Why can t you divide one corner into thirds, and is it okay to divide any angle into thirds

This cannot be proved impossible in plane geometry. So for more than 2,000 years, it will attract countless people to try the ** decision. Among them are some of the world's top mathematicians.

Euclid, Archimedes, Gauss, Euler, etc., all tried and failed. And no one has been able to prove that this is impossible. It was not until later after the invention of analytic geometry that people used analytic geometry to prove that this diagram was impossible.

You go to look up the three major ruler drawing problems: the third division angle, the circle into a square, and the cube times problem.

There is a library. In 1837, the French mathematician Vantel gave a proof (the problem of trinomical angles and the problem of cubic multiples).

The page above has it, and it's good.

Summary. Hello dear! I will be glad to answer for you, first of all, it is not that there is no solution to the third part of the angle, but only the ruler and the compass without a scale are used for any part of the third part of the angle without a solution.

If you can use other tools, or special corners, you can still do it by dividing a corner in three equal parts. Second, the three equal parts of arbitrary angles, the double cube, and the circle into a square are known as the three major problems of ruler and gauge diagramming, and their impossibility has long been proven by mathematicians. Third, the process of proof is complex.

Hello dear! I will be glad to answer for you, first of all, it is not that there is no solution to the third part of the angle, but only the ruler and the compass without a scale are used for any part of the third part of the angle without a solution. If you can use other tools, or special corners, you can still do it by dividing a corner in three equal parts.

Second, the three equal parts of arbitrary angles, the double cube, and the circle into a square are known as the three major problems of ruler and gauge diagramming, and their impossibility has long been proven by mathematicians. Third, the process of proof is complex.

The main reason is that the ruler can make all the lines or things with quadratic root numbers, or multiple quadratic root numbers. And the third angle needs to use the third root number, which cannot be made with a ruler.

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 basis of this problem is still the Liang's three-point angle fixed operation mentioned earlier. The principle is that the angle of the middle part after the quintile is both 1 3 in the upper part and 1 3 in the lower part. Thus, the quintile arbitrary angle can be seen as a continuation of the trisect arbitrary angle.

Fig. 1 is a schematic diagram of the quintile, and Fig. 2 is a schematic diagram of the operation of the Liang's three-point angle setting.

It is impossible to draw with a ruler. It has been proven by Vantis.

However, relaxing the drawing method can be done as well.

It is also impossible to draw 5 equal parts of general angles with rulers.

For more information, you can refer to some books on abstract algebra.

It is impossible to draw a ruler and a ruler to divide any angle into three equal parts. This is mathematically proven! The problem of trisecting angles is one of the three major geometric drawing problems proposed by the ancient Greeks 2,400 years ago, that is, using a compass and a ruler to divide an arbitrary angle into threes.

The difficulty lies in the limitations of the tools used in the drawing. The ancient Greeks demanded that geometric drawings should be made only with straightedges (rulers without scales, only straight lines) and compasses. This is a question that has attracted many people to study, but none of them have succeeded.

In 1837 Van Zier (1814-1848) used algebraic methods to prove that this is an impossible problem of ruler drawing.

In the process of studying the trisectal angles, special curves such as mussel lines, heart lines, and conic curves were discovered. It was also found that as long as the precept of ruler drawing was abandoned, the trichotomy was not a difficult problem. The ancient Greek mathematician Archimedes (287 BC, 212 BC) found that the problem could be solved by fixing a ruler a little.

The method is as follows: add a little p to the edge of the ruler, and the end of the ruler is oLet the angle to be trisected be acb, with c as the center of the circle and op as the radius as the intersection of the corners of the semicircle at a,b; Make point O move in the CA extension **, point P move in the circumference, and when the ruler passes through B, connect OPB

Since OP PC CB, COB AC B 3The tools used here are not limited to rulers, and the drawing methods are not in line with the common designation.

But there are many ways to use other tools, which are described here:

Archimedes' three-point method.

Plotting: 1 Set any acute angle AOB;

2 Take O as the center of the circle and make the circle O, AOB and the circle intersect at points A and B;

3 Extend the bo, to a considerable distance;

4 Intersect the ruler with the circle O, one point is A, and the other point is P;

5. At the same time, the extension line of the ruler and the BO intersect at point C;

6. Adjust the position of the ruler appropriately so that PC=AO;

7 with ac, then acb=(1 3) aob

Proof: It can be proved by the relationship that the outer angles of the triangle are equal to the sum of the two inner angles that are not adjacent.

There is another mechanical drawing method to guess the hidden base third angle of the drawing, which is briefly described as follows:

As shown in the figure on the right: ABCD is a square, let AB move evenly to CD in parallel, AD turns to DC in a clockwise direction with D as the center, if AB arrives at DC when DA also happens to arrive at DC, then the trajectory of their intersection point AO is called the Qusui line.

Let A be any point on the AC arc, we want to divide the ADC in thirds, let DA and the three-point line AO intersect R, pass R as the parallel line of AB to cross AD, BC in A, B, let T and U be the third equal points of AD, and cross T and U as the parallel lines of AB to cross the three-point line AO in V and W, then DV and DW must be ADC in three equal parts.