有关“三等分任意角”的探究（尺规作图）

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　　 你好，我是一名学生。但请不要小看我，事先申明：

　　 ①我的论文你们未必能看懂，但你们可以按文中所说的方法检验，毕竟实践是检验真理的唯一标准，②我还有许多还不懂，有事给我发电子邮件，我定期查看。

　　 以下是论文的正文部分：

　　 关于“尺规三等分角”的探究

　　 众所周知，三分角从提出至今已有三千多年了，古今多少学者对其望而生叹，在严格的尺规作图的限制下，至今尚无人能解。这给科学工作带来不便。因此，古往今来人们不断探索解决尺规三等分角问题，曾今三等分角被数学家万彻尔用数学方法证明——尺规作图法无法解出；阿基米德得出直尺三分角法。更有许多种方法解决三等分任意角。但至今三分角还未被人们在尺规作图的限制下解决，所谓尺规作图，就是使用无刻度的直尺和几何圆规作图。经本人长期对三分角研究后，得到尺规作图解决三分角的方法，以下是我的解题思路；（我暂定为中华定理或C1定理）。

　　 ⒈如图做任意角∠AOB的角平分线，OC，所以∠AOC=∠BOC；

　　 ⒉以OA为半径O为圆心作圆O，圆O交∠AOB于点A，B，所以OA=OB；

　　 ⒊连接AB交OC于C点，所以AC=BC；

　　 ⒋以AC为半径C为圆心作圆C，用弦长AC将弧ADB三等分，三等分点为D，E。所以AD=BE，

　　 ⒌连接DC，EC，所以DC=EC，AC=BC，所以△ADC为等腰三角形，且△ADC≌△BEC，所以∠DCA=∠ECB，令AB交OD，OE于点H，I，

　　 ⒍因为AO=BO，所以∠OAB=∠OBC，又因为∠AOC=∠BOC（由1得）所以∠ACO=∠BCO且都为直角，所以∠HCO=∠ICO；

　　 ⒎因为∠DCA=∠ECB，∠ACO=∠BCO。所以∠DCO=∠ECO

　　 ⒏因为DC=EC，∠DCO=∠ECO，共OC边，所以△DCO≌△ECO，所以DO=EO，∠DOC=∠EOC；

　　 ⒐因为AD=BE（由5得）AO=BO（由2得）DO=EO（由8得），所以△DAO≌△EBO，所以∠AOD=∠BOE，所以∠AOG=∠BOF；

　　 ⒑命圆O分别交OD，OE于点G，F，所以AO=OG=OF=OB；

　　 ⒒因为∠DOC=∠EOC（由8得），∠HCO=∠ICO（由6得）CO=CO，所以△HCO≌△ICO所以OH=OI，HC=CI

　　 ⒓以OH为半径O为圆心做圆O，圆O交∠AOB于J，K，两点，所以OJ=OH=OI=OK，又因为AO=OG=OF=OB，同理得AJ=GH=FI=BK；

　　 ⒔连接BG，HK，因为HO=OK，所以∠OHK=∠OKH，同理∠OGB=∠OBG，所以∠OHK=∠OGB，且GH=BK，所以四边形KBGH为等腰梯形，HB为四边形KBGH的对角线，且HB交OF于I点。

　　 ⒕连接AF，JI同13理四边形IFAJ为等腰梯形，因为∠AOG=∠BOF（由9得）AO=GO=FO=BO，JO=HO=IO=KO，∠AOF=∠AOG+∠GOF，∠BOG=∠BOF+∠GOF，所以∠AOF=∠BOG，所以弧AGF等于弧BFG，得AF=BG，JI=HK；所以四边形AJIF≌四边形BKHG，又因为AI是四边形AJIF的对角线，且AI交OD于H点，又因为HC=IC，HCO=ICO=90度，所以H关于OC的对称点为I，让等腰梯形AJIF以O为圆心，OJ为半径旋转与等腰梯形GHKB重合，因为AI为等腰梯形AJIF的一条对角线，且AI交OD于点H，同理在等腰梯形GHKB中GK为对角线且GK交OE于点I，在等腰梯形GHKB中，HB，GK分别两条对角线，且交点为I，所以HI=IK，又因为HO=IO=KO，所以△HIO≌△KOI，所以∠HOI=∠IOK，且∠AOG=∠BOF[由9得]，所以∠JOH=∠HOI=∠IOK，

　　 综上所述，三分角被完全解决。以上所述解决为小于180度角，对于大于180角可将其分为平角和锐角再解决。三分角后可部分等分圆周。（如：9，13等分圆周）

　　 以下是我个人信息：

　　 姓名：谢光超 民族：汉 出生：1990年3月5日

　　 住址：安徽省合肥市长丰县罗塘乡双合村吴小村民组

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　　 手机：13295514155（我的） 父亲手机：13170018085

　　 The following is the translation I use a computer, all the main Chinese version.

　　 Written in the pre-reading:

　　 Note to Editors and reviewers are:

　　 Hello, I am a student. But please do not look down on me, and I'll submissions, because I think I have a paper weight.

　　 ① my thesis that you may not be able to read, you can be the method described in the text test, after all, practice is the sole criterion for testing truth, ② the first time I Received, there are many still do not know, something happens to me e-mail, I periodically. ③ I hope that the papers previously published in the month of 6. I graduated.

