Simple Geom problem.

[davieddy]

The coincidence is that the intersection of
V(n-5) and V(n+5) lies the same distance from
the centre as that between V(n-6) and V(n+6).
That this occurs on a diagonal is unsurprizing.

[davieddy]

My terminology needs elaboration, but I'm sure you are up to it!

[davieddy]

Originally Posted by davieddy

My terminology needs elaboration, but I'm sure you are up to it!

On second thoughts, I think it needs much elaboration!

[mfgoode]

Originally Posted by davieddy

On second thoughts, I think it needs much elaboration!

Frankly your one liners are getting boring as they lack adequate info content as you yourself have said above.

A hundred thousand welcomes !

Mally

[mfgoode]

Originally Posted by cheesehead

In Mesopotamia the Sumerians had, by 2400 BC, a calendar of 12 months of 30 days each. Apparently they valued the arithmetic niceties of the number 360 more than they were irritated by the five-day yearly discrepancy. They also invented the 360-degree circle, but not subdivisions of degrees, which came later.
.

Not exactly. The Egyptians were well aware of the discrepancy and always added the 5 days at the end of their year.
The Great Pyramid was constructed over 5000 years ago (Cheops circa 4000 B.C.)
These and other secrets were encrypted in the Kufru including the accurate values of Pi , epsilon, phi. and Tau, i , 0 , infinity the seven Royal numbers.
Recommended reading "Pyramid Prophecies" by Max Toth

Mally

[philmoore]

Originally Posted by davieddy

The coincidence is that the intersection of
V(n-5) and V(n+5) lies the same distance from
the centre as that between V(n-6) and V(n+6).
That this occurs on a diagonal is unsurprizing.

If you mean that it occurs on a diameter is unsurprising, then what you say is correct. Returning to my earlier notation, we have the diameter V₃V₁₂, and the two pairs V₂V₉ and V₄V₁₅, and also V₁V₇ and V₅V₁₇. We can either prove that the diameter and a single segment from each pair are concurrent, or we can prove that one segment from either pair is concurrent with both members of the other pair.

[davieddy]

Yep.

[davieddy]

Originally Posted by mfgoode

Frankly your one liners are getting boring as they lack adequate info content as you yourself have said above.
Mally

I think your penchant for verbosity is widely understood :)

[davieddy]

Originally Posted by philmoore

Thanks for posting that, Mike. I'll try to explain what I found interesting. It was provoked by David's observation of reflection. Suppose we label the vertices consecutively as V₁, V₂, V₃, ... V₁₈. Then what we notice is that the following line segments are all concurrent: V₁V₇, V₂V₉, V₃V₁₂, and the reflections of the first two segments across the last: V₄V₁₅ and V₅V₁₇. My guess is that there is some underlying symmetry that explains why these lines are concurrent, but I haven't done much research on it. Could something related to Pascal's theorem be at work here?

With your notation, Mally's constuction line is V₂V₅.
This must help somehow.

Remind me of Pascal's theorem please.

David

[davieddy]

Thanks for the history lessons.
Approx 360 days in a year was (I would hope) an
unimportant distraction. But a hexagon consisting
of 6 equilateral triangles allied to base 60 arithmetic
seems eminently sensible.

David

[davieddy]

How did the Babylonians (or whoever) represent their 60 digits?