The butterfly theorem was first published in a popular magazine "Diary of a Man" in 18 15 as a proof. Because of its peculiar geometry, it looks like a butterfly, hence the name. Theorem content: If the midpoint M of the chord PQ on the circle O is two chords AB and CD, and the chords AD and BC intersect PQ on X and Y respectively, then M is the midpoint of XY.

There have been many beautiful and strange solutions, the earliest of which should be Horner's proof in 8 15. As for the proof method of elementary mathematics, in foreign materials, it is generally believed that Steven, a middle school teacher, first put forward the proof method of area, in which the area formula is applied: S= 1/2 BCSINA. 1985, in the first issue of Mathematics Teachers in Henan Province, Comrade Du Xilu introduced the butterfly theorem to China with the title "Famous Problems in Plane Geometry and Their Wonderful Solutions", and since then the butterfly theorem has spread all over China.

This paper introduces a simple method to prove elementary mathematics.

It is proved that the center O is the vertical line between AD and B, and the vertical foot is S and T, connecting OX, OY and OM. SM .MT .

∫△SMD∽△CMB and SD= 1/2ADBT= 1/2BC,

∴DS/BT=DM/BM and ≈ D = ∠ B.

∴△MSD∽△MTB,∠MSD=∠MTB

∴∠msx=∠mty； And o, s, x, m and o, t. Y .m are a four-point * * * circle,

∴∠XOM=∠YOM

∵OM⊥PQ∴XM=YM

As shown in figure 1, the major axis A 1A2 of the ellipse is parallel to the X axis, and the minor axis B 1B2 is on the Y axis with the center of M(o, r) (b > r > 0).

(1) Write the equation of ellipse, and find the focal coordinates and eccentricity of ellipse;

(2) the straight line y=k? X passes through the ellipse at two points C (X 1, Y 1) and D (X2, Y2) (Y2 > 0); The straight line y=k2x intersects the ellipse at two points G(x3, y3) and H(x4, y4) (y4 > 0).

Verification: k1x1x 2/(x1+x2) = k2x3x4/(x3+x4)

(Ⅲ) For C, D, G and H in (Ⅱ), let CH intersect with X axis at point P and GD intersect with X axis at point Q. ..

Proof: | OP | = | OQ |.

(The proof process does not consider the case that CH or GD is perpendicular to the X axis. )

2. Answer: The reference answers given by experts from the Admissions Examination Office of Beijing Education Examinations Institute in the "2003 National Unified Entrance Examination for Colleges and Universities" are as follows:

(18) This topic mainly examines the basic knowledge of straight lines and ellipses, and examines the ability to analyze and solve problems. 15.

(1) solution: the elliptic equation is x2/a2+(y-r)2/b2= 1.

The focal coordinates are

(2) prove the equation y=k of straight line CD? Substituting x into the elliptic equation gives b2x2+a2(k 1x-r)2=a2b2.

Pack up, take it

(B2+a2k 12)x2-2k 1a2rx+(a2r 2-a2 B2)= 0

According to Vieta's theorem, you must

x 1+x2 = 2k 1a2r/(B2+a2k 12)，x 1 x2 =(a2r 2-a2 B2)/(B2+a2k 12)，

So x1x2/(x1+x2) = (R2-B2)/2k1R1.

Substituting the equation y=k2x of the straight line GH into the elliptic equation, the same can be obtained.

x3x4/(x3+x4)=( r2-b2)/2k2r ②

K1x1x2/(x1+x2) = (R2-B2/2r = K2X3x4/(x3+x4) from ① and ②.

So the conclusion is established.

(3) Proof: Set point P(p, o) and point Q(q, o).

Through the c, p, H*** lines, we get

(x 1-p)/(x4-p)= k 1x 1/k2x 4

The solution is p = (k1-k2) x2x4/(k1x1-k2x4).

Through the D, Q, G*** lines, it can also be obtained.

q =(k 1-k2)x2x 3/(k 1x 2-k2x 3)

From k1x1x2/(x1+x2) = k2x3x4/(x3+x4) to:

x2x 3/(k 1x 2-k2x 3)= x 1x 4/(k 1x 1-k2x 4)

Namely: (k1-k2) x2x3/(k1x2-k2x3) = (k1-k2) x1x4/(k1x1

So |p|=|q|, that is |OP|=|OQ|.

