABC and A'B'C' be given triangles of the Euclidean plane.
Determine whether A'B'C' can be inscribed inside of ABC. That means,
determine whether there is a triangle T' congruent to A'B'C' such that
T' is a subset of ABC".

Right now, I only have a few necessary conditions and a few sufficient
conditions. But they don't come close to "closing the gap". Here is
list of conditions I found so far:

Necessary:

1. max(a',b',c') <= max(a,b,c) (Condiion on the diameters of the
triangle)
2. a'b'sin(gamma') <= absin(gamma) (Condition on the areas + similar
conditions for different pairs of sides and enclosed angles).

Sufficient:

1. Radius of circumscribed circle of A'B'C' <= Radius of inscribed
circle of ABC.

I'm not much further than this. It would be nice to know whether the
following is true:

(*) Suppose A'B'C' \subset ABC, then there is a congruent copy T of
A'B'C' such that T \subset ABC and T shares one or more vertices with
ABC.

If (*) were true one could reduce the problem considerably.

Thanks for any kind of help.

--

Thomas Nordhaus
From: Achava Nakhash, the Loving Snake on 7 May 2010 07:29
On May 7, 2:54 am, Thomas Nordhaus <thomas.nordh...(a)googlemail.com>
wrote:
> Hi.
> I'm trying to solve the following (self-posed ;-) ) problem:
>
> "Let ABC and A'B'C' be given triangles of the Euclidean plane.
> Determine whether A'B'C' can be inscribed inside of ABC. That means,
> determine whether there is a triangle T' congruent to A'B'C' such that
> T' is a subset of ABC".
>
> Right now, I only have a few necessary conditions and a few sufficient
> conditions. But they don't come close to "closing the gap". Here is
> list of conditions I found so far:
>
> Necessary:
>
> 1. max(a',b',c') <= max(a,b,c) (Condiion on the diameters of the
> triangle)
> 2. a'b'sin(gamma') <= absin(gamma) (Condition on the areas + similar
> conditions for different pairs of sides and enclosed angles).
>
> Sufficient:
>
> 1. Radius of circumscribed circle of A'B'C' <= Radius of inscribed
> circle of ABC.
>
> I'm not much further than this. It would be nice to know whether the
> following is true:
>
> (*) Suppose A'B'C' \subset ABC, then there is a congruent copy T of
> A'B'C' such that T \subset ABC and T shares one or more vertices with
> ABC.
>
> If (*) were true one could  reduce the problem considerably.
>
> Thanks for any kind of help.
>
> --
>
> Thomas Nordhaus

Your definition is inscribed is atypical, since usually it means that
whatever inscribed touches the edge of the figure in which it is
inscribed. Thus for one triangle to be inscribed within another
should mean that the vertics of the inscribed trinagle are on the
perimeter of the triangle in which it is inscribed. A terminalogy
nitpick only - your question is still interesting. The question as I
described inscribed is also interesting.

I don't see how your conjecture can be true. You can take a tiny
triangle with an angle at some vertex that has a larger angle than any
on the inscribee. That angle cannot be placed at any vertex of the
inscrobee (by which I mean the triangle in which it is inscribed, and
I am not using a real English word). I have no idea of what the
answer is to either of our questions.

Regards,
Achava
From: Thomas Nordhaus on 7 May 2010 08:27
On 7 Mai, 13:29, "Achava Nakhash, the Loving Snake"
<ach...(a)hotmail.com> wrote:
> On May 7, 2:54 am, Thomas Nordhaus <thomas.nordh...(a)googlemail.com>
> wrote:
>
>
>
> > Hi.
> > I'm trying to solve the following (self-posed ;-) ) problem:
>
> > "Let ABC and A'B'C' be given triangles of the Euclidean plane.
> > Determine whether A'B'C' can be inscribed inside of ABC. That means,
> > determine whether there is a triangle T' congruent to A'B'C' such that
> > T' is a subset of ABC".
....

>
> > I'm not much further than this. It would be nice to know whether the
> > following is true:
>
> > (*) Suppose A'B'C' \subset ABC, then there is a congruent copy T of
> > A'B'C' such that T \subset ABC and T shares one or more vertices with
> > ABC.
>
> > If (*) were true one could  reduce the problem considerably.
>
> Your definition is inscribed is atypical, since usually it means that
> whatever inscribed touches the edge of the figure in which it is
> inscribed.

OK - I'll call it "weakly inscribed" then. ;-) I think one could fill
a whole thread on that topic.

>
> I don't see how your conjecture can be true.  You can take a tiny
> triangle with an angle at some vertex that has a larger angle than any
> on the inscribee.  That angle cannot be placed at any vertex of the
> inscrobee (by which I mean the triangle in which it is inscribed, and

Well, not all of the angles of the tiny triangle can be larger than
the largest angle of the containig triangle. Why not place the small
triangle at that vertex and see if you can rotate it cleverly that it
stays inside? Or maybe I don't understand your point.

