Between 5000 BC and 600 BC, Greek mathematicians had already thought of the method of bisecting any angle, just as we learned in geometry textbooks or geometric paintings: take the vertex of a known angle as the center, use an appropriate radius as the intersection point of both sides of the arc, and then draw an arc with an appropriate length as the center, and the intersection point of these two arcs is connected with the corner top to divide the known angle into two halves. Since it is so easy to bisect a known angle, it is natural to change the question slightly: How about bisection? In this way, this problem naturally arises.
In the 4th century BC, Ptolemy I Soter made Alexandria its capital. By virtue of the superior geographical environment, we will develop maritime trade and handicrafts and reward academics. He built a large-scale "art shrine" as an academic research and teaching center; He also built the famous Alexandria Library with a collection of 750,000 books. Ptolemy I Soter deeply understands the importance of developing science and culture. He invited famous scholars to Alexandria, when many famous Greek mathematicians came to this city.
There is a round villa on the outskirts of Alexandria, where a princess lives. There is a river in the middle of the round villa, and the princess's bedroom is built in the center of the circle. A door was opened on the north and south walls of the villa and a bridge was built over the river. The location of the bridge and the location of the south and north gates are exactly in a straight line. Every day, the items given by the king are brought in from the north gate, first put in the warehouse at the south gate, and then the princess sends someone to retrieve the room from the south gate.
One day, the princess asked her entourage, "How far is it from the north gate to my bedroom or from the north gate to the bridge?" The attendants didn't know, so they hurried to measure it, and the result was that the two roads were as far apart.
After a few years, the princess's sister, the little princess, grew up, and the king wanted to build a villa for her. The little princess suggested that her villa should be built like her sister's villa, with rivers, bridges and north and south gates. The king promised that the little princess's villa would be started soon. When the south gate is built and the positions of the bridge and the north gate are determined, a question arises: How can the distance from the north gate to the bedroom be the same as the distance from the north gate to the bridge?
It is known that the position of the south gate is p, the bedroom (center of the circle) is o, the position of the north gate is q, and the bridge is k,
The third division of an angle
The key to determine the location of the North Gate Bridge is to make ∠OPQ and the angle between PO and the river α.
Author QK=QO,
Get ∠QKO=∠QOK
But ∠QKO=α+∠KPO,
∠OQK =∠ OPK again
So in △QKO,
∠QKO+∠QOK+∠OQK
=(α+∞KPO)+(α+∞KPO)+∞KPO
=3∠KPO+2α= 180
That is, ∠KPO=( 180-2α)/3.
As long as the angle of 180-2α is divided into three parts, the position of the bridge and the north gate can be determined. The key to solving the problem is how to divide an angle equally.
But there is no pure ruler method that can bisect any given angle.
The craftsmen tried to determine the position of the bridge by drawing with a ruler, but it took them a long time to solve the problem. So they asked Archimedes.
Archimedes solved the problem of bisecting an angle by making a fixed mark on the ruler, thus determining the position of the north gate. Just as everyone praised Archimedes for his greatness, Archimedes said, "This method of determining the position of the north gate is feasible, but it is only a stopgap measure. Is flawed. " Archimedes' so-called flaw is to mark on the ruler, which is equivalent to marking, which is not allowed in the ruler drawing method.
This story raises a mathematical problem: how to bisect any known angle with a ruler, which even Archimedes did not solve.
In order to explain the necessary and sufficient conditions for ruler drawing, it is necessary to first transform geometric problems into algebraic language. For a plane drawing problem, the premise is always to give some plane graphics, such as points, lines, angles, circles and so on. But a straight line is determined by two points, an angle can be determined by its vertex and three points on each side, and a circle is determined by the center of the circle and a point on the circumference, so the problem of plane geometric drawing can always be summed up as given n points, that is, n complex numbers.
(and, of course, z0= 1). The drawing process of ruler and ruler can also be regarded as using compasses and rulers to get new complex numbers, so the problem becomes: give a reply number.
And z0, where can we get information?
Starting with a ruler, we can get the complex number z we want in advance. For the convenience of discussion, the following recursive definition is given: [1]?
Definition: let S={Z0= 1, Z 1, ... Zn} be n+ 1 complex numbers, and then
(1) Z0= 1, Z 1, ... Zn is called point S;
(2) A straight line passing through two different S points is called an S line, and a circle with one S point as the center and the distance between any two S points as the radius is called an S circle;
(3) The points where the line S intersects with the line S, the line S intersects with the circle S, and the circle S intersects with the circle S are also called S points.
The above definition completely describes the process of ruler drawing. If p represents the set of all S points, then P is defined by S={Z0= 1, Z 1, ... Zn}.
Theorem: Let Z 1, ... Zn(n≥0) be n complex numbers. Let F= Q(Z 1, ... Zn, Z 1', ... Zn'), (z' stands for * * * yoke complex number), then a complex number z can be given by S={Z0= 1, Z 1. In other words, z contains a quadratic radical extension of f.
System: let S={Z0= 1, Z 1, ... Zn}, F= Q(Z 1, ... Zn, Z 1', ... Zn'), and z is point S, then [f]
It is proved that any angle can't be bisected, and ruler drawing can't bisect an angle of 60 degrees:
It is proved that giving an angle of 60 degrees is equivalent to giving a complex number z1=1/2+√ 3/2 i. Therefore, S={Z0= 1, Z 1}, F=Q(z 1, z/. If you can make an angle of 20 degrees, you can certainly get cos20, but cos20 satisfies the equation 4x3-3x- 1/2=0, that is, 8x3-6x- 1=0. Since 8x3-6x- 1 is irreducible in Q[x], [q (cos 20): q] = 3, then
6=[ Q(cos20，√-3):Q]=[F(cos20):Q]=[F(cos20):F][F:Q]
Since [f: q] = [q (√-3): q] = 2, [f (cos20): f] = 3, according to the above system, cos20 is not an S point, so it is impossible to divide 20 degrees equally.
I hope it can help you solve the problem.