Step 3 make a brief comment

This small topic mainly examines basic knowledge such as straight lines and ellipses, and examines the ability to analyze and solve problems. It's easy to get started The first question (1) examines the elliptic equation, undetermined coefficient method, coordinate translation and elliptic properties: focus coordinates, eccentricity, and talking with pictures can solve problems, but all problems are the focus.

Problem (Ⅱ) is a typical problem of the positional relationship between a straight line and an ellipse. The formulas to be proved include x 1x2, x 1+x2, X3X4, x3x4 and other symmetric formulas. The formula is symmetrical and beautiful, harmonious and balanced, which makes people easily associate with Vieta's theorem of the relationship between the roots and coefficients of a quadratic equation with one variable, and inspires the method of proving the problem. The most basic ideas and methods of analytic geometry are used here. Two simultaneous binary quadratic equations are solved, and the univariate quadratic equation is obtained by substitution elimination method. Vieta's theorem gives the expressions of separation coefficient about x 1x2, x 1+x2, x3x4, x3+x4, and the purpose of proof is achieved by substituting them into both sides of the equation to be proved. In the process of proof, "the same reason can be obtained" is applied through the structural similarity of two simultaneous equations, and the whole proof process is also pleasing to the eye, feeling the charm of logical proof and smooth and concise expression.

In the proof of problem (ⅲ), the necessary and sufficient conditions of three-point * * * line and the slope formula of straight line passing through two points are used. After solving p and q respectively, |OP|=|OQ| is equivalently converted into p= -q (or p+q=0. ) At this time, the key to analyzing the prerequisite (Ⅱ) and the conclusion to be proved p= -q is to communicate k1x1x 2/(x1+x2) = k2x3x4/(x3+x4) and x1x4/(k/).

Let k 1 x1x 2/(x1+x2) = k2x3x4/(x3+x4) be the formula (1), and both sides are reciprocal.

1/k 1x 2+ 1/k 1x 1 = 1/k2 x4+ 1/k2x 3①'

Let x1x 4/(k1x1-k2x4) =-x2x3/(k1x2-k2x3) be of type ②, and the two sides are reciprocal to obtain k1/x4-k2/.

①' Multiply both sides by k 1 k2 to get the product.

k2/x 1+k2/x2 = k 1/x3+k 1/x4

It is exactly the same as ②'. Here, it is easier to achieve the goal by using the method of simultaneous deformation of two formulas, with analysis, synthesis, thinking and operation. The choice of thinking depends on the observation and association of the characteristics of the formula.

Looking at the characteristics of this problem and the process of solving it, we see the function and power of using algebraic equation but method to deal with geometric problems.

4. Appreciate:

From the above, we can see that the structure and answers of the test questions are pleasing to the eye. At this point, we can't help but ask: how did the test questions come out? What is its background? What are the implications for our math learning and teaching, and for reviewing for the exam in senior three?

There is an interesting theorem about circles:

Butterfly theorem assumes that AB is the chord of circle O and M is the midpoint of AB. The two chords CD, EF, CF and DE that pass through M to make a circle O pass through AB and G respectively. Then MH=MG.

The geometric figure drawn by this theorem is like a fluttering butterfly, so it is called the butterfly theorem (Figure 2).

Look carefully at the picture of the test question 1. Is it like a butterfly dancing on an ellipse?

Yes, very much. The proving process and results of the test questions tell us that the butterfly theorem in the ellipse still holds, and it is proved by analytical method. If the major axis and minor axis of an ellipse are equal, that is, a=b, then the ellipse becomes a circle, and the butterfly theorem in the ellipse becomes the butterfly theorem on the circle. The above proof also applies. Since an ellipse can also be regarded as a circle obtained by "compression transformation", the butterfly theorem on the circle can also be changed into the butterfly theorem on the ellipse by "compression transformation". "Dancing butterflies and dancing ellipses fly away from the flower of mathematics in the college entrance examination." Readers appreciate this. Do you realize the "ingenuity" and good intentions of the experts in mathematics proposition geometry to point out the college entrance examination questions?

