Who can give me some questions that have answers (not too simple, the harder the better) and the pigeon hole principle is applied to life (not life)?

Example 1. Put 13 points into a square with a side length of 1 at will. It is proved that there must be four points, so that the area of the quadrilateral with these four points as vertices does not exceed.

Proof: As shown in the figure, if a square is divided into four rectangles with equal areas, then the four points of 13 must fall within the same rectangle, and the area shall not exceed.

Example 2. Seven points are randomly placed on a circle with a radius of 1, which proves that there must be two points, and the distance between them is not greater than 1.

As shown in the figure, if the circle is divided into six equal sectors, two of the seven points must fall in the same sector, so it is easy to know that their distance is not more than 1.

Example 3. Place 6 points arbitrarily in a 3×4 rectangle. It is proved that there must be two points, and the distance between them is not greater than.

As shown in the figure, if the rectangle is divided into five pieces, two of the six points will fall on the same piece, so it is easy to know that their distance is not greater than.

Two. The problem of numbers

Example 4. Give 7 different integers at will. It is proved that there must be two integers whose sum or difference is a multiple of 10.

Proof: Divide integers into 10 class according to the remainder divided by 10, and divide this 10 class into the following six groups: {0} (representing all integers divided by10); { 1}、{9}; {2}、{8}; {3},{7}; {4},{6}; {5}.7 There must be two numbers from the same group, and if they are of the same kind, the difference is a multiple of 10; If they are different, the sum is a multiple of 10.

Example 5. It is proved that there is a positive integer whose digit is 0 or 1 and is a multiple of 1993.

Proof: Consider the following numbers of 1993: 10, 1 10,10, …. If there is a multiple of 1993, the proof is completed; Otherwise, the remainder of their division by 1993 can only be 1, 2, …, 1992. There must be two numbers divided by 1993 with the same remainder, and their difference is a multiple of 1993. Obviously, the number of digits of this difference is 0 or 1.

Example 6. Write a 30-bit number composed of 1, 2, 3 at will, and intercept three adjacent numbers from this 30-bit number to form a 3-bit number. It is proved that two identical 3-digit numbers can be obtained by the above method.

A total of 28 3-digit numbers can be intercepted, and 33=27 3-digit numbers consist of 1, 2,3, and the two numbers must be the same.

Example 7. Given any n+ 1 different positive integers less than 2n, it is proved that three numbers can be selected from them so that the sum of two numbers is equal to the third number.

Proof: Let n+ 1 positive integer be A0.

Three. Dyeing problem

Example 8. Each square of a 3×7 chessboard is dyed in one of two colors: red and blue. It is proved that there is a rectangle composed of several squares, and the squares at the four corners of the rectangle have the same color.

Proof 1: The two squares in each column are the same color. Connect the centers of two squares with a line segment of the same color to get seven line segments, four of which must be of the same color and set to red. Because there are only three ways to connect (two out of three squares), two red line segments must be connected in the same way, and the corresponding four squares form a rectangle with four corners all red.

Proof 2: There are at least four squares of the same color in the first row. We assume that the first four squares are red. If there are two red squares in the first four squares in the second row, find a rectangle with all four corners red. Otherwise, at least three squares are blue, let's say the first three squares. At this point, the first three squares in the third row must have two squares of the same color. If they are red, they will form a rectangle with four corners in the same column as the first row. If it is blue, it and two blue squares in the same column in the second row form a rectangle with all four corners blue.

Example 9. There are six points on the plane, any three points are not collinear, and a red line segment or a blue line segment is connected between any two points, which proves that there must be a triangle with the same color (triangle with the same color on three sides).

Proof: Three of the five line segments starting from a certain point A must have the same color. Let AB 1, AB2 and AB3 be red. Consider line segments B 1B2, B 1B3 and B2B3. If there is a red line segment BiBj, then △ABiBj is a red triangle. If they are all blue, △ B 1B B 3 is a blue triangle.

Comments: If a point is regarded as an element, red dye is regarded as the relationship between elements, and blue dye is regarded as the relationship between elements, then this problem can be expressed as: given six elements, any two elements are related or not, then three elements can be selected, and they are related or not.

For example, changing elements into adults and changing the relationship between two elements into mutual understanding can lead to the following interesting propositions:

Choose six people at random in the world, which proves that three people can be found out. Do they know each other?

Four. The problem of "continuity"

Example 10. A student spent 1 1 week reviewing mathematics. He does at least one problem every day and at most 12 problems every week. It proved that there must be several days in a row, and he just did the 2 1 problem. (Tutorial P295/7)

Certificate: Suppose this student did the xi problem (i= 1, 2, …, 77) the day before, then X 1

Example 1 1. A worker in the TV repair department repaired at least one TV set every day during March 1 day, and repaired a total of 56 TV sets, which proved that he must have bought five TV sets for several consecutive days (including1day). (For details, see P 167/3)

Proof: Suppose he built the xi station the day before (I = 1, 2, …, 3 1), then x1< x2 <; …& lt； X3 1=56, let Yi =xi+2 1, then y 1

V miscellaneous issues

Example 12. There are 12 pairs of chopsticks, including 4 pairs of red chopsticks, 4 pairs of white chopsticks and 4 pairs of black chopsticks (two chopsticks in the same pair have the same color). To take out some chopsticks, two pairs of chopsticks with different colors are required. How many chopsticks should you take out at least?

Solution: First, taking out 10 chopsticks is not guaranteed, such as 8 red chopsticks and 2 white chopsticks. Secondly, taking out 1 1 chopsticks can be guaranteed, because there must be four pairs of chopsticks with the same color in 1 1 chopsticks, and there is already a pair of red chopsticks. Because there are only eight red chopsticks, at least three chopsticks are of other colors.

Comments: To solve this kind of problem, we usually find the untenable maximum number through the "worst case", and then prove that this number+1 must meet the requirements.

Example 13. There are 48 students in Class A, and each student has some friends in the class (if A is a friend of B, then B is also a friend of A). Prove that there are at least two students who have as many friends in the class.

Proof: Everyone's number of friends in the class can only be 0, 1, …, 47, but 0 and 47 cannot be obtained at the same time, so there must be two people with the same number of friends in the class.

Example 14. Around a rotatable round table, there are eight chairs evenly, and there are eight business cards on the table facing the chairs. After eight people sat down, they found that no one was looking at their business cards. Facts have proved that turning the table properly can at least make two people face their business cards.

C: The table rotates every 45 degrees, including the starting position, eight times. If no two or more people point to their business cards in these eight times, and they notice that everyone points to their business cards once in these eight times, then only 1 person points to their business cards in these eight times, but no one points to their business cards at the beginning, which is contradictory.