The Art of geometric constructions
Angel Garcia
We have been discussing the "simplest construction of 1/18 and its sqrt.
We have reached already a final construction for it (the sqrt(1/18) which
I have posted as nice drawing in my URL (see below): just after all "monthly
mathematical comments".
The "preview" shown in there is no trifle: sure that the sqrt(1/18) = (Od, Ad)
is a trivial construction, but it is probably "the simplest"; just a
semicircle Od, Ad, td of radius 1/6 (half of given 1/3) which gives for
segment
(Od,Ad) = sqrt(2)/6 = sqrt(1/18)
Generally the ART of geometric constructions is very intricate and not trivial
at all. One can distinguish many facets to it. For istance: the FIRST
construction ever given for the famous "Reichenbacher's conundrum",
namely, that 10-digit approximation to Pi as presented in my math. month. of
dec.-1996, was presented in an obscure german journal in year 1900 (just a
page: I have photocopy and can post it if interest)... that FIRST construction
was cumbersome and analytic, rather than ELEGANT. A few years later came a
young genius Ramanujan and rediscovered such unique conundrum (perhaps the
grandest ever found on Earth) and, not knowing about publication of
Reichenbacher's publication, he thought to be the FIRST to discover it and
Published his own construction for it: this one of Ramanujan is really
ELEGANT and much shorter than that of Reichenbacher. Recently Lahoz has
published a 3rd. construction for the same as mere 'ornament' in front cover
of his book "The name CANADA in historic perspective" (1992): this time
the construction is ARTISTIC and ELEGANT: it portrays a Christ hanging
(in profile) from sort of Cross: yet it is valid mathematical construction
for the same famous Reichenbacher's conundrum,
Elegance in geometric constructions is incompatible with:
a) too many details.
b) disproportionate length of lines (one very short others very long).
Elegance mixed with Art is always a bonus which rarely has been
cultivated (just that case of the Cross by Lahoz) on Earth; but obviously
that is the main trend at Cydonia: just watch "The Gladiator": insuperable
work of Art with just few master strokes via angle beta; its FIST at D&M
is really astonishing in mathematical precision and artistic depiction
of a human closed hand with a dagger.
--
Angel, secretary of Universitas Americae (UNIAM). His proof of ETI at
Cydonia and complete Index of new "TETET-96: Faces on Mars.." by Prof.
Dr. D.G. Lahoz (leader on ETI and Cosmogony) can be studied at URL:
http://www.ncf.carleton.ca/~bp887 ***************************
Angel Garcia
Angel Garcia (bp...@FreeNet.Carleton.CA) writes:
> We have been discussing the "simplest construction of 1/18 and its sqrt.
> We have reached already a final construction for it (the sqrt(1/18) which
> I have posted as nice drawing in my URL (see below): just after all "monthly
> mathematical comments".
> The "preview" shown in there is no trifle: sure that the sqrt(1/18) = (Od, Ad)
> is a trivial construction, but it is probably "the simplest"; just a
> semicircle Od, Ad, td of radius 1/6 (half of given 1/3) which gives for
> segment
> (Od,Ad) = sqrt(2)/6 = sqrt(1/18)
------
>
> Elegance in geometric constructions is incompatible with:
>
> a) too many details.
> b) disproportionate length of lines (one very short others very long).
As it happened historically with Reichenbacher's conundrum (3 independent
published geometric constructions), it is very desirable to find (or try it
at least) some geometric constructions for these SIMPLE numerical conundrums
which are of large size.
In the case of Angel's conundrum:
3^(-4) * sqrt [ 1 - 3^(-4) ] = 0.01226 9 2344444432 9 ....
we have already found a very reasonable construction for it... well actually
we followed the given hints at Cydonia (which are geometric constructions
in germinal way) and we are satisfied about it: one can see part of such
construction in that mentioned "preview" at end of my math. comments (see
URL); from 3 and 1/3 one gets first the length (Md, hd) and from that the
rest is trivial. Note that such incipient construction is "reinforced" via
ancillary conundrum: angle (qd, hd, pd) = 0.111 341 0 143 ...
