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The Proof of Impossibility is Unpredictable

Possible Trisection Method of an Angle

2021.1

Dr. Takao Yamada

SEA Seismic Engineering Associates

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Preface

This is a rewritten version of “Note on the Trisection of an Angle” published half a year ago.[17]

Since circa 300 B.C. of Greek, there is one of the three famous construction problems of how to trisect of an angle in Euclid geometry. This is the geometrical problem of how to trisect the angle by ruler and compass. The existing area of trisected angle is not necessarily restricted.[1,2,3]

Trying to make construction precisely on a sheet of paper must have been difficult through ages.. German mathematician Wantzel concluded in 1837 using the simple algebraic figure model of included angle 3θ that the trisection is impossible because the cubic equation could not be solved by ruler and compass.

French Academy of Sciences announced that the thesis on the trisection could not be accepted in the middle of the nineteenth century based on Wantzel theory. The official views of Mathematical Society of Japan is the same and may be due to Kentaro Yano’s “Trisection of an Angle” following Wantzel. [4,5] K.Yano showed specific impossible angles as 120°,60°, 30°,24° and 15° .The author is extremely disappointed that there are many expressions in this book to call the amateur scholar interested in the trisection as “Trisector” and this derogatory term has been fermented.

It is worthy of notice that, there is the following description in “Encyclopedic Dictionary of Mathematics” edited by MSJ: It is not necessary the most suitable to apply the analytic geometry for all geometry problem. But there is no description on trisection.[8]

Kodaira Kunihiko stated as follows in the book “Invitation to Geometry” : As the teaching material of mathematics for elementary education, if the plane geometry is strict as the science of figures, that’s enough. On the contrary, analytic geometry beyond the student’s academic ability gives them a vague impression.[9]

TorahikoTerada stated as follows in the essay “the novelist Saikaku and Science” : First thing to notice, I think that Saikaku had a clear awareness of the breadth of the world of knowledge and the unpredictability of the of limit of possibility. Scientist doesn’t affirm anything without proof, as well as, doesn’t deny the possibility of anything that doesn’t prove impossible. But it’s not uncommon for some scientists ashamed later who denied the possibility from shallow dogma.[10]

The proposed trisection method is derived from the sum of internal angles of a triangle is equal to the external angle, as shown ACO∆ in the cover, CAO ACO θ 2θ AOB 3θ∠ +∠ = + = ∠ = . The trisected angle θ is formed out of the narrow region of 3θ . The presented trisection method satisfies the algebraic relation of triplex formula of cos3θ .

To draw the figure exactly, geometry software “Cinderella” is used. It is written that the trisection is impossible there, but it is actually possible as shown in the figures of example.

January, 2021 the author

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Contents page 1 Fundamental Matters 1 1-1 Fundamental Condition of Trisection 1-2 Existing Area of the Trisected Angle θ 2 1-3 Trisection Method Satisfies the Triplex Formula of cos3θ 3 2 Example of Trisection 4 2-1 Explanation of Trisection Step 2-2 Example of Trisection and Proof 5 2-2-1 The case of Angle 36° 2-2-2 The case of Angle 48° 7 2-2-3 The case of Angle 60° 9 2-2-4 The case of Angle 72° 11 3 Appendix 13 3-1 On the book of Kentaro Yano：“Trisection of Angle” 3-2 On the home page of the Mathematical Society of Japan 14 3-3 How to open Cinderella on Windows 15 4 Reference 16

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1 Fundamental Matters

1.1 Fundamental Condition of Trisection

Let a point, line and circle satisfy “Fundamental Condition of Trisection” shown in Fig-1-1.

(1) Draw the line a and b through the point O with given angle 3θ . (2) Get the point B on a . Draw the circle C-1 with center O-radius OB. (3) Get the point C reversing to B with respect to O at the intersection of C-1 and a. (4) Draw the line d perpendicular to b through O. (5) Draw the line c through C, then get the point A at the intersection of b and c, point E at the

intersection of d and c , and point D at the intersection of C-1 and c. (6) Draw the vertical bisector e which passes through the midpoint R of OE. (7) If e passes through D, then DE=DO=DA , where A,E andO are on the circle C-2 with center

O. Since OC=OD=DA , OCD∆ and DOA∆ are isosceles triangles. If CAO=θ∠ , then ACO=2θ∠ . AOB= ACO+ CAO∠ ∠ ∠ , then CAO∠ becomes the trisected angle of AOB∠ .

Fig-1-1 Fundamental Condition of Trisection

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1-2 Existing Area of the Trisected Angle θ

The point, line and circle are shown in Fig-1-2 “Existing Area of the Trisected Angleθ ”

(1) Get the point F on C-1 so that FOB 6θ∠ = . (2) Get the point I and J at the intersection of b and C-1. Get the point G and H on C-1 at the

symmetrical point D and C with respect to b. The point G and H are on the line h passing through the point A. The lines c and h are symmetric with respect to the line b. ∴ , ,CH b DG b⊥ ⊥ CF//BH//b, CFB BHC FCH HBF R∠ = ∠ = ∠ = ∠ = ∠

DAI DOI GAI GOI FCD BHG θ, FOD BOG 2θ∠ = ∠ = ∠ = ∠ = ∠ = ∠ = ∠ = ∠ = FOI BOI OFC OCF OBH OHB 3θ∠ = ∠ = ∠ = ∠ = ∠ = ∠ =

(3) d is the vertical bisector of FC. Get N and M at the intersection of d and C-1. (4) Get the point L and K on C-1. L and D are symmetric with respect to O. K and G are

symmetric with respect to O. Get the point E at the intersection of d and FK. Draw the vertical bisector e passing through

the midpoint R of EO. Move A on b so that e passes through D, then KFH FKG FHG KGH 2θ∠ = ∠ = ∠ = ∠ = , FK GH= , FK//DL//GH, CDK θ∠ =

Fig-1-2 Existing Area of the Trisected Angle θ

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2-2-3 The case of 60°

(1) step-1

(2) step-2

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(3) step-3

DAI DOI= GAI= GOI=θ∠ = ∠ ∠ ∠