[On the "Butterfly on the Ellipse", Academician Zhang Jingzhong made an incisive exposition in the section "Clever Thinking and Wonderful Solution" in his book "The Eyes of Mathematicians", which is a gift for middle school students. Interested readers please refer to P54-59].

Step 5 enlighten

The butterflies flying on the ellipse landed in the Baihua (Grass) Garden of Beijing Mathematics College Entrance Examination, which was delightful. Although there is a background of competition mathematics, affine transformation and famous mathematical problems, here we only use the three-point * * * line problem and slope formula repeatedly mentioned in the textbook to prove it, using the most basic method of analytic geometry. Quan Yi Volume (compulsory) of Plane Analytic Geometry, a textbook for senior high school, mentions three problems of * * * line in several places, such as P 13, exercise 1, problem 14: three points A( 1,-1) are known. Proof: Three points are on a straight line: P 17 Exercise 4: Proof: Three points A, B and C are known. If the slopes of straight line AB and AC are equal, then these three points are on the same straight line; P27 Exercise 2 Question 9: Prove that A (1, 3), B (5, 7) and C (10, 12) are on the same straight line; P47 Review Reference Question 1 and Question 3: Prove by two methods: A (-2, 12), B (1 3) and C (4 4,6) are on the same straight line. You see, the exercises, exercises and review reference questions in the textbook repeatedly mention the proof of three lines, emphasizing the proof in different ways. Why? Did you (teacher, student) pay attention?

In fact, the different proofs of the three-point line can fully mobilize and organize the key basic knowledge of the first chapter of analytic geometry. You can use the basic formula-the distance formula between two points on the plane.

Prove | AC | =| AB ∣+∣ BC ∣; It can also be proved that λ = (x 1+λ x2)/( 1+λ) and Y = (y 1+λ y2)/( 1+λ). The slope formula KP1p 2 = (y2-y1)/(x2-x1) can be used to prove that KAB = KAC；; ; You can also establish the equation f (x, y)=0 of straight line AB, and then verify that the coordinates of point C are suitable for the equation of straight line AB, that is, f(x, y) = 0; You can also use the distance formula from point to line after establishing the equation of straight line AB.

Prove that DC-ab = 0；; You can also prove that s △ABC = 0 by calculating the area of △ ABC. You see, there are five or six solutions to the same problem, of course, the difficulty is high or low. The selection method and optimization method in multiple solutions to one problem are also the embodiment of ability (insight and observation), and the advantages and disadvantages of the methods can be identified by comparison. It is said that after the exam, some top students in key middle schools regret that they can't solve the third question. Some teachers also said that this topic is "too much to finish"! I don't know if readers appreciate this, but can you find out where the crux of the above problem occurs? There are many teachers and students in key middle schools in Beijing, who are dismissive of the exercises in high school mathematics textbooks, and rarely delve into the examples and exercises in the textbooks to seek and discover the internal relationship between knowledge and summarize the principles, ideas and laws of solving problems. Various review materials and dozens of sets of simulated test papers from all over the country make senior three students jump into the sea of questions and find it difficult to extricate themselves. How to talk about quality education and cultivating ability? We should appreciate the important role of "choosing topics and making full use of the nutritional value of topics" in mathematics teaching from the appreciation of Dancing Oval Butterfly, so as to emancipate our minds and boldly abandon the "sea tactics". If students want to jump out of the ocean of problems, teachers should first jump into the ocean of problems, "find pearls from the sea" and understand the true meaning of mathematics education reform. -Pay attention to foundation, understanding, connection and ability.

At first, the perpendicular lines intersecting with O of AD and BC intersect with st respectively (not the perpendicular lines of AD and B), then obviously the triangle AMD is similar to BMC, and then SD= 1/2AD BT= 1/2BC, ∴DS/BT=DM/BM and ≈d =∞. And O, S, X, M and O, t. Y .m is a four-point * * * circle, ∴∠XOM=∠YOM ... Details A 61903481909-30/KLOC-0.