MY 11 TRISECTION APPROXIMATION METHODS
or
MY TRISECTION ENCYCLOPEDIA

 
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AUTHOR:  Klaas Van Fleet

 

B.A.Sc. 1969 at UNIVERSITY of WATERLOO

 

Mechanical Engineering: La Place & La Grange

  pg. i

ABOUT THE AUTHOR: (cont'd)

i
 
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Born of Dutch parents during World War II of 1943. Walked away from
 
an annihilated bunker that was too full to let us in. Became landed
  Pg. i.a
immigrants and Canadian citizens in 1953. Doctors removed shrapnel
 
from my chest when I was 15 and again when I was 60 scraped shrapnel
 
from my cranium. Owned different cars since I was 16, almost lost my
 
left ear in my first car accident and knows Ontario like the back of my
 
hand. Almost drowned in my sailboat on Lake Erie in bright but wavy
 
weather and almost got run over by a Laker before that. Enjoyed
 
meccano sets, radio control models and crafts. Donated a large box of
 
Comic Books (only @10¢) by a concerned invalid, because I was suspect
 
-ed of having Lukemia and given a year tto live. An Empire Loyalist at
 
heart who graduated from UNIVERSITY OF WATERLOO and other
 
scholastic institutions around the GTA; and worked as a Consultant
 
Engineer all over Ontario and in Nova Scotia. A Torontonian who
 
played and won chess with almost every computer in downtown
 
Toronto. Was a ZX81 computer club member who lectured machine
 
code. Loves designing computer programs like kvvchess.com and
 
JavaScript. And published a 'Machine Code Training For The
 
IBM-PC Using Debug.exe' book through Xlibris.com Corp.
  Pg. i.b
My RÉSUMÉ can be found at:
 
http://www.stnick.faithweb.com/r/resume.gif
 
My WEB PAGES can be found at:
 
/klaus_vanv/index.htm
 
http://www.geocities.com/klaus_vanv/nick.htm#mnu
 
http://www.geocities.com/klaus_vanv/art7.htm
 
http://www.stnick.faithweb.com/javagames/games.htm
 
http://www.geocities.com/klaus_vanv/javagames/games.htm
 
http://www.klausvanv.4mg.com/javagames/games.htm
 
http://www.klausvanv.50megs.com/javagames/games.htm
 
My EMAIL address is:
 
nickgvanv@aol.com
 
My POSTAL address is:
 
4-375 Sackville St.,
 
TORONTO, ON
  Pg. i.c
M5A 3G5
 
CANADA
 
416-921-4653
 
I am eager to read all the comments about this book from any
 
mathematician around the globe - including revisions and additions!
 

EXPLANATION OF THE COVER:

 
3 sailsurfers or the QUEEN's sloop, 'THE QUEBEC', above, in low
 
white fog made from the 3 complete color plots of FIG. 5.
 

MY CONTRIBUTION TO MATHEMATICS:

 
1. I think that my THE ALLEYWAY PROBLEM solution is another
   
  addition to mathematics using trisection as a solution! CLICK HERE
   
2. I feel that the LOGO TRISECTION solution is accurate enough for
   
  most mathematical applications! - also it seems to be a part of
   
  reality in the PEACE SYMBOL. CLICK HERE
    pg. i.d
3. I am proudest of my SAIL SURFER TRISECTION for its complex-
   
  ity and focus ability to become a unique computer solution! CLICK HERE
   
4. My TRISECTION GADGET would be useful in getting the FIELD
   
  COIL current directly by trisecting STEPPER and 3-PHASE
   
  resultant current diagrams. CLICK HERE
   
5. I remain hopeful for my solution to 'THE PLANIMETER FORMULA SOLVES
   
  GOAT'S PROBLEM' since I had so much fun with it. CLICK HERE
 

MY THOUGHTS ON MATHEMATICS:

 
1. Nothing is parallel in MATH; because all lines come to a point, the
   
  distance to this point is unknown!
   
  - the distance might as well be infinity = 8" - from ear to ear!
   
  Also: space looks like a saddle!
   
  - how hard is it to measure a saddle!
   
