São Paulo , March 5, 2014 .
Dear Reader.
I recently decided to propose a solution to the problem called trisection of the angle is to divide an angle into three equal parts . Divide an angle is the same as dividing an arc , the arc has concreteness , the angle is like two sticks surrounding a void . So I gave the first steps , a new problem arose : do not know the proof of the theorem TY1 regard to figure FY1.
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| Figure FY1 |
I learned in my high school , in the first half of the last century , this theorem to accept as true and use it . Now I realize that it is very important for the trisection of the angle and do not know to show it. Before, I thought you knew . I do not know the name of the author of the statement . I'll call him Mr. Y. Initial TY mean theorem of Mr. Y in Figure FY1 , the thesis argues that , over the same arc AB , the AV1B e AV2B and angles are equal . An immediate consequence of this theorem is that the arc need not be common to both angles . We imaginardois separate arcs of the same circle. If these arcs are equal angles are equal.
Consider also the TY2 and Figure FY2 theorem, where (AC) is a diameter , OBC is an isosceles triangle the angles OBC and OCB being equal . Here we have the relation between the angles 180 – AOB = BOC and also 180 – (OBC + OCB) = BOC, therefore AOB = OBC + OCB ou OCB = AOB/2 ou AOB = 2OCB , because OBC = OCB. This relationship is very important to demonstrate the theorem of trisection . The figure makes clear that this case is FY2 below.
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| Figure FY2 |
Figures and properties seen so far belong to the high school. The reader can find this feature in the textbooks of children and grandchildren . What must be emphasized here is that the relationships between arcs and angles refer to the angles whose vertices are the center of the circle. Let's call them central angles . Angles whose vertices are points on the circle itself, let's call them peripheral angles . They are in the ratio of 1 to 2 . There are two more types of angles : internal and external , whose vertices are neither central nor peripheral . The vertices are internal and external points of the circumference . You, my acquaintance , textbooks did not mention this issue . If there is , unknown to the author's name.
Figure FY3 , below, shows the division of 180 degrees into three equal parts because the OAC equilateral triangle is therefore AOC angle is 60 degrees. This point is contained in textbooks.
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| Figure FY3 |
Chapter I: Theorem trisection , TT.
First part , 1TT1.
1FT1 Figure below shows a process of division of A1AA2 arc into three equal parts , it is possible to prove that AA2 = 2AA1 then AA1 = A1A2/3. Demo: the theorems TY1 and TY2 , Figures FY1 , FY2 and FY3 , worth the relationship between arcs AA2 = 2AA3 , however, by symmetry, AA3 = AA1 then AA2 = 2AA1. So A2A1 = 3AA1. Note that when any divide an arc into three equal parts the method also divides the complement of that arc into three equal parts ie splits the difference 180 – A1A2.
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| Figure 1FT1 |
Note : The above conclusion does not solve the problem . It's just part of the solution because the arc A1A2 is unknown . Not choose the length of that arc. It was designed with the aid of three points: OC , which are known points and the point P is arbitrary , so the arc is arbitrary , ie unknown. We face the paradox of the bow : we can divide by three unknown arc. We can not divide by three a known arc. Can not fit a known arch in the correct position for the point A to divide into three equal parts This paradox is similar to the uncertainty principle of quantum mechanics . Like and do not like because there is a fundamental difference between the two sciences . The physical progress and making measurements from them , drawing conclusions . These measures she admits a margin of error , which in classical mechanics e'um crash course in quantum mechanics and one fatality. So physics has analog nature. The geometry acmite no margin of error . Not a theorem about right there , just as there is not a more or less pregnant woman. The geometry is digital . Perhaps , because of this difference , it is possible to circumvent the barrier of paradox.
Theorem trisection, TT.
Second part , 2TT1.
The second part is to propose a method , a same circumference , delimiting equal arcs. To do this we consider Figure 2FT1 , below where delimit one arc A1A2 of our free choice . Then we draw the line segments crossed (CA1) and (BA2) determining the point V1, A1V1A2 vertex of the angle.
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| Figure 2FT1 |
Note that this figure the reader is the first step in the study of angles internos.A example of peripheral angles whose measure does not depend on the position of the vertex , since it is a point of the circumference , here we also find the locus of the vertices whose angles are constant. At first glance , this finding locus seems very difficult , it is actually very easy . For this we draw the circle that passes through the points BV1C and thus any angle with vertex on such a circle formed by lines passing through the points B and C will be equal to BV1C.
Show below the figure 2FT2.que is a copy of the picture which was 2FT1 acrecentado A3V2A4 the angle of A1V1A2 theorem due to TY1 . Now we have to prove that the two arcs A1A2 and A3A4 are equal.
Demo: Figure 2FT2 is below where we put the two angles and the two arcs:
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| Figure 2FT2 |
Also we trace segments (A2C) and (A4C) , so doing we train and V1A2C V2A4C triangles . these triangles and the BA2C and BA4C angles equal to 90 degrees , so they are equal. On the other hand, if A1V1A2 and A3V2A4 and angles are equal for the construction and A2V1C and A4V2C angles are also equal for the two previous additions are compared to 180 degrees. Then the V1A2C and V2A4C triangles are similar and V1CA2 and V2CA4 and angles are equal and the A1A2 and A3A4 arcs are equal because of the theorem TY1.
Trisection theorem , TT.
3TT1 third party.
To draw the 3FT1 figure below , the reader should draw a circle to its diameter any ( BC ) and choose it and mark the arc of X and Y extremes to be divided by 3. That done , trace the BY and CX segments getting the point Z. Now draw the arc by BZC points and also the bow radius OC by points A and The . Segments (OD1) and ( AD ) are perpendicular to the diameter ( BC ) . The V1 point , that solves the problem , is the intersection of the arc with the line BZC (AD1).
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| Figure 3FT1 |
Proof: by Theorem TY1 and Figure FY1, the arc A1A2 to be divided is equal to the arc XY chosen by the player. By Theorem TY2 and Figure FT2 A2OC the angle is twice the angle A2BC and A3A2 segment, parallel to (BC) makes A3O arc equal to the arc A2C . Yet by the theorem TY2 A1YB arc is twice the arc A2C. therefore the segments (OA2) and (CA1) intersect at point P belonging to the segment (AD1) perpendicular to (BC).
The third part is demonstrated due to the first part of the theorem TT .
Antonio Carli
Engineer POLI USP
carlisp@terra.com.br
