Measures-size and shape

The book is a recent PhD in mathematics at Columbia University, Paul Lockhart, a professor at the University of California and Pericles, who has a very unique aspect of the idea of elaborating mathematical issues, in accordance with the chapters of the book, according to reality and imagination, on mathematical problems, size and shape, Time and Space Four section to introduce this part of the content.

Size and shape This chapter records the author's unique introduction to geometry, and the greatest feature of the book is that there is no textbook-like seriousness, no seemingly rigid definition, the same question is full of interest and exploration of the spirit of inquiry and thinking, his purpose is only one, let the reader feel the happiness of research mathematics.

Then we start the text, mainly on the book to give the corresponding discussion and analysis of the problem.

The beautiful pentagram:

At first glance this problem seems to be very difficult to think Ah, to seek the degree of each angle of n-angle star, in fact, is equivalent to We ask n angle star n sharp angle of the degree and.

So how do you ask for a positive n angle star angle?

Obviously first of all we have to figure out what is N-angle star, we can easily draw a positive pentagram, hexagonal star it? Obviously we need a simple definition of what a positive n-angle star is.

Definition: We make a graph that is composed of N-and N-isosceles triangle that is called a positive n-angle star. (This may involve some of the awn concepts, but simplifying the process is what we define here)

So the question is, from this definition, we need to draw some figures, whether the pentagram is this kind of drawing? What about the six-point star?

If you think there is only one kind of painting, it is too too young too simple, please see the following photos.

Obviously, for the N-angle star that satisfies the definition, we can draw a variety of patterns, which tells us that the scream for the N-angle star is not just an answer, but a row of answers.

So for the continuation of the study, we have to look like a way to draw a positive n-angle star, we observe the nature of its drawing, it is not difficult to generalize the law, for each positive n-angle star, it is from a point of n vertex, the same number of points spaced, clockwise (or counterclockwise) connected, and finally back to the starting point Of course, if there is a connection that comes back to the starting point, then proceed from the starting point to continue to connect, such as the above figure on the left of the second figure.

So we are more algebraic to represent all the cases of n-angle stars, we start from the interval points, set X is the positive n-Star formation process of the two adjacent points between the interval between the point, taking into account the CIS-pointer and anti-clockwise symmetry, that is, x = i and x = n-2-I formed a positive n-angle star is equivalent,

When n is odd, x=0,1,2,.., (n-3)/2. There is (n-3)/2+1 species.

When n is an even number, x=0,1,2,... (n-2)/2, that is (n-2)/2+1 species.

So the following is a discussion, we start with a few simple examples, when n=5,x = 0, 1. For x=0, there is no number analysis, is a positive 5-side shape. For X=1, if we do its circumscribed circle, and then use the circumference of the central angle is inscribed angle twice times, can easily get at this time the internal angle and is 180°.

So now we want to generalize this mechanism to a more general situation, and we are more abstract and see that the reason we can get 180° is because X=1 decides that the two points of the adjacent access are exactly separated by a point, which results in the next vertex of the two edges of each sharp angle being adjacent ( Note that this is adjacent instead of adjacent access), which causes it to be in the corner of the center of the time, do not repeat, so you can get a 180° answer. To generalize, that is, for n is odd, x= (n-3)/2 of the situation, all meet this situation, screaming and all are 180°.

So further inquiry, x= (n-3)/2-1, which leads to the result is that each sharp corner on the side of the other 2 vertices, separated by 2 vertices (this can be calculated by simulating a stroke process), which will lead to the center of the corner, just the center of the corner is counted 3 times, that is, more than 2 circumference, That is, the sharp angle and the 180°+360°=540° are the case.

Similarly, when x= (n-3)/2-2, screaming and adding 360 °.

And eventually when X=0, we were able to see according to the above method we would get screaming and for the N-3 + (n-2)/2 * * * *, in fact, exactly corresponds to the x=0 formed by the positive n-side shape.

Similarly, when n is an even number, we can do similar processing, but for x= (n-2)/2, we can see that the other 2 vertices along the edge of one of its sharp corners are separated by a vertex, that is to say, the circumference of this case will get two circumference, The inner angle of the sharp angle and should be 360°.

As a result, the problem has been satisfactorily resolved.

When n is odd, the sharp angle and is 180°, 540° 、...、 + (n-3)/2 * 360°.

When n is an even number, the sharp angle and the 360°, 720° 、...、 (n-2)/2 * 360°.

How many regular polyhedron are there:

For the tetrahedron, positive hexahedral everyone is very familiar with, we give regular polyhedron next definition, each face each other congruent, any one vertex of the plane expansion diagram are the same multi-faceted body called regular polyhedron, then, how many regular polyhedron in the space?

We consider using the method of partial algebra to solve this problem.

Set regular polyhedron vertex number is V, the number of polygons is F, the number of edges is E, each vertex has M-bar, each polygon is a positive n-sided shape.

It is easy to see NF = 2e①

MV = 2e②

That is f= 2e/n,v=2e/m.

With Euler formula V+F-E = 2, with 2e/m+2e/n-e=2, finishing can be obtained by:

1/m+1/n=1/2+1/e. ③

Since E is an integer so 1/e>0, considering the n≥3, there is 1/m>1/2-1/3 = 1/6, which is m<6, and m>2, so the value of M is 3, 4, 5.

Symmetric, because E is an integer so 1/e>0, considering the m≥3, there is 1/n>1/2-1/3 = 1/6, which is n<6, and n>2, so the value of N is 3, 4, 5.

Based on the integer pair (m,n), the feasible e based on the ③ is obtained.

Finally, we can see that there are only 5 kinds of regular polyhedron in space: tetrahedron, positive hexahedral, positive octet, positive 12, positive 20-body.

are triangles equal in each corner?

This problem seems to be a bit retarded, because we have "two horns equal two or three angle similar" the Similarity triangle judgment theorem, but how it comes, here we may wish to trace back.

First of all, to give the concept of similarity, we will each side of the corresponding to a constant proportion of the two graphs called is similar. From this definition, it is easy to prove that the two triangles parallel to each other on the three sides are similar.

So here we just need to convert three angles equal two triangles to a three-side parallel, we might as well, according to the triangle of equal nature to try to simulate the generation of two triangles, this method in solving a lot of proof problems are very useful.

First we are based on two equal angles, we let the two equal corners of an edge coincident, then the two corners of the remaining two sides must be parallel, and then add two triangles of their respective third side, because the angle must be guaranteed equal, then by the nature of the parallel line, then the third side of the two triangle is also a parallel relationship. The coincident edges of the first two equal angles can obviously be considered parallel to each other. Evidence.

Measures-size and shape