  - it's a good thing that MATH has a point!
    pg. i.e
2. MATH should be converted to a new numbering system, such as:
   
  pi = 3.141592653589793 @ 7*0.44880 @ 22/7 would be written as pi = X + (Y,  Theta):
   
  pi = 3 + (0.1415927,  2° 47" 0.2305') or %ERROR = -2.0707E-16
   
  or pi = 3 + (0.141593,  7° 36" 15.7432') or %ERROR = 8.2827E-17
   
  or pi = 3 + (0.1416,  35° 1" 6.3619') or %ERROR = 1.4495E-16
   
  This formula makes me think of the 5th DIMENSION
   
  - where most of these unique points mighht as well be.
   
3. Until points turn into squares and cubes replace spheres, MATH will
   
  always be full of imaginary points!
   
4. Optical illusions own MATH since it represents an 'overalignment' of
   
  lines to please the eyeball instead of being unique in '3D'!
   
5. Parallel lines meeting at a point is an eyeball pleaser and has nothing
   
  to do with reality!
   
  - see my optical illusions at the bottomm of each Art#.htm example!
    pg. i.f
  - if you draftsmen are having a problem with drawings, it's because
   
     you are trained not to draw optical illusion points!
   
6. A square can easily be filled by a coil, giving whole numbers to any
   
  imaginary point by describing it as:
   
  Length + Rotation (of coil) = X + Theta
   
7. And what about the points outside the square but still on the coil;
   
  they probably define the 5th DIMENSION!
   
  - the only thing missing is the time, juust 1 second!
   
  - a crescent cubby hole that might define the VAN ALLEN BELT!
   
     making us a MATH planet!
   
  - In 3D this spiral would look like a saddle or snail shell!
   
8.

NOTE:

   
  Nominate me for TRISECTION for the Nobel Peace Prize in OSLO, NORWAY!
   
  Nobel Foundation Mailing address:
    pg. i.g
  Box 5232, SE-102 45 Stockholm
   
  Street Address: Sturegatan 14
   
  Phone: +46 8 663 09 20
   
  Fax: +46 8 660 38 47
   
  Web Address:
   
  http://www.nobel.se/index.htm
   
 
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  pg. ii  

ACKNOWLEDGEMENTS:

ii
 
Thanks to all the individuals who helped to make this book possible!
 
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  pg. iii  

DEDICATIONS:

iii
 
Dedicated to my dad and mom who witnessed my LOGO TRISECTION
 
and PERPETUAL CALENDAR, my brother who kept me focused on
 
computers to write this book, and all the kids who are creating their
 
own methods of trisection.
 
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  pg. iv

TABLE OF CONTENTS:

 
ABOUT THE AUTHOR ........................................................................ i
 
ACKNOWLEDGEMENTS .................................................................... ii
 
DEDICATIONS ..................................................................................... iii
 
  1. SLIDER TRISECTION fig. 1 .............................................................. pg.     1
 
  2. LOGO TRISECTION fig. 2 ................................................................ pg.     8
 
  3. TRISECTION GADGET fig. 3/5 ......................................................... pg.   16
 
  4. THE ALLEYWAY PROBLEM fig. 6/7 .................................................. pg.   36
 
  5. SQUARING THE CIRCLE fig. 8/9 ...................................................... pg.   40
 
  6. TRISECTION CHART fig. 10 ............................................................. pg.   50
 
  7. TRISECTION USING M/3 & M/pi fig. 11 ............................................... pg.   54
 
  8. TRISECTION RELATIONSHIPS fig. 12/13 .......................................... pg.   65
 
  9. CUPID'S BOW fig. 14 ..................................................................... pg.   78
 
 10. SIMPLICITY fig. 15 ...................................................................... pg.   85
 
 11. BASIC TRISECTION FORMULAS fig. 16 ............................................. pg.   87
 
 12. SAIL SURFER TRISECTION fig. 17/26 ............................................. pg.   90
  pg. v
 13. MEASUREABLE TRISECTION fig. 27 ................................................ pg. 118
 
 14. PLANIMETER & GOAT PROBLEM fig. 28/32 ...................................... pg. 121
 

APPENDIX A (11 fig.s) ........................................................

iv
 
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  pg. iv

APPENDIX A

iv
 

OTHER FAMOUS TRISECTION METHODS:

 
  1.

