MY 11 TRISECTION APPROXIMATION METHODS         Pg. 8

 

LOGO TRISECTION

 
tri2c.gif


FIG. 2

 

CONSTRUCTION:

 
1.  Draw any suitable 'right angled' D OAB with Ð OAB = 90o
   
  and Ð ABO = 3Æ, as shown in FIG. 2 above.
   
2. Extend line AB a suitable distance to pt D, and extend line OB a
   
  suitable distance to pt E, see FIG. 2 above.
    Pg. 9
3. Extend line OA to pt C so that line OA = AC and OB = BC.
   
4. Locate 2 suitable pts F & G about ¼ inch above and below
   
  estimated center pt M'. Try BE = AB.
   
 

NOTE:

   
  All steps 1 to 12 should be done at least twice to locate center
   
  pt M' on line AM'D produced.
   
5. With center pt F and a radius of length R1 = OF = FC draw a
   
  complete circle cutting line AHD produced at pt H.
   
6. With center pt H and a radius of length r = OC cut an arc to inter-
   
  sect the previous circle OHJC at pt J near line OE, see FIG. 2.
   
7. Similarly with center pt G and a radius of length R2 = OG = GC
   
  draw a complete circle cutting line AID produced at pt I.
   
8. With center pt I and a radius of length r = OC cut an arc to inter-
    Pg. 10
  tersect the previous circle OIKC at pt K near line OE, see FIG. 2.
   
9. Join pts K & J with a straight line and call its intersection with
   
  line OLE produced pt L.         [Idealy this gives: M'L = M'O = M'C]
   
 

NOTE:

   
  Graph KL'J' is really a thin hysteresis or SINE curve!
   
10. Draw the 'right bisector' of line OBL to cut line AM'D at pt M'.
   
 

NOTE:

   
  Idealy this leads to: R = ML = MO = MC & Ð MOL = Ð MLO
   
  and Ð MLO = Æ trisects Ð OBA = 3Æ.
   
  TO CONTINUE WITH THE DRAWING.
   
11. With center pt M' and radius R' = OM' = M'C draw a complete
   
  circle cutting line APD at pt P.
   
12. With center pt P and a radius of length r = OC, cut an arc to inter-
   
  sect the previous circle OPL'C at pt L' near pt L
    Pg. 11
 

NOTE

   
  When pts L' & L coincide, the ideal condition exists:
   
  r = OC = PL = DJ = IK and R = MP = ML = MO = MC.
   
  Isosceles D OLP = D POC, Ð POL = Ð OPC = 2Æ and Ð MOL = Æ
   
13. Join and bisect pts L' and L at pt L'' and redo steps 1 to 12.
   
14. Assuming pt L is ideal now: bisect line PNL at pt N.
   
15. Join pts O and N with a straight line and call the intersection with
   
  line AMD pt M. Line ON 'right bisects' line PL, see proof below.
   
16. [Arcs and chords of the same length on the same circumference subtend the same
   
  angle at the opposite circumference - theorem]
   
  Let Ð POL = ÐOPL = 2Æ.
   
  In D OPL & D POL:
   
  DOPL is isosceles since sides OP = PL, thus in DOPL: since side
    Pg. 12
  OP is common, Ð POL = Ð OPL = 2Æ, and r = OC = PL it is
   
  isosceles too. Thus AP and ON are both bisectors and
   
  thus Ð OPA = Ð APC = Ð PON = Ð LON = Æ.
   
17. Thus Ð LON = Æ and trisects Ð OBA = 3Æ, as required.
   
  NOTE: it is amazing how accurate this approximation formula is.
 

PROOF:

 
1.  Assuming all pts O, C, P & L are precisely located.
   
2. [Arcs of the same length on the same circumference subtend the same angle at the
   
  opposite circumference - theorem]
   
  In DOPC & DPOL:
   
  since r = OC = PL thus Ð OPC = ÐPOL.
   
  Let Ð OPC = ÐPOL = 2b.
   
  since Ð OAD = 90o and OA = AC then AD is the 'right bisector'
    Pg. 13
2. of OC and thus D OPC is isosceles.
   
  Since OP is common, r = OC = PL and Ð OPC = Ð POL = 2b
   
  then D OPC = D POL and thus D POL is isosceles too.
   
3. since OA = AC and Ð OAP = Ð PAC = 90o, by construction,
   
  therefore AD is the 'right bisector' of the base OC of D OPC.
   
  Then in D OPA & D CPA:
   
  since PA is common, therefore DOPA = DCPA
   
  thus Ð OPB = Ð BPC = Ð POL/2 = 2b/2 = b.
   
4. In D OPM:
   
  since radius R = MP = MO it is an iscoceles D,
   
  and therefore Ð MOP = Ð OPM = Ð POL/2 = b, and
   
  the exterior Ð OMA = Ð MOP + Ð OPM = b + b = 2b
   
5. In D OPB:
    Pg. 14
  the exterior Ð OBA = Ð OPB + Ð OMA = b + 2b = 3b
   
  but Ð OBA = 3Æ = 3b and thus Æ = b.
   
6. [Arcs or chords of the same length on the same circumference subtend the same angle
   
  at the opposite circumference and again twice this angle at the centre - theorem]
   
  In D OPC & D POL: since arc or chord OC = OL,
   
  ÐOPC = ÐPOL = 2Æ and ÐOMC = ÐPML = 2ÐOPC = 4Æ
   
7. In D MNP & D MNL:
   
  since MN is the 'right bisector' of PL, PN = LN = PL/2 = OC/2,
   
  and Ð PNM = Ð LNM = 90o, by construction MN is common,
   
  and R = MP = ML; therefore, they are simimlar Ds.
   
  Thus ÐPMN = ÐLMN = ÐPML/2 = 4Æ/2 = 2Æ
   
8. In D POL & D PMO:
   
  ÐPMO = 180o - ÐMPO - ÐMOP = 180o - 2Æ = 2i - 2Æ.
    Pg. 15
8. Line ON: ÐPMO + ÐPMN = 2i - 2Æ + 2Æ = 2i = 180o.
   
  Thus line ON is a straight line and is the right bisector of line PL
   
  of D POL, thus making it an isosceles triangle with ÐBOM =
   
  ÐMOP = Æ trisects ÐOBA = 3Æ as required.
 
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