MY 11 TRISECTION APPROXIMATION METHODS Pg. 1
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FIG. 1
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NOTE: angle º Ð
; triangle º D
; pi º p
; pt º pt
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1.
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Draw any suitable straight line JOB
with a suitable perpendicular,
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drawn up from pt O on line JOB to pt C,
as shown in FIG. 1 above.
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| Pg. 2
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2.
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Draw a straight line OD so that ÐDOC = 3Æ,
see FIG. 1 above.
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3.
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Starting with pt A'
and a large enough radius draw an arc F'H'
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clockwise from pt F' on line OF'D across line JOB, see FIG. 1.
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4.
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With the same radius R = A'F' slide the center suitably forward,
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| along line JOB, to create centres A'' and A''' with arcs F''H'' and
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| F'''H''',
respectively, so that they cross line OG''G'D at pt G' and
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| and G'', respectively, as shown in FIG. 1 above.
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NOTE:
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This might take a few attempts to get it right.
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Ignore pt G''' it does not exist, since arc F'''H''' is below the line
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OD. Ideally pt G lies more than half way up the line FGO: since
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GO2 = GE2 + EO2 + 2*GE*EO*cosÐAEF [from CRC TABLES], thus GO > GE = FO/2
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One could start with G'O = F'G' producing A' reducing a few steps.
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| Pg. 3
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5.
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Join pts A' andG' with a straight line.
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6.
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Similarily draw straight lines A''G'' and A'''G''', etc., see FIG. 1.
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7.
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Drop a perpendicular from pt F' to meet line A'I'G' at pt I'.
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8.
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Similarily drop a perpendicular from pt F'' to meet line A''I''G''
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at pt I'', as shown in FIG. 1 above.
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9.
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With centre G' and radius r' = G'F' draw a clockwise arc F'E' from
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pt F' on line DF'G'O to cut arc F'E'H' at pt E', see FIG. 1 above.
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10.
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Similarily with centre G'' and radius r'' = G''F'' draw a clockwise
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arc F''E'', from pt F'' to cut arc F''E''H'' at pt E''.
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NOTE:
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The ideal condition for the pt A to be perfect: the pt E must lie
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on the line JEOB and the radius r = GF = GE.
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By inspecrion pts E', E'' and E''' are well off from line JOB.
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| Pg. 4
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Graph I'''I''I' must pass thru pt I & E'''E''E' must pass thru pt E.
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If the ideal pt G has not been found, steps 1 to 10 must be redone.
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11.
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With centre pt G and radius r = FG = GE draw a circle to cut line
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JEOB at pt E.
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12.
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Join pts A and F with a straight line.
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13.
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Join both pts F & E and pts A & G with a straight line and call
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the intersection pt I, see FIG. 1 above.
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| NOTE:
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Line AIG is the right bisector of chord FIE because of similar
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isosceles triangles D AFG and D AEG: same radius R = AF = AE
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= AG and radius r = FG = GE, thus ÐGAF = ÐGAE.
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14.
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Let ÐGAF = ÐGAE = 2b. Thus ÐFAE = ÐGAF + ÐGAE = 4b
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| Pg. 5
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15.
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Since Ð EFG is the exterior
angle of the isosceles triangle D EAF
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subtended at the center pt A of the circle FGE; it is 1/4 of ÐFAE,
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thus ÐEFG = ÐFAE/4 = b.
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16.
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In isosceles triangle D EAF:
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ÐAFE = ÐAEF = (180o - 4b)/2 = 90o - 2b = i - 2b.
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17.
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In triangle D EFO:
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Since ÐFOE = ÐAEF - ÐEFO = i - 2b - b = i - 3b = i - ÐDOC.
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but ÐDOC = 3Æ thus b = Æ.
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18.
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Thus ÐEFG = Æ trisects ÐDOC = 3Æ.
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1.
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[Arcs and chords of the same length on the same
circumference subtend the same
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angle
at the center of the circle - theorem]
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Since chords FG = GE, R = AF = AG = AE, thus Ð GAF = Ð GAE.
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| Pg. 6
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Let Ð GAF = Ð GAE = 2b and Ð FAE = Ð GAF + Ð GAE = 4b.
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2.
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Since D AEF
is isosceles with the same radius R = AF = AG = AE
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then Ð AEF
= (180o - Ð FAE)/2
= (180o - 4b)/2
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= 90o - 2b = i - 2b
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And in D AIE & D AIF: Ð AIE
= 180o - Ð EAG
- Ð AEI
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= 180o - 2b
- ( 90o - 2b) = 90o = i = Ð AIF.
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Therefore D AIE & D AIF are 'right angled' Ds.
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3.
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[The exterior tangential angle
at the base of an isosceles triangle, subtended at the centre
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of a circle,
is 1/4 the size of the interior angle - theorem.]
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Thus Ð EFO = Ð
FAE/4 = 4b/4 = b.
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4.
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In D AFO: the exterior Ð FOB = Ð FAO + Ð AFE + Ð EFO
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= 4b
+ ( 90o - 2b) + b = 90o + 3b = i + 3b.
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But Ð FOB = Ð COB + Ð DOC = 90o + 3Æ = i + 3Æ
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| Pg. 7
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Thus Ð DOC = 3Æ = 3b and thus Æ = b and
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thus Ð EFO = Æ trisects Ð FOC = 3Æ as required.
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