Of course we are open to new constructions for it (if they are really
simple and elegant).
In the case of Kevin's conundrum:
1/809 * sqrt( 1 - 1/402 ) = 0.0 123455555555 ...
(the largest so far found with remarkable simplicity) a geometric construction
for it is not easy (within reasonable elegance) due to the prime 809; some
preliminary re-writing of it in terms of smaller numbers seems to be required.
Curiously one can write it also as:
sqrt ( 401 / 263101362 )
which shows a palindrome in denominator. Any TENTATIVE constructions for it ?.
Angel Garcia
Angel Garcia (bp...@FreeNet.Carleton.CA) writes:
> Angel Garcia (bp...@FreeNet.Carleton.CA) writes:
>> We have been discussing the "simplest construction of 1/18 and its sqrt.
>> We have reached already a final construction for it (the sqrt(1/18) which
>> I have posted as nice drawing in my URL (see below): just after all "monthly
>> mathematical comments".
>>
>> Elegance in geometric constructions is incompatible with:
>>
>> a) too many details.
unless, of course, some 'artistic drawing' is intended.
c) too many discontinuous 'transfers' of lengths from place to place.
> As it happened historically with Reichenbacher's conundrum (3 independent
> published geometric constructions), it is very desirable to find (or try it
> at least) some geometric constructions for these SIMPLE numerical conundrums
> which are of large size.
>
> In the case of Angel's conundrum:
> 3^(-4) * sqrt [ 1 - 3^(-4) ] = 0.01226 9 2344444432 9 ....
> we followed the given hints at Cydonia (which are geometric constructions
> Of course we are open to new constructions for it (if they are really
> simple and elegant).
Let us put forward some comments about two open problems:
A) A reasonably elegant (even if tentative) construction of Kevin's conundrum:
sqrt(1- 1/402) / 809 = 0.0123455555555 .. (8 fives in a row)
B) To prove (or disprove) Lahoz's conjecture: "It is not possible to build a
quinsector compass with less than 7 hinged rods".
To A): some preliminary re-writing is needed to avoid nonsensical
construction of large prime 809.
To B): Using 2-dimensional (hinged) tools, like a T (say), then it is possible
to build quinsector compass with five (or less ?) pieces. I did such
thing and is published in my URL at jan-feb-1997 file of month. math.
comments. The conjecture refers to exclusive use of 1-dimensional
Archimedes' tools (unmarked hinged rods as in Lahoz's 7-rod quintrisector).
Angel Garcia
Regarding B) above there is a related construction due to Newton in his
"Arithmetica Universalis". He describes several "trisections" with due
emphasis in the "master of all", namely the insuperable 4-rod trisector
of Archimedes. Newton describes his own device (a sex-tri-sector compass)
as follows:
A
| \
\ \
| \ Semicircle centered at O, with diameter DB, has arc
\ \ (DFG) to be trisected. Lines OA and OG give the
| \ solution: angle AOG = DOG / 3. Can you prove it ?.
\ \ G
F | * / \ *
* \ / | \ * As a "compass" the hinges are at
* /|K . \ * D,O,B,K,G and A. It has 5 rods:
* / \ | \ * one more than in Archimedes' trisector
/ | | \ * but the 'bonus' is that such device is
* / \ . \ actually a sextrisector. In fact
/ | | \ * as such compass opens its arms
* / \ | \* the angle at A is continuously:
D-----------------O---------------B angle OAB = DOG / 6 . Proof ?
Therefore this unique device of the great Newton is not only "trisector"
but also "sexsector" (shall we name it 'sixsector' to avoid Latin ?).
It remains the nagging Lahoz's conjecture:
"Not possible to build quinsector with less than 7 hinged rods". Proof ?.