Mathematical proof of the angle of thirds, how to prove the third division of any triangle?

In order to illustrate the sufficient conditions for the possibility of ruler drawing, it is first necessary to translate geometric problems into the language of algebra. The premise of a plane drawing problem is always given some plane figures, for example, points, lines, angles, circles, etc., but the straight line is determined by two points, an angle can be determined by its vertex and a point on each side, a total of three points, and a circle is determined by one point at the center and circumference of the circle, so the plane geometry drawing problem can always be reduced to a given n points, that is, n complex numbers (of course, z0=1). The process of drawing a ruler can also be seen as using a compass and a straightedge to constantly get new complex numbers, so the problem becomes:

Given a plurality of plurality and z0, can you get the pre-desired plural number z by using the ruler from the starting point? For the sake of discussion, the following recursive definition is given:

Definition: Let s= be n + 1 complex number, will.

1) z0=1，z1，..zn is called the s-point;

2) A straight line passing through two different S-points is called an S-line, and a circle with an S-point as the center and the distance between any two S-points as the radius is called an S-circle;

3) The point where the s-line intersects the s-line, the s-line and the s-circle, and the s-circle and the s-circle are also called the s-point.

The above definition completely describes the process of drawing a ruler, if p represents the set of all s-points, then p is all the complex numbers obtained from the drawing of the ruler and gauge from s=.

Theorem: Let z1 ,..zn(n 0) is n complex numbers.

Let f= q(z1,..zn，z1'，..zn')，(z'represents the conjugate plural), then, the sufficient and necessary condition for a complex number z to be made by s= is that z belongs to f(u1,..

un)。where u12 belongs to f, and ui2 belongs to f(u1,..ui-1)。

In other words, z contains a 2nd root expansion of f.

Department: Let s=,f= q(z1,..zn，z1'，..zn'), z is the s-point, then [ f(z) :f] is the power of 2.

The following proves the impossibility of dividing any angle into thirds, and proves that the ruler and gauge drawing cannot be divided into 30 degrees at an angle of 60 degrees:

Proof: The so-called giving an angle of 60 degrees is equivalent to giving the complex number z1 = 1 2 + 3 2 i. Thus s=,f=q(z1, z1')=q(√-3)。

If you can make an angle of 20 degrees, of course, you can also get cos20, but cos20 satisfies the equation 4x3-3x-1 2=0, i.e., 8x3-6x-1=0. Since 8x3-6x-1 is irreducible in q[x], thus [q(cos20):q]=3, thus.

6=[ q(cos20, √3)：q] = [f(cos20)：q]=[f(cos20)：f] [f：q]

Since [f:q]=[q( -3):q]=2, so [f(cos20):f]=3, according to the above system, we can know that cos20 is not an s-point, so that 20 degrees cannot be divided into thirds. Certification.

The third angle is proved wrong by mathematical and other methods. The logical theory is even more wrong. Geometry has its own uniqueness, and the thirds of the angle can be solved, using the absolute knowledge of geometric theorems.

The tripartite division of lines, the bisection of angles, the parallel theorem, some theorems of triangular isosceles, etc. There is also the relationship between chord length and arc, and parallelograms. I can say with certainty that any angle less than or equal to 180 degrees can be solved, there is only one way to make it, more than 180 degrees just add the length of the half-warp that is, the chord length can be proved, the above can be proved, the theory is provable, there is no near value, and the tangent radian of the chord is 100% true.

Wang Chengyou.

Under the premise of drawing a ruler, there is no solution to this problem. Trisected angles were one of the three major geometric problems of ancient Greece. The problem of trisecting arbitrary angles may have appeared earlier than the other two geometric problems, and there is no record of them in history.

But there is no doubt that it will appear naturally, and we ourselves can imagine it now. It has been proved that there is no solution to this problem under the premise of ruler drawing.

Definition. In order to illustrate the sufficient and necessary conditions for the possibility of ruler and gauge diagramming, it is first necessary to convert geometric problems into algebraic languages. The premise of a plane drawing problem is always given some plane figures, for example, points, lines, angles, circles, etc., but a straight line is determined by two points, an angle can be determined by three points on each side of its vertices and frugals, and a circle is determined by one point at the center and circumference of the circle.

This is a good proof :

If E is a third near point A, then af:fd=1:1 If E is a third point near point B, then AF:

FD=3:1 proves the first, let the third division point R near point B and connect Dr, then Dr Ce, i.e., EF is the median line of Ard, so F bisects AD with a ratio of 1:1

Prove the second, let the third division point near point A be r, connect CR, and CR intersects AD with Q, then de cr, then CF:EF=AQ:DQ=DF:QF=1:1

That is, q is the midpoint of AD, f is the midpoint of Qd, and F is the quartile point of AD close to D, and the ratio is 3:1

Trisected angles were one of the three major geometric problems of ancient Greece. The trisect angle is a famous problem in the ancient Greek geometric ruler diagram, and the problem of square and double cube is one of the three major problems of ancient mathematics, and now it has been proved that this problem is unsolvable. The full description of the issue is:

A given angle is divided into three equal parts using only a compass and an ungraduated ruler. Under the premise of ruler drawing (ruler drawing refers to drawing with a ruler and compass without scale), there is no solution to this problem. If the conditions are relaxed, such as allowing the use of graduated rulers, or if they can be used in conjunction with other curves, a given angle can be divided into thirds.

The latest method is the segmented angular method, which can arbitrarily divide any angle. The key point is to the m power where the longitudinal height is set to 2.

Hello classmates, there are 5 ways to prove the congruence of triangles The methods of triangle congruence: 1. The three sides correspond to two triangles that are equally congruent. (sss) 2, the two sides and their angles correspond to the congruence of two triangles that make the ages equal.

SAS) 3, the two corners and their edges correspond to two equal triangle congruences. (asa) 4, there are two corners and the opposite side of one of the corners corresponds to two equal triangle congruence (AAS) 5, the hypotenuse and one right-angled side correspond to the equal two right-angled triangle congruence. (hl) Hope it helps, thank you

The latest method is the segmented angular method, which can arbitrarily divide any angle.

According to the median line theorem of triangles, 1) Triangle median line definition: The thick line segment connecting the midpoints on both sides of the triangle is called the median line of the triangle.

2) Median theorem of triangles: The median line of a triangle is parallel to the third side and equal to half of it.

As a result, the three sides of each small triangle are equal to half of the original triangle sides.

So the four small triangles are equal in area, that is, the original triangle is divided into four equal parts.

Satisfied with Danner.