Current location - Training Enrollment Network - Mathematics courses - How to make a regular 17 polygon with an uncalibrated ruler and compass?
How to make a regular 17 polygon with an uncalibrated ruler and compass?
1796 One day, at the University of G? ttingen in Germany, a 19-year-old young man was extremely talented in mathematics. After dinner, he began to do three routine math problems assigned to him by his tutor.

The first two questions were successfully completed in two hours. The third question is written on another small piece of paper: it is required to draw a polygon of positive 17, with only one ruler and no scale ruler.

He felt very tired. A minute passed, and the third question made no progress. Young people rack their brains, only to find that all the mathematics knowledge they have learned doesn't seem to help solve problems.

Difficulties aroused his fighting spirit: I must do it! He picked up the compass and ruler, and while thinking, he drew on the paper, trying to find the answer with some unconventional ideas.

When the dawn appeared, the young man breathed a sigh of relief, and he finally finished the difficult problem.

Seeing the tutor, the young people felt guilty and blamed themselves. He said to his tutor, "The third question you assigned me, I worked all night and failed to live up to your cultivation ..."

When the tutor took over the student's homework, he was shocked at once. He said to the young man in a trembling voice, "Did you make this yourself?" The young man looked at the tutor doubtfully and replied, "I did it." However, I spent the whole night. "

The tutor asked him to sit down, took out the compass and ruler, spread the paper on the desk, and asked him to make another regular 17 polygon in front of him.

The young man quickly made a polygon with vertex 17. The tutor excitedly said to him, "Do you know? You solved a math unsolved case with a history of more than two thousand years! Archimedes didn't solve it, Newton didn't solve it, you solved it in one night. You are a genius! "

It turns out that my tutor has been trying to solve this problem. That day, because of a mistake, he handed the note with this topic to the students.

Whenever the young man recalls this scene, he always says, "If someone tells me that this is a math problem with a history of more than 2,000 years, I may never have the confidence to solve it."

This young man is Gauss, the prince of mathematics.

Gauss is solved by algebraic method. He also regarded it as a masterpiece of his life and told him to carve the regular heptagon on his tombstone. But later, his tombstone was not engraved with a heptagon, but with a 17 star, because the sculptor in charge of carving thought that the heptagon and the circle were too similar, and everyone must be confused.

On the drawing of regular heptagon (I didn't mean to copy Gauss's ideas _);

Here is a theorem to be used:

If line segments with lengths | a | and | b | can be geometrized, so can line segments with lengths |c|.

Where c is the real root of the equation x 2+ax+b = 0.

The above theorem is actually to make a line segment with the length sqrt (a 2-4b) when there are line segment lengths |a| and |b|.

You can draw this step, right? )

To make a regular heptagon in the unit circle is mainly to make a line segment with the length of cos(2pai/ 17).

Next, I will give the proof that cos(2pai/ 17) can be proved by Gauss, and also give the concrete method.

Let a = 2[COS(2 factions/17)+COS(4 factions/17)+COS(8 factions/17)+COS( 16 factions/17)] >

a 1 = 2[cos(6pai/ 17)+cos( 10 pai/ 17)+cos( 12 pai/ 17)+cos( 14 pai/ 17)]& lt; 0

Then a+a 1 =- 1, and a * a 1 =-4, that is, a, a 1 is the root of the equation x 2+x-4 = 0, so it can be made that the lengths are |a| and | a/kloc.

Let b = 2 [cos (2pai/17)+cos (8pai/17)] > 0b1= 2 [cos (4pai/17)+cos (16pai. 0

c = 2[cos(6pai/ 17)+cos( 10 pai/ 17)]& gt; 0 c 1 = 2[cos( 12 pai/ 17)+cos( 14 pai/ 17)]& lt; 0

Then b+b1= ab * b1=-1c+c1= a1c * c1=-1.

In the same way, line segments with lengths of | b |, | b 1 |, | c | and | c 1 | can be made.

Then 2cos (2pai/17)+2cos (8pai/17) = b [2cos (2pai/17)] * [2cos (8pai/17)] = c.

Thus 2cos(2pai/ 17) is the larger real root of the equation x 2-bx+c = 0,

Obviously, it can also be done, and the drawing method has been given above