Which way did the bicycle go (below)

 

23. Some coins do not overlap on the table. The four-color theorem tells us that if we want to dye a coin and make the color of the coin together different, we only need four colors at most. Are there at least four colors required?

 

Answer: Yes ., If only three colors are allowed, the color of a must be the same as that of all shadow coins, and the color of B must be the same as that of all shadow coins. A and B will be the same.

 
 

24. Use a matchstick to create a plan where the degrees of all vertices are 3. Note that the vertex of the graph can only be composed of the endpoint of the matchstick.

 

Answer :.

 
 

25. There are 2n equal points in the circumference, and an edge is connected between two points to form a complete graph k_2n. Is there a Hamilton loop that makes the passing edge of the graph parallel to each other?

 

Answer: there is no such loop. The 2n vertices are sequentially numbered from 1 to 2N, and the direction value of an edge is defined as the sum of the numbers of the points at both ends. Obviously, when two sides are parallel and only when their direction values are 2n same. Assume that there is a Hamilton loop so that any two sides on the way are not parallel, then the direction values of these sides may only be 1, 2 ,..., 2n (mod 2n), whose sum is 1 + 2 +... + 2n = 2n ^ 2 + n ≡ N (mod 2n ). In addition, because this is a Hamilton loop, the sum of the direction values of the edge passing by the entire path is 2*(1 + 2 +... + 2n) = 4n ^ 2 + 2n limit 0 (mod 2n), which is a conflict.

 
 

26. Proof: For any positive integer N, the total number of numbers from 1 to 10 ^ N is equal to the number 0 displayed from 1 to 10 ^ (n + 1.

 

Answer: for each number from 1 to 10 ^ N, list all the numbers that can be obtained after inserting a 0 in the middle to form a new series:

10, 20, 30, 40, 50, 60, 70, 80, 90,100,100,101,110,102,120 ,...

For a K-digit number, there are K ways to insert a 0 in the middle, so the total number of items in the series is equal to the total number in 1 to 10 ^ n. Next, we only need to explain that the total number of items in the series is equal to the number 0 in 1 to 10 ^ (n + 1. This is because the number of all columns obviously does not exceed 10 ^ (n + 1), and it is clear that a number has a few zeros, which appears several times in the series.

 
 

27. How many spaces are divided by N planes passing through a common point (but any three planes are not in the same straight line?

 

Answer: We want to create a sphere based on this public point. What we need is the number of areas split from the sphere. Using the Euler's formula v-e + F = 2 on this sphere, we can quickly find the value of F. Because each pair of big circles has two intersections, V = 2 * C (n, 2); E = 4 V/2 = 2 V because the degree of each vertex is 4. So f = E-V + 2 = 2 V-V + 2 = V + 2 = n ^ 2-N + 2.
Using the Euler formula, you can instantly kill the following classic question: how many blocks can a plane be divided into at most N circles? Because each circle has two intersections, the number of vertices v = n (n-1). Since each circle is cut into two (n-1) arcs, therefore, E = 2n (n-1 ). Therefore, F = E-V + 2 = 2n (n-1)-N (n-1) + 2 = n ^ 2-N + 2.
Note that the answers to the above two questions are the same. In fact, this is not a coincidence. If you project the sphere in the previous problem from the Arctic point to the plane tangent to the South Pole, you will find that these two problems are actually one thing.

 
 

28. Proof: When m and n are mutually reinforcing, the diagonal lines of the m × n board exactly pass through the m + n-1 lattice.

 

Answer: Because m and n are mutually qualitative, it is clear that the diagonal lines do not pass through the intersection. From the lower left to the upper right corner, you must go through m-1 vertical line and n-1 horizontal line. Therefore, there must be m + n-2 intersections between the diagonal line and the chessboard, that is, it passes through the m + n-1 lattice. The conclusion can be expanded to: the diagonal lines of the m × n board just pass through the m + n-gcd (m, n) lattice.

