Using analogy to solve problems, the basic process can be represented by the block diagram as follows:
It can be seen that the key to using analogy method is to find a suitable analogy object. According to the different angles of looking for analogy objects, analogy methods are often divided into the following three types. If the object in three-dimensional space is reduced to the object in two-dimensional (or one-dimensional) space, this comparison method is called dimension reduction analogy.
In Example 2, each side of a regular tetrahedron with a side length of 1 is regarded as a sphere, and S is the intersection of six spheres. It is proved that there is no point pair with a distance greater than 1 in S.
Analyze and consider the analogy proposition on the plane: "A regular triangle with a side length of 1 is a circle with each side as its diameter, and S' is the intersection of three circles". By exploring the similar nature of s', we can seek the argumentation idea of this question. As shown in the figure, it is easy to know that s' is contained in a circle with the center of gravity of a regular triangle as the center and the radius as the center. Therefore, the distance between any two points in s' does not exceed 65438.
Proof: As shown in the figure, in the regular tetrahedron ABCD, m and n are the midpoint of BC and AD, g respectively.
Is the center of △BCD, Mn ∩ Ag = O. Obviously, O is the center of regular tetrahedron ABCD. It is easy to know that OG = Ag =, and it can be deduced that the distance between any two points on a sphere with O as the center and OG as the radius is not greater than, and its sphere O must contain S, which is proved as follows.
According to the symmetry, we might as well look at the tetrahedron OMCG in the spatial region. Let p be any point in tetrahedral OMCG, and it is not in sphere O, which proves that p is not in S either.
If the ball o passes through OC at t point. △TON，ON=，OT=，cos∠TON=cos(π-∠TOM)=-。 According to cosine theorem:
At =,∴TN=. ，tn2 = on2+ot2+2
In Rt△AGD, n is the midpoint of AD, ∴GN=. From GN= NT=, OG=OT, ON=ON, we get △ gon △ ton. ∴∠TON=∠GON, they are all very slow.
So obviously, any point P in △GOC that does not belong to the ball O has ∠ PON >; ; ∠TON, that is, there is PN & gtTN=, except for the sphere with the center of n and the diameter of AD, the point p does not belong to the region S.
It can be seen that the ball O contains the intersection point S of six balls, that is, there are no two points in S, so that the distance is greater than. For some problems to be solved, there is no ready-made analogy, but we can find analogy problems by observing and relying on structural similarity, and then transform the original problems into analogy problems through appropriate replacement.
Example 3 gives any seven real numbers xk (k = 1, 2, …, 7). It is proved that two xi and xj satisfy the inequality 0 ≤≤
The analysis shows that if any two of the seven real numbers are equal, the conclusion is obviously valid. If the seven real numbers are not equal, it is difficult to start. However, after careful observation, we can find that the tangent formula of the difference between the two angles is very similar in structure, so we choose the latter as an analogy and turn it into an analogy problem through proper substitution. For substitution, xk = tanαk(k = 1, 2, ..., 7), which proves that it must exist.
Prove xk=tanαk(k =l, 2, …, 7) and αk∈(-,), then the original proposition is transformed into: prove that there are two real numbers αi, αj∈(-,), and satisfy 0 ≤ tan (α i-α j) ≤
According to pigeonhole principle, αk must have four in [0,] or in (-0), so it is better to set four in [0,]. Note that tan0=0, tan=, and in [0,], tanx is increasing function, so we only need to prove the existence of αi and αj, so 0.
Example 4 shows that Xi ≥ 0 (I = 1, 2, …, n) and x 1+x2+…+xn = 1.
Verification: 1 ≤++…+≤
Analysis, we can compare it to a simple analogy problem: "xl≥0, x2≥0, and xl+x2 = 1, verification 1 ≤+≤". The proof idea of this analogy problem is: ∫2≤XL+x2 = l, ∴ 0 ≤ 2 ≤ 60.
It is proved that if the basic inequality has 0≤2≤xi+xj, then
0≤2 ≤( n- 1)(XL+x2+…+xn)= n- 1
∴ 1≤xl+x2+…+xn +2≤n, that is,1≤ (+…+) 2 ≤ n.
∴ 1≤++…+≤.
The so-called induction refers to a form of reasoning that draws general conclusions through the analysis of special cases. It consists of two parts: the premise is a number of known individual facts, which are individual or special judgments and statements, and the conclusion is a guess derived from the premise through reasoning, which is a general statement and judgment. Its thinking mode is: let MI (I = 1, 2, ... yes.
If =M, induction at this time is called complete induction. The conclusion is correct and reliable because it exhausts all the special cases of the studied object. Complete induction can be used as an argument method, also called enumeration induction.
If it is proper subset of M, the induction at this time is called incomplete induction. Because incomplete induction does not exhaust all the research objects, the conclusion can only be regarded as conjecture, and the correctness of the conclusion needs further proof or counterexample.
This section mainly introduces how to obtain conjecture by incomplete induction. For the complete induction, it will be explained in combination with related contents (such as classification) in the future.
Example 5 proves that the sum of the perimeter of any convex quadrilateral with an area equal to 1 and the length of two diagonals is not less than 40.
It is difficult to analyze the mixture of quadrilateral perimeter and diagonal length. We can start with a special case study: first, consider a square with an area of 1, whose circumference is exactly 4 and the sum of diagonal money is 2, and then consider a diamond with an area of 1. If the two diagonals are marked as l 1 and l2, then the area of the diamond is S = L 1 L2.
L 1+ l2≥2=2=, diamond perimeter: l=4≥2=4.
It can be guessed that the perimeter and diagonal length of a general convex quadrilateral can also be considered separately.
It is proved that let ABCD be an arbitrary convex quadrilateral with an area of 1, and its related line segments and corner marks are shown in the figure.
SABCD= (eg+gf+fh+he)sinα
≤ (e+f)(g+h)≤，
∴e+f+g+h≥2, that is, the sum of diagonal lengths is not less than.
∴a+b+c+d≥4, that is, the circumference is not less than 4.
To sum up, the conclusion is that,