DICTIONARY OF MATHEMATICS: 1966

   
  by: T & W MILLINGTON:
   
    i. TRISECTION pg 246 is defined by the following formula:
   
    4cos3Æ - 3cosÆ + cos3Æ = 0
   
    ii. CONCOID OF NICOMEDES pg. 51
    NICOMEDES GIF
    r = asecÆ + b
   
   
    NICOMEDES' value of 1<R<2
   
    is disproved in my book:
   
    MY 11 TRISECTION METHODS pg 23 Click Here!.
   
    iii. EUCLIDEAN TRISECTION pg. 246
   
    was disproved by Wantzel in 1847
    pg. iv.a
    iv. LIMAÇON OF PASCAL pg. 133
    PASCAL GIF
    r =2acosÆ + b
   
   
   
   
   
   
   
    v. TRISECTRIX OF MACLAURIN pg. 246
    TRISECTRIX GIF
    x3 - 3ax2 + xy2 + ay2 = 0
   
    Also p = r + r2Æ
   
    And p = Æ2rcosÆ
   
   
    pg. iv.b
 2. GALOIS TEORY ON TRISECTION
   
  Theorem 440 in algebra vol1 rede1 qa 3 15 v1.91 c.4 Waterlo U
   
 3. THE SURPRISE ATTACK IN MATHEMATICAL PROBLEMS pg. 75
   
  by: L A GRAHAM
   
  LITTLE EUCLID NICOMEDES pg. 51
   
  by: W B Anderson euclid.gif
   
   
   
   
   
   
4. ENCYCLOPEDIA BRITANICA:
   
    i. ARCHIMEDES' TRISECTION THRU CHEATING
     
    archimed.gif
      pg. iv.c
    ii. KEMPE'S LINKAGE for trisecting any angle.
     
    kempe.gif
     
    iii. UNITY TRIANGLE
     
    UNITY GIF
     
    Any triangle as shown with all angles trisected. Every
     
    triangle contains a smaller central triangle on trisecting all 3
      pg. iv.d
    angles. This triangle is known as the unity triangle for an
     
    equilateral triangle with all angles (60o) and sides equal.
     
    Perhaps the WANKEL engine should be called the UNITY
     
    engine!
     
    orchidmy.gif
      pg. iv.e
5. FAMOUS TRISECTION METHODS found in TORONTO LIBRARY, Yonge & Asquith
     
  i. TORONTO LIBRARY'S microfishe 02754
     
    TRISECTION OF ANY RECTILINEAL ANGLE 1980 ch. 1
     
    EXTRAORDINARY GEOMETRICAL DISCOVERY:
     
    by: ANDREA II DOYLE
     
    DOYLE.GIF
     
    The angle to be trisected is ĞBOC = 3Æ
     
    Draw equilateral D BCZ with BC=BZ=CZ=r:
      pg. iv.f  
    with centre Z and radius BZ=BC=r draw the semicircle DBCE
     
    Join pts D&B, B&C, C&E, D&Z and Z&E with a straight line.
     
    D DZB = D CZE = D BZC equals an equilateral D of 60o.
     
    Line DZE is a straight line since ÐDZE = 3ÐBZC = 180o.
     
    Extend lines DB & EC above line BC to meet at pt T and
     
    join pts B & T and pts C & T with a straight line.
     
    Ð TBC = ÐTDZ = ÐTEZ = ÐTCB = 60o .
     
    Bisect Ð TCB to intersect line DT at pt a.
     
    Similarily bisect Ð TBC to intersect line TE at pt b.
     
    Thus Ba = ab = bC, by construction.
      pg. iv.g
    The trapezium BabC is a semicircle as DBCE.
     
    The circle BabC cuts line TLO at pt L below line BC.
     
    Join pts B & L, a & L, c & L and C & L with a straight line,
     
    crossing arc DBCE at pts r&l respectively.
     
    All equal arcs or cords subtend the same angle at the circum-
     
    ferance, thus ÐBLa = ÐaLb = ÐbLC trisects Ð BLC.
     
    Join pts r & O and pts l & O with a straight line.
     
    Thus Ð BOr = Ð rOl = Ð = lOC trisects Ð BOC!
     
    The author indicates farther that Br = rl = lC = 1/3 arc BC!
      pg. iv.h
  ii. THE TORONTO LIBRARY'S microfishe 78258
     
    THE TRISECTION OF ANY RECTILINEAL ANGLE 1911 ch. 3
     
    A GEOMETRICAL PROBLEM:
     
    by: GEORGE GOODWIN
    GOODWIN GIF
     
    The angle to be trisected is ÐCBA = 3Æ.
     
    With centre pt B draw the semicircle CAD, not shown,
     
    with radius BC = BA = BD = r.
     
