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-   -   Simple Geom problem. (https://www.mersenneforum.org/showthread.php?t=7093)

davieddy 2007-03-07 13:24

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 2007-03-07 13:27

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

davieddy 2007-03-07 14:04

[quote=davieddy;100137]My terminology needs elaboration, but I'm sure you are up to it![/quote]
On second thoughts, I think it needs much elaboration!

mfgoode 2007-03-07 16:45

[QUOTE=davieddy;100138]On second thoughts, I think it needs much elaboration![/QUOTE]

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 2007-03-07 17:09

[QUOTE=cheesehead;100066]
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.
.[/QUOTE]

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 2007-03-07 19:08

[QUOTE=davieddy;100136]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.[/QUOTE]

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 V[sub]3[/sub]V[sub]12[/sub], and the two pairs V[sub]2[/sub]V[sub]9[/sub] and V[sub]4[/sub]V[sub]15[/sub], and also V[sub]1[/sub]V[sub]7[/sub] and V[sub]5[/sub]V[sub]17[/sub]. 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 2007-03-07 22:18

Yep.

davieddy 2007-03-07 23:42

[quote=mfgoode;100150]Frankly your one liners are getting boring as they lack adequate info content as you yourself have said above.
Mally[/quote]

I think your penchant for verbosity is widely understood :)

davieddy 2007-03-08 10:11

[quote=philmoore;100095]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 V[sub]1[/sub], V[sub]2[/sub], V[sub]3[/sub], ... V[sub]18[/sub]. Then what we notice is that the following line segments are all concurrent: V[sub]1[/sub]V[sub]7[/sub], V[sub]2[/sub]V[sub]9[/sub], V[sub]3[/sub]V[sub]12[/sub], and the reflections of the first two segments across the last: V[sub]4[/sub]V[sub]15[/sub] and V[sub]5[/sub]V[sub]17[/sub]. 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?[/quote]

With your notation, Mally's constuction line is V[sub]2[/sub]V[sub]5[/sub].
This must help somehow.

Remind me of Pascal's theorem please.

David

davieddy 2007-03-08 10:38

360 degrees
 
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 2007-03-08 11:06

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


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