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　　 致編輯，審稿者：

　　 你好，我是一名學(xué)生。但請不要小看我，事先申明：

　　 ①我的論文你們未必能看懂，但你們可以按文中所說的方法檢驗，畢竟實踐是檢驗真理的唯一標(biāo)準(zhǔn)，②我還有許多還不懂，有事給我發(fā)電子郵件，我定期查看。

　　 以下是論文的正文部分：

　　 關(guān)于“尺規(guī)三等分角”的探究

　　 眾所周知，三分角從提出至今已有三千多年了，古今多少學(xué)者對其望而生嘆，在嚴(yán)格的尺規(guī)作圖的限制下，至今尚無人能解。這給科學(xué)工作帶來不便。因此，古往今來人們不斷探索解決尺規(guī)三等分角問題，曾今三等分角被數(shù)學(xué)家萬徹爾用數(shù)學(xué)方法證明——尺規(guī)作圖法無法解出；阿基米德得出直尺三分角法。更有許多種方法解決三等分任意角。但至今三分角還未被人們在尺規(guī)作圖的限制下解決，所謂尺規(guī)作圖，就是使用無刻度的直尺和幾何圓規(guī)作圖。經(jīng)本人長期對三分角研究后，得到尺規(guī)作圖解決三分角的方法，以下是我的解題思路；（我暫定為中華定理或C1定理）。

　　 ⒈如圖做任意角∠AOB的角平分線，OC，所以∠AOC=∠BOC；

　　 ⒉以O(shè)A為半徑O為圓心作圓O，圓O交∠AOB于點A，B，所以O(shè)A=OB；

　　 ⒊連接AB交OC于C點，所以AC=BC；

　　 ⒋以AC為半徑C為圓心作圓C，用弦長AC將弧ADB三等分，三等分點為D，E。所以AD=BE，

　　 ⒌連接DC，EC，所以DC=EC，AC=BC，所以△ADC為等腰三角形，且△ADC≌△BEC，所以∠DCA=∠ECB，令A(yù)B交OD，OE于點H，I，

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

　　 ⒎因為∠DCA=∠ECB，∠ACO=∠BCO。所以∠DCO=∠ECO

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

　　 ⒐因為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命圓O分別交OD，OE于點G，F(xiàn)，所以AO=OG=OF=OB；

　　 ⒒因為∠DOC=∠EOC（由8得），∠HCO=∠ICO（由6得）CO=CO，所以△HCO≌△ICO所以O(shè)H=OI，HC=CI

　　 ⒓以O(shè)H為半徑O為圓心做圓O，圓O交∠AOB于J，K，兩點，所以O(shè)J=OH=OI=OK，又因為AO=OG=OF=OB，同理得AJ=GH=FI=BK；

　　 ⒔連接BG，HK，因為HO=OK，所以∠OHK=∠OKH，同理∠OGB=∠OBG，所以∠OHK=∠OGB，且GH=BK，所以四邊形KBGH為等腰梯形，HB為四邊形KBGH的對角線，且HB交OF于I點。

　　 ⒕連接AF，JI同13理四邊形IFAJ為等腰梯形，因為∠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；所以四邊形AJIF≌四邊形BKHG，又因為AI是四邊形AJIF的對角線，且AI交OD于H點，又因為HC=IC，HCO=ICO=90度，所以H關(guān)于OC的對稱點為I，讓等腰梯形AJIF以O(shè)為圓心，OJ為半徑旋轉(zhuǎn)與等腰梯形GHKB重合，因為AI為等腰梯形AJIF的一條對角線，且AI交OD于點H，同理在等腰梯形GHKB中GK為對角線且GK交OE于點I，在等腰梯形GHKB中，HB，GK分別兩條對角線，且交點為I，所以HI=IK，又因為HO=IO=KO，所以△HIO≌△KOI，所以∠HOI=∠IOK，且∠AOG=∠BOF[由9得]，所以∠JOH=∠HOI=∠IOK，

　　 綜上所述，三分角被完全解決。以上所述解決為小于180度角，對于大于180角可將其分為平角和銳角再解決。三分角后可部分等分圓周。（如：9，13等分圓周）

　　 以下是我個人信息：

　　 姓名：謝光超 民族：漢 出生：1990年3月5日

　　 住址：安徽省合肥市長豐縣羅塘鄉(xiāng)雙合村吳小村民組

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　　 手機：13295514155（我的） 父親手機：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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