 
 

29. Place eight points in the space, so that for any three points, the three line segments between them have at least two same lengths. In other words, all triangles identified by these eight points (including the triangle that degrades to a line) are isosceles triangles.

 

Answer: Make a unit circle on the x-y plane and mark its five-point. The other three points are placed at (0, 0, 0), (0, 0,-1), and (0, 0, 1.

 
 

30. Proof: each side is an acute angular triangle in the same pair of edges.

 

Answer: think of the triangle △pqr and △psr as two triangle boards connected by the PR axis, and QS as a rubber band. If you rotate △srs so that △srs and △pqr are on the same plane, the rubber band QS is extended to D. In this case, the entire graph becomes a parallelogram. Because of the diagonal D> C, the ∠ pqr is less than 90 °. Similarly, we can see that every angle in a single triangle is less than 90 °.

 
 

31. For point set E in space, the point set composed of all the points and the straight lines identified by all the points is recorded as L (E ). If V is the four vertices of a positive triangle, L (v) is the point set composed of the straight lines of the six edges of the triangle. Is L (V) included in space?

 

Answer: No ., Imagine a positive triangle formed by the diagonal lines of six cubes. P points are not in L (v. First, take points on the adjacent edge of the triangle, and the determined straight line is always on the surface of the triangle without passing through P. The line formed by the points on the edge will not pass through P, for example, the line between P and the vertices in the red line can only be on the top of the cube, and there cannot be a common point with the blue line.

It can be proved that the only point not in L (V) is the other four vertices in the cube that are not on the triangle.

 
 

32. There are 1001 wires at the bottom of a building, which extend to the roof of the building. You need to determine the relationship between the underlying 1001 wire headers and the 1001 wire headers on the roof. You have a battery, a light bulb, and many very short wires. How can we determine the ing between the wire and cable headers only once?

 

Answer: The following method can be used in the case of an odd number of wires. Connect the two wire headers downstairs, and the other one is not connected. Mark the single line header as #1. Find the wire that does not constitute a loop with anyone connected to the roof. It is #1. In the remaining wires, determine the wire pair that can constitute a loop and mark it as (#2, #3), (#4, #5 ),...... . Next, connect #1 and #2, #3 and #4, and so on. Remove the original connection downstairs, and then start from #1 to determine all the corresponding relationships: And #1 can constitute the loop of the wire is #2, the original and #2 is #3, and #3 can constitute the loop #4, and so on.

 
 

33. There are 15 sheets of arbitrary paper on the table, covering the entire desktop. The paper may overlap or extend out to the desktop. Proof: five sheets of paper are always taken away, so that the rest of the paper still covers at least 2/3 of the desktop.

 

Answer: Let's say that paper is filled with red paint before it is placed on the desktop. After the paper is covered, some paper (all or some parts) is colored. Because the entire desktop is completely covered, the total area of all red areas on all paper is exactly equal to the desktop area. If we take away the five sheets of paper with the least paint, the area of the taken paint will obviously not exceed 1/3, that is, the sum of the red area on the paper will be at least 2/3, that is, at least 2/3 of desktops are covered.

 
 

34. Even if the hour hand and the minute hand are as long as the minute hand, the correct time can be read most of the time. For example, if one of the two needles points to 12 and the other points to 6, the former can only be a split needle, and the latter can only be a clockwise hand. But sometimes, after the interchange between the hour hand and the minute hand, the point of time still makes sense, so we can say that the pointer position is ambiguous. During the 12 hours from to, how many time periods will the pointer position produce ambiguity?

 

Answer: There are 132. Suppose there are two clocks A and B stacked together, a runs at a normal speed, and B runs at a speed of 12 times. Therefore, the hour hand of B will always overlap with the minute hand of. Whenever the minute hand of B is the hour hand of A, the time that a points is ambiguous. The minute-hand of B is 144 times faster than the hour-hand of A. Therefore, after the hour-hand of a turns around, the minute-hand of B turns around 144 times, therefore, the sub-needle of B overlaps with the hour hand of a for 143 times. However, 11 of these times are the same hour and minute hands, which will not lead to ambiguity. Therefore, 143-11 = 132 moments will actually lead to ambiguity.

