MY 11 TRISECTION APPROXIMATION METHODS       Pg. 78

 

CUPID'S BOW

 

tri14c.gif
FIG. 14

 

METHOD:

 
NOTE: similar to SLIDER TRISECTION, FIG. 1 on Pg. 1.
    
1.  In FIG. 14 above ÐKOA = 3Æ and is the angle to be trisected.
   
2.  With a suitable radius R = OK and pt O as center draw an arc KI,
   
  see FIG. 14, above.
   
3.  Slide side JD of CUPID'S BOW'S arm along the X-axis or arm
    Pg. 79
  OA of ÐKOA until its two stationary pts C & B lie on and
   
  concur with arc KI.
   
4.  Slide slider F along HC until pt E lies on and concurs with
   
  arc KI, as with pts C & B.
   
5.  Assume the 1st attempted intersected line HFC with line OA.
   
  is pt O', to the left of pt J, which meant that the arc KI was too
   
  flat and a smaller radius was required.
   
 

NOTE:

   
  Else a new pt K'' must be choosen so that OK'' > OK on the arm
   
  of ÐKOA and all the previous steps 3 to 6 must be repeated.
   
  Remember that pt E must be on the line OK unless line O''E is parallel to line OK.
   
6.  Create pt O'' to the right of pt O; and with the smaller radius
   
  O''K and center pt O'', draw arc KI'' as shown in FIG. 14 above.
    Pg. 80
 

NOTE:

   
  If a smaller radius is needed after step 8 than create pt O'''
   
  halfway between pt O and pt O'', etc.
   
7.  Repeat steps 4 to 7.
   
8.  When pts C, B, and E lie on the current arc "KI" and line HOFC
   
  and line JOA intersect at the current center pt "O" (not necessarily
   
  pt O or O' or O'' or O''') all conditions are met and thus:
   
  ÐEOA = 3Æ is trisected by ÐCOA = Æ, as required.
   
  NOTE:
   
i.  If one drew a circle center O'' with radius R'' = O''B = O''C = O''E
   
  = O''K the arc KI would be KECB, as shown in FIG. 14 above.
   
ii.  Each of the pts E, C, B, and F are hinges forming a RHOMBUS
   
  of solid wires, which form a machine linkage; hinges or pts C and
    Pg. 81
  B are fixed to CUPID'S BOW; and the hinge at pt F also slides
    
  along the solid wire HFC, which must pass through the current
   
  center pt O of arc KI giving radius R = OK = OE = OC = OB.
    
iii. The black line JOGDA id considered the x-axis of the angle
   
  being trisected.
 

CONSTRUCTION:

 

NOTE:

   
  Assuming the current center pt O (not necessarily pt O or O' or O''
   
  etc.) and the current arc KECBI are ideal and the line HOFC and
   
  the line JOA intersect at the ideal pt O; all fitting exactly on
   
  CUPID'S BOW; as shown in FIG. 14 above (with O = O'').
   
1.  Draw any suitable right angled DODC on straight line OD and
    Pg. 82
  let Ð COD = Æ.
    
2.  Duplicate DODC about side OD to create DODB with Ð BOD =
   
  ÐCOD = Æ, as shown in FIG. 14 above.
    
3.  Duplicate DCOB about side OC to create DEOC with
   
  ÐEOC = ÐBOD + ÐCOD = 2Æ, see FIG. 14, above.
    
4.  Drop a perpendicular from pt E to intersect the line OGA at pt G,
   
  and call its intersect with line OFC at pt F.
    
5.  Join pts F & B, pts E & C, and pts C & B, with a straight line.
 

PROOF:

 

NOTE:

   
  Current center pt O'' shown as pt O for legibility, as shown in
   
  FIG. 14 above.
   
1.  In similar 'right angled' Ds, DODC & DODB:
    Pg. 83
  Since Ð ODC = 90o = i and Ð DOB = Æ
   
  then Ð DBO = 90o - Ð DOB = i - Æ
   
  DOCE & DOEC are isosceles since sides OK = OE = OC = OB = R
   
  thus ÐDBO = i - Æ = ÐOCD = ÐOCE = ÐOEC
   
2.  [When an isosceles D subtends another D at its base, the 2nd. D is
   
  also isosceles and a similar D with both sides equal to the width
   
  of its base and both apex Ðs are equal to each other - THEOREM].
    
  In similar 'right angled' Ds, DOGF & DODC:
   
  Since Ð OGF = Ð ODC = 90o = i
   
  and Ð FOG = Ð COD = Æ = Ð DOB
   
  and thus Ð COB = Ð COD + Ð DOB = 2Æ
   
  also Ð OFG = 90o - Ð FOG = i - Æ = Ð EFC [construction & intersection]
   
  And since Ð EFC = Ð OCE = i - Æ
    Pg. 84
  Therefore DEFC & DFCB are isosceles.
   
  also all sides are equal CE = CB = CF = FE, construction
   
  of CUPID'S BOW, and form a parallelogram or rhombus machine
   
  linkage at all times, as shown in FIG. 14 above.
    
3.  In DFEC:
   
  Ð FEC = 180o - 2Ð ECF = 2i - 2(i - Æ) = 2Æ
   
  And similarly in DFBC:
   
  Ð FBC = 180o - 2Ð BCF = 2i - 2(i - Æ) = 2Æ
   
  Thus making both triangles similar triangles.
   
  and Ð EOB = Ð EOC = Ð FEC = Ð FBC = 2Æ
    
4.  In figure Ð EOBA:
   
  Ð EOD = Ð EOC + Ð COD = 2Æ + Æ = 3Æ.
    
5.  Thus Ð EOA = 3Æ is trisected by Ð COD = Æ as required.
 

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