　　 The following is the body of paper:

　　 On the "Ruler Angle trisection" of the inquiry

　　 As we all know, three-point corner from 3000 has been raised for years, the number of scholars, ancient and modern look of their born sigh, in the strict Geometric Construction of the constraints, since no one can answer it. This gives the scientific work inconvenience. Therefore, the ages people are constantly finding ways to solve Ruler Angle trisection problem, had this trisection of an angle is a mathematician 10000 Archer proved mathematically - Ruler mapping method can not be solved; Archimedes derived Ruler third angle method. Moreover, many kinds of solutions to an arbitrary angle trisection. But so far people have not yet been third angle Ruler mapping constraints to solve the so-called ruler mapping, is the use of non-scale ruler and compass geometry mapping. After my long-term study on the third angle obtained Ruler third angle method of mapping solution, the following is my problem-solving ideas; (I am tentatively scheduled for China theorem).

　　 ⒈ do ∠ AOB in Figure bisector angle, OC, so ∠ AOC = ∠ BOC;

　　 ⒉ with OA as radius of circle O as the center of the circle for O, Yuan O cross ∠ AOB at point A, B, so OA = OB;

　　 ⒊ Connect AB isosceles trapezoid cross-OC at C point, so AC = BC;

　　 ⒋ with AC as the radius of C as the center of the circle to circle C, with the chord AC arc ADB will be three equal portions, trisection points D, E. Therefore, AD = BE,

　　 ⒌ connected DC, EC, therefore DC = EC, AC = BC, so for the isosceles triangle △ ADC and △ ADC ≌ △ BEC, so ∠ DCA = ∠ ECB, so that AB cross-OD, OE at the point H, I,

　　 ⒍ make AB cross-OD, OE at points H, I. Because AO = BO, so ∠ OAB = ∠ OBC, and also because ∠ AOC = ∠ BOC (by a too), so ∠ ACO = ∠ BCO and all the right angles, so ∠ HCO = ∠ ICO;

　　 ⒎ because ∠ DCA = ∠ ECB, ∠ ACO = ∠ BCO. Therefore, ∠ DCO = ∠ ECO

　　 ⒏ because DC = EC, ∠ DCO = ∠ ECO, a total of OC side, so △ DCO ≌ △ ECO, so DO = EO, ∠ DOC = ∠ EOC;

　　 ⒐ since AD = BE (from 5 too) AO = BO (by 2 get) DO = EO (from 8 too), so △ DAO ≌ △ EBO, so ∠ AOD = ∠ BOE, so ∠ AOG = ∠ BOF;

　　 ⒑ Ming Yuan O were cross-OD, OE at point G, F, so AO = OG = OF = OB;

　　 ⒒ because ∠ DOC = ∠ EOC (from 8 too), ∠ HCO = ∠ ICO (from 6 too) CO = CO, so △ HCO ≌ △ ICO so OH = OI, HC = CI

　　 ⒓ with OH as the radius of circle O as the center of the circle to do O, Yuan O cross ∠ AOB in the J, K, 2 o'clock, so OJ = OH = OI = OK, and also because AO = OG = OF = OB, empathy have AJ = GH = FI = BK;

　　 ⒔ connections BG, HK, because the HO = OK, so ∠ OHK = ∠ OKH, empathy ∠ OGB = ∠ OBG, so ∠ OHK = ∠ OGB, and GH = BK, so quadrilateral KBGH for isosceles trapezoid, HB for the quadrilateral KBGH the diagonal, and the HB cross-OF in the I-point.

　　 ⒕ connection AF, JI quadrilateral with 13 reasons for the isosceles trapezoid IFAJ, because ∠ AOG = ∠ BOF (from 9 to get) AO = GO = FO = BO, JO = HO = IO = KO, ∠ AOF = ∠ AOG + ∠ GOF, ∠ BOG = ∠ BOF + ∠ GOF, so ∠ AOF = ∠ BOG, is why AGF is equal to arc-arc BFG, too AF = BG, JI = HK; so AJIF ≌ quadrilateral quadrilateral BKHG, and also because AI is a quadrilateral AJIF the diagonal, and the AI cross-OD in the H-point, and also because HC = IC, HCO = ICO = 90 degrees, so the symmetry point H on the OC as I, so that isosceles trapezoid AJIF to O as center of a circle, OJ for the radius of rotation coincides with the isosceles trapezoid GHKB because the AI for the isosceles trapezoid AJIF of a diagonal, and the OD at the point of AI cross-H, empathy in the isosceles trapezoid GHKB in GK and the GK for the diagonal cross-OE on the point I, in the isosceles trapezoid GHKB in HB, GK 2 diagonal, respectively, and the intersection of I, therefore HI = IK, and also because HO = IO = KO, so △ HIO ≌ △ KOI, so ∠ HOI = ∠ IOK, and ∠ AOG = ∠ BOF [by the 9 were], so ∠ JOH = ∠ HOI = ∠ IOK,

　　 In summary, the third angle has been fully resolved. The above address for less than 180-degree angle, for more than 180 angle can be divided into straight angle and an acute angle to resolve the issue. After the third angle can be part of the circle equal portions.

　　 The following is my personal information:

　　 Name: Xie supra-national: Hans Born: March 5, 1990

　　 Address: Changfeng County, Anhui Province, Hefei, Lo Wu Tang Xiang pairs of co-villagers in the village group

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