■ Angle trisection problem: bisect any angle;
■ Cube problem: Make a cube, so that its volume is twice that of the known cube;
■ The problem of turning a circle into a square: make a square so that its area is equal to the area of a known circle.
These problems were raised in ancient Greece 2400 years ago, and it was not until 1 837 that the French mathematician Mancer first proved that "angle bisection" and "bicubic" could not be used for ruler drawing. After German mathematician Lin Deman proved that π is a transcendental number in 1882, "turning a circle into a square" has also been proved to be a problem that neither a ruler nor a ruler can draw.
Another method to solve the problem that the ruler can't be drawn
■ overview
After repeatedly failing to solve the three major geometric drawing problems with a ruler, people doubt whether they can do it from the opposite side; On the other hand, it is natural to consider whether these problems can be solved by jumping out of the box of ruler drawing, that is, using curves or tools other than ruler instead of ruler.
It is found that once we jump out of the box of ruler drawing, the problem will be solved. Many people have done this work and made great achievements. The following entries should be introduced.
■ On the problem of bisecting any angle
★ Practice 1
Nicomedes method (around 250 BC) takes any point B on one side of the known acute angle ∠O as the perpendicular of OB, and the other side of the angle intersects with ∠O at point A, with O as the fixed point, BA as the fixed line, 2OA as the fixed length, as the right branch of mussel line C, and point A as the perpendicular of BA and intersecting with mussel line C at point S..
★ Practice 2
Pascal (B. 1623- 1662) method, for ∠AOB, take any long OA as the radius of one side and make a circle with point O as the center (Figure 12). Extend AO, intersect with circle o at point c, take circle o as a circle and take it as a circle.
★ Exercise 3
Pascal (B. 1623- 1662) method, for ∠AOB, take any long OA as the radius of one side and make a circle with point O as the center (Figure 12). Extend AO, intersect with circle o at point c, take circle o as a circle and take it as a circle.
★ Exercise 4
Rose line method: ∠AOB two sides intersect at point A and point B, draw an arc with O and A as the center, A as the radius, and the two arcs intersect at point S, so there is ∠BOS= 1/3∠BOA.
■ On the cubic product problem
★ Practice 1
Plato's (427-347 BC) method: make two mutually perpendicular straight lines, intersect at point O, and intercept OA=a on one straight line and OB=2a on the other straight line, where A is the known cube side length. Take points C and D on these two straight lines respectively, so that ∠ ACD = ∠ BDC. The side of the other square passes through point B, the other sides of the two square rulers overlap, and the right-angle vertices of the two square rulers are on two straight lines respectively. At this time, the right-angled vertices of the two rulers are points C and D). The length of line segment OC is one side of the cube.
★ Practice 2
Menaechmus's method (about 375-325 BC): It can be obtained from a: x = x: y = y: 2a.
Y2=2ax，x2=ay。 Therefore, draw two parabolas corresponding to the above two quadratic equations on the rectangular coordinate plane (figure 16). These two parabolas intersect at O and A, so the distance from the projection of point A on the X axis to the origin is the side length of the cube.
★ Exercise 3
Apollonius de Pergi (about 260-200 BC) Method: Make a rectangular ABCD, where AB=a and AD=2a. Take the diagonal intersection point G of this rectangle as the center, make a circle of appropriate length as the radius, and the extension lines of AB and AD intersect at E and F respectively, so that E, C and F are * * * lines, then AB:1
■ The problem of turning a circle into a square
★ Practice: For the known circle O, make its circle product line ①l in the first quadrant. Connect the two endpoints B and F of this circle product line, draw the vertical BG of BF through point B, intersect the X axis at point G, and take a point H on OA, so that HA= 1/2GO. Draw an arc with H as the center and HG as the radius, intersect the Y axis at point K, and then take OK as it.
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Anaquel Sa Gollers was a famous scholar in ancient Greece. In astronomy, he is famous for his achievements in explaining solar and lunar eclipses, and realized that the moon itself does not shine. It was his excellent research results that brought him misfortune. When he was about 50 years old, he fell from the sky and suffered unjust imprisonment. The cause of this disaster is that he thinks the sun is a hot stone. Because the religion at that time had asserted that the sun was a god, the scholar ignored it.
Although the time in prison is not too long, the grievances, depression and boredom in the days in prison really make people feel like years. In the dark and humid cell, Anacker Sa Gollers can't see the sunrise outside, and there is only a short time every day, and the sunlight can enter the room through the narrow square window of the cell. Whenever sunlight enters the cell and projects on the wall, it will always arouse his association as a scholar.
One day, when he was staring at the light of the square given to him by the round sun, his thinking mind suddenly had a whim: Can you make a square (using only rulers and compasses) so that its area is exactly equal to the area of a known circle? In this way, a world-famous problem-"turning a circle into a square" was born, which is related to the problem of "cubic product". The problem of "bisecting any angle" is called three difficult problems of geometric drawing in ancient Greece by later generations. Anacker Sa Gollers was very excited when he thought about turning a circle into a square, because he didn't have books and pens around, so it was difficult to study other problems. However, this problem is different, just drawing on the ground with a straw stick, and there are a lot of straw sticks in his cell.
Before entering the high wall, he never dreamed that in his most painful time, mathematics ruled out some of his troubles. However, the problems he raised in his life have never been solved.