    With centre B and radius 2*BD = 2*BA = 2*r draw an
      pg. iv.i
    arc to intersect the line CBDE produced at pt E.
     
    Join the pts A & E with a straight line, crossing the
     
    perpendicular BHF produced, from pt B, at pt H.
     
    With centre H and radius 2*BA=2*r draw an arc to
     
    intersect the line CBDRE produced at pt R.
     
    Join pts A & R with a straight line, crossing the
     
    perpendicular BTHF, from pt B, at pt T.
     
    Thru pts T & H draw lines TT' & HH' parallel to line CBDE
     
    cutting line BT'H'A at pts T' & H' respectively.
     
    Extend the line BT'H'AA' so that section AA' = T'H'.
      pg. iv.j
    With centre pt B draw the semicircle C'A'D', not shown,
     
    with radius A'B = C'B = D'B = r'.
     
    With centre pt H and radius A'B = r' draw an arc to
     
    intersect arc C'A'XD' at pt X.
     
    Join pts B & X with a straight line.
     
    Thus Ð XBD trisects Ð CBA!
     
    The author indicates farther that Ð XBD = 1/3*Ð CBA
     
    and that the semicircle C'A'D', not shown, is ideal!
     
    NOTE: Another variation measures H'T' and AA'
     
    along AE instead of AO.
      pg. iv.k
    See 4.i: ARCHIMEDES' TRISECTION THRU CHEATING, APPENDIX A above
     
    or my SQUARING THE CIRCLE SOLUTION Fig. 8:
     
    H'E = c = r in figure below:
     
    SQUARING THE CIRCLE Fig. 8
      pg. iv.l
  iii. THE TORONTO LIBRARY'S microfishe 45769
     
    GEOMETRICAL SOLUTIONS of the length and divisions of CIRCULAR ARCS 1851
     
    TRISECTION OF THE ANGLE:
     
    by: PETER FLEMING
     
  FLEMING GIF
     
    Angle to be trisected is arc HCF = 3Æ.
     
    This book is hard to read; n is obtained through a finite series;
     
    In theory arcs F1' = 2*H1; arcs 1'2' = 2*12; etc.
      pg. iv.m
    until arcs Fn = 2*Hn and thus Hn = HF/3.
     
    Let Hn = x + x/(n - 1) + x/(n - 1)2 + x/(n - 1)3 + ... + C1
     
    Let Fn = x(n - 1) + x + x/(n - 1) + x/(n-1)2 + x/(n-1)3 +...
     
    + C1 or Fn = Hn(n - 1)
     
    NOTE:
     
    I would like to disagree here since Fn = x(n - 1) + Hn =
     
    2*Hn thus x(n - 1) = Hn or the 1st term = x(n - 1) or
     
    arc F1 must contain the Ð Æ already to work.
     
    See my magenta & red circles, above!
     
    And the author does not show how he located pts x, y, z
      pg. iv.n
    to get line abc...t.
     
    TO CONTINUE...
     
    If one lets x =1 and n = 3
     
    Then Ht = 1' + 1'/2 + 1'/4 + 1'/8 + C1 = 2 units
     
    And Ft = 2' + 1' + 1'/2 + 1'/4 + 1'/8 + C1 = 4 units
     
    Thus Hn = 1/3 arc HF
     
    Thus taking a suitable distance chord H1 = 1 unit on arc
     
    HF, see figure sbove, keep bisecting the chord so that
     
    chord 12 = 1/2*chord H1; chord 23 = 1/2*chord 12;
     
    chord 34 = 1/2*chord 23; etc.
      pg. iv.o
    Similarily chord F1' = 2*chord H1 = 2 units; chord 1'2' =
     
    2*chord 12; chord 2'3' = 2*chord 23; chord 3'4' =
     
    2*chord 34; etc.
     
    Pt B bisects arc HB1; with centre B and radius BC cut arc
     
    AIHBDF at pts I & D, shown above.
     
    Using pts B & D as centres cut arcs through pts 2'3'4'..t' and
     
    pts x, y, z...t creating the intersections a, b, c..t'', draw
     
    throught these intersections the straight line abc..t''
     
    meeting the arc HnF at the pt n.
     
    As before arc Hn = arc Fn/2 or arc Hn = 1/3 arc HF.
      pg. iv.p
    Thus Ð HCn = Ð nC1 = Ð nCE = Ð ECF = Æ trisects
     
    Ð HCF = 3Æ.
     
    aurora.gif
     

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