--

Thomas Nordhaus
From: Achava Nakhash, the Loving Snake on 7 May 2010 08:48
On May 7, 5:27 am, Thomas Nordhaus <thomas.nordh...(a)googlemail.com>
wrote:
> On 7 Mai, 13:29, "Achava Nakhash, the Loving Snake"<ach...(a)hotmail.com> wrote:
> > On May 7, 2:54 am, Thomas Nordhaus <thomas.nordh...(a)googlemail.com>
> > wrote:
>
> > > Hi.
> > > I'm trying to solve the following (self-posed ;-) ) problem:
>
> > > "Let ABC and A'B'C' be given triangles of the Euclidean plane.
> > > Determine whether A'B'C' can be inscribed inside of ABC. That means,
> > > determine whether there is a triangle T' congruent to A'B'C' such that
> > > T' is a subset of ABC".
>
> ...
>
>
>
> > > I'm not much further than this. It would be nice to know whether the
> > > following is true:
>
> > > (*) Suppose A'B'C' \subset ABC, then there is a congruent copy T of
> > > A'B'C' such that T \subset ABC and T shares one or more vertices with
> > > ABC.
>
> > > If (*) were true one could  reduce the problem considerably.
>
> > Your definition is inscribed is atypical, since usually it means that
> > whatever inscribed touches the edge of the figure in which it is
> > inscribed.
>
> OK - I'll call it "weakly inscribed" then. ;-) I think one could fill
> a whole thread on that topic.
>
>
>
> > I don't see how your conjecture can be true.  You can take a tiny
> > triangle with an angle at some vertex that has a larger angle than any
> > on the inscribee.  That angle cannot be placed at any vertex of the
> > inscrobee (by which I mean the triangle in which it is inscribed, and
>
> Well, not all of the angles of the tiny triangle can be larger than
> the largest angle of the containig triangle. Why not place the small
> triangle at that vertex and see if you can rotate it cleverly that it
> stays inside? Or maybe I don't understand your point.
>
> --
>
> Thomas Nordhaus

My point is a mistaken. I was thinking that vertex with the largest
angle of the weakly inscribe triangle must be able to be coincident
with one of the vertices of the inscribee, but this was not what you
said. If I have a chance, I will think about your interesting
problem. It is the middle of the night for me here -5:47 AM as I
write, this, and I am finally getting sleepy.

Regards,
Achava
From: spudnik on 8 May 2010 16:58
spatially, there are "mutaully inscribed tetrahedra,"
meaning that the vertices of one lie on the faces
of the other, and vise versa.

thus:
the formalism of relativity isn't needed, if
one does not presume that Pascal's vacuum was perfect
(and still is) a la "Newtonian optics" or ray-tracing, and
the calculus-launch problemma of the brachistochrone.

thus:
how about this:
show us that your theory agrees with Sophie Germaine; then,
tackle the remaining primes.

thus:
NB, Lanczos used quaternions in _Variational Mechanics_
for special relativity, and it's just "real time" and
"three ('imaginary') axes of space;" but,
this is just the original "vectors."

compare Lanczos' biquaternions
with the "Cayley-Dickerson doubling" procedure,
to go from real to complex to quaternion to octonion.

"wroldlines" are just the crappola in Minkowski's "pants,"
totally obfuscatory outside of a formalism --
time is not a dimension; time is awareness & mensurability
(of dimensionality !-)

thus:
try a search on Gauss & Ceres. or
"go" to wlym.com.
> This problem and its solution are found in a paper by Ceplecha, 1987,

thus:
the problem appears to be,
"some observers measure the angle to the marker,
relative to the other observers,"
which would not give you the distance *on a plane*,
because of similar trigona. Gauss meaasured the curvature
of Earth with his theodolite *and* a chain measure
of distance (working for France in Alsace-Lorraine,
triangulatin' that contested area .-)

thus:
notice that no-one bothered with the "proofs" that I've seen, and
the statute of limitation is out on that, but, anyway,
I think it must have been Scalia, not Kennedy,
who changed his little, oligarchical "Federalist Society" mind.

thus:
sorry; I guess, it was Scalia who'd "mooted" a yea on WS-is-WS, but
later came to d'Earl d'O. ... unless it was Breyer, as I may
have read in an article about his retirement.

> I know of at least three "proofs" that WS was WS, but
> I recently found a text that really '"makes the case,"
> once and for all (but the Oxfordians, Rhodesian Scholars, and
> others brainwashed by British Liberal Free Trade,
> capNtrade e.g.).
> what ever it says, Shapiro's last book is just a polemic;
> his real "proof" is _1599_;
> the fans of de Vere are hopelessly stuck-up --
> especially if they went to Harry Potter PS#1.
> http://www.google.com/url?sa=D&q=http://entertainment.timesonline.co....

--Light: A History!
http://wlym.com

--Waxman's capNtrade#2 [*]:
"Let the arbitrageurs raise the cost of your energy as much as They
can ?!?"
* His first such bill was in '91 under HW on NOx & SO2 viz acid rain;
so?
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