MY 11 TRISECTION APPROXIMATION METHODS Pg. 40
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THE FOCUS FORMULA IN SQUARING THE CIRCLE
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FIG. 8
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1.
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Draw any suitable straight line AOB, as shown in FIG. 8 above,top
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2.
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At center pt O and a suitable radius OA draw a complete circle,
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| Pg. 41
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| crossing the line OB, as shown in FIG. 8 above.
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3.
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Draw any straight line OC so that ÐAOC = 6Æ [60o was used
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for clarity] so that pt C lies on the
circumferance of the previous circle
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at pt C, as shown in FIG. 8 above.
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4.
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Draw 2 straight lines DA and DO so that ÐDAO = 4Æ [40o] and
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ÐDOA = 90o - Æ = i - Æ [80o] above the line AOB
so that the
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2 lines intersect at pt D, as shown in FIG. 8, above.
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5.
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Draw the line DF parallel to line AOB.
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6.
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With center pt D and radius of length DA draw a circle to cut arc
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ACE produced at pt E, just above line OB, see FIG. 8, above.
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Thus let DA = DE = [b - c] = r1 and OC = OE = c = r.
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7.
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Join pts O & E
with a straight line and extend the line past pt E to
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intersect the line DF extended at pt F so that OF = DF = b,
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| Pg. 42
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see proof below.
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8.
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With center pt E and radius EO = c = r cut line AOB
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extended at pt B
and let the distance OB = a.
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9.
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Join pts E & D
with a straight line and extend the line ED to
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meet arc ACE at pt C, as shown in FIG. 8 above, see proof below.
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10.
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Join pts D & F and pts B & F with straight lines.
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1.
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In DDOA & DDOE:
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Line DO is common and radii OA = OE and DA = DE; therefore,
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they are similar and equilateral Ds.
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Thus ÐOAD = ÐOED = 4Æ and ÐODA = ODE =
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= 2i - ÐDAO - ÐDOA = 2i - 4Æ - (i - Æ) = i - 3Æ.
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and since DF is parallel to OB then ÐAOD = ÐODF = i - Æ
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| Pg. 43
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and thus ÐFDB = ÐODF - ÐODB = (i - Æ) - (i - 3Æ) = 2Æ
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and also ÐFDB = ÐOBD = 2Æ
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2.
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In isosceles DCOE:
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Since 2 sides OC = OE = r = c and ÐOCE = ÐOEC = 4Æ
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therefore ÐCOD = ÐDOA - ÐCOA = (i - Æ) - 6Æ = i - 7Æ.
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3.
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In DEOB:
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Since 2 sides EO = EB = r = c
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therefore DEOB is isosceles and ÐEOB = ÐEBO = 2Æ
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and since DF is parallel to OB then ÐDFO = ÐFDB = 2Æ
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4.
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In DFOD:
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Since ÐFDO = ÐDOA = i - Ø
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and ÐFOD = 2i - ÐAOD - ÐBOE = 2i - (i - Æ) - 2Æ = i - Æ
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then DFOD is isosceles and thus DF = OF = b.
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| Pg. 44
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5.
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In isosceles DDEF & DOEB:
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Since ÐEDF = ÐEFD = ÐEOB = ÐEBO = 2Æ
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and since ED = EB = c = r
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and since DE = EF = DA = OF - OE = (b - c) = r1
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therefore the ratio DF/OB = EF/ER = DE/OE = b/a = (b-c)/c or
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1/c = 1/a + 1/b - which is known as the FOCUS FORMULA.
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FIG. 9
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NOTE
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D ADB, FIG. 9 was reproduced from D ADB, FIG. 8, above.
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1.
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Let b = pi = p = 3.141593
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2.
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Let a = pi - 1 = p - 1 = 2.141593
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| Pg. 46
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3.
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1/c = 1/a + 1/b
= 1/3.141593 + 1/2.141593 = 0.785252025
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c = ab/(a + b) = 1.2734765400
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4.
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d = 2b*sinÆ
[from ISOSCELES DFOD]
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5.
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d/sin2Æ = a/cos(i - 3Æ)
[from SINE LAWS for DODB]
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d = a*sin2Æ/cos3Æ
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d = 2a*sinÆ/(1 - 4sin2Æ)
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d = 2b*sinÆ = 2a*sinÆ/(1 - 4sin2Æ)
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a = b(1 - 4sin2Ø)
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4sin2Æ = (b - a)/b = (p - (p - 1))/p = 1/p
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(.5cscÆ)2 = p
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6.
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a = 2c*cos2Æ
[from ISOSCELES DEOB]
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cos2Æ = a/2c = 0.840845
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2Æ = 32o 46' 13.12" and Æ = 16o 23' 6.57"
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| Pg. 47
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pr2 = (0.5r2cscÆ)2 = A2
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A = 0.5r*cscÆ
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p = .5csc2Æ
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| NOTE
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With the angle Æ = 16° 23' 6.57" one can square any any
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circle, but I will leave that to you the mathematician to prove!
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DADB, Fig. 8, above, is reproduced as DADB, Fig. 9, below.
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FIG. 9a
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NOTE
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Figure DDEBOA, Fig. 9a,
above, is reproduced from DDEBOA, Fig. 8 above.
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(2cscÆ)2 = 1 + b/a and (2sinÆ)2 = 1 - a/b
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The circumferance C = 2*pi*r = 2pr = pDia. = pi*Dia.
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pi = p
= b = a + 1 = Cir./Dia. = 3.141593653589
[from CRC TABLES]
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| Pg. 49 NOT REQUIRED AN EXTRA PAGE SEE PAGE 77A
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