 
 

35. Alice and Bob play the quiz game. Alice writes n integers x1, x2,..., xn behind Bob. Then Bob selects n positive integers A1, A2,..., and an tells Alice that Alice has the A1 · X1 + A2 · X2 +... + an · XN value. Next, Bob selects another n positive integers b1, b2,..., bn, and obtains the Σ Bi · Xi value. So on, until Bob can release the N number written by Alice. How many times does Bob need to guess all N numbers?

 

Answer: twice. First select a1 = a2 =... = An = 1 to know the sum of N numbers S. Note that because the numbers written by Alice are all positive integers, all I have Xi ≤ S. Then, select b1 = 1, b2 = S + 1, B3 = (S + 1) ^ 2 ,..., BN = (S + 1) ^ (n-1), then Alice will tell him a number

N = X1 + x2 · (S + 1) + X3 · (S + 1) ^ 2 +... + Xn · (S + 1) ^ (n-1)

Next, Bob only needs to write N in the S + 1 hexadecimal notation, and the numbers on each digit are the N numbers of Alice in turn.

Next we will prove that the two queries are already the best, and they cannot be guessed only once. Suppose Bob's first query result is A1 + A2 +... + An + A1 · A2, then he cannot know that Alice's n numbers are A2 + 1, 1, 1 ,..., 1, or 1, A1 + 1, 1, 1 ,..., 1.

 
 

36. There are two boxes on the table, 51 coins in one box and 101 coins in one. Alice and Bob take turns to drop the coins in one box, and then separate the coins in the other box into two boxes. In the end, no one can continue the operation and no one will lose. If Alice leaves, who has a winning strategy?

 

Answer: Bob wins. After Alice leaves, the number of coins in the two boxes must be odd. Bob dumped the box with an odd number of coins and divided the remaining coins into 1 and an odd number. In this way, Bob always has an even number of coins in front of him.

 
 

37. If a plane vector satisfies y ≥0, we can say that this vector is facing up. Two face-up unit vectors, the vectors and the possible values are very small. Proof: the length of the vector and the unit vector cannot be less than 1.

 

Answer: Assume that n vectors are V1, V2,..., and VN. Note that V1 + V2 +... + VN = V1 + (V2 +... + VN), and the angle between the two vectors is reduced, so V1 is changed to one of (1, 0) or (-1, 0, then the sum of the n vectors will become smaller. We can see that when all n vectors are horizontal vectors, the vectors are the smallest and the smallest. The minimum value of an odd number of horizontal vectors is 1.

 
 

38. If n polynomial p (x) is satisfied, p (x) is greater than or equal to 0 for all X. It is proved that all X has p (x) + p '(x) + P ''(x) +... + P ^ (n) (x) ≥ 0.

 

Answer: because all X has p (x) ≥ 0, the number of times N of P (x) must be an even number, and the maximum coefficient of occurrence must be greater than 0. So that Q (x) = p (x) + P' (x) +... + P ^ (n) (x). Obviously, the number of Q (x) and the maximum coefficient are the same as p (x). Therefore, Q (x) has a minimum value, for example, Q (x0 ). We only need to explain that Q (x0) is greater than or equal to 0.
Q (x0) is the minimum value, so Q' (x0) = 0. Q' (x) = P' (x) +... + P ^ (n) (x) = Q (x)-P (x), so q (x0)-P (x0) = 0, that is, Q (x0) = P (x0) is greater than or equal to 0.

 
 

39. Is it possible to draw an irrevocable 8 non-Intersecting number on the plane?

 

Answer: No. For any 8-shaped image, take a rational point p and q in each of the two holes (this is always possible because the rational points on the plane are dense ), the 8-shaped circle contains the rational point (p, q ). Note that because the 8-shape cannot overlap, the two 8-shape cannot enclose the same pair of rational points. Since there are only a few rational points on the plane, there are only a few 8-shaped points.