The butterfly theorem and its proof

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Butterfly the ForexrebateforExness The butterfly theorem was first publ cashbackforexexness"https://www.luckysteeltraders.com">forexrebateshed as a problem for proof forexcashrebate a popular magazine "Mens Diary" in 1815 because of its strange geometric figure, resembling a butterfly, so named, theorem content: circle O in the middle point of the string PQ M, through the point M any cashback forex chords AB, CD, chord AD Forex rebate for Exness BC intersect PQ in X, Y, then M is the middle point of XY There have been many beautiful and strange solutions, the earliest of which should be the first one given by Horner in 1815 as for the proof of elementary mathematics, in foreign sources, it is generally believed that it was first proposed by a secondary school teacher, Sturgeon, which gave the area proof, which applied the area formula: S = 1/2BCSINA1985, in Henan Province, "Mathematics Teacher In 1985, in Henan Province, "Mathematics Teacher" inaugural issue, Comrade Du Xilu to "plane geometry in the name of the problem and its wonderful solution" as the title, contains an article to the domestic introduction of the butterfly theorem, from the butterfly theorem in the land of God everywhere spread here to introduce a more simple elementary mathematical proof Proof: over the center of the circle O for AD and BC vertical line, the vertical foot for S, T, connecting OX, OY, OM, SM, MT ∵ △AMD ∽ △CMB&there4 AM/CM=AD/BC ∵ SD=1/2AD, BT=1/2BC∴AM/CM=AS/CT also ∵∠A=∠C∴△AMS∽△CMT∴∠MSX=∠MTY∵∠OMX=∠OSX=90& deg;∴∠OMX+∠OSX=180°∴O,S,X,M four points co-circle Similarly, O,T,Y,M four points co-circle∴∠MTY=∠MOY,∠MSX=∠MOX∴∠MOX =∠MOY, ∵OM⊥PQ∴XM=YM This theorem also holds in the ellipse, as shown in Figure 1. The long axis A1 and A2 of the ellipse are parallel to the x-axis, the short axis B1B2 is on the y-axis, and the center is M(o, r) (b〉r〉0) (I) Write the equation of the ellipse and find the coordinates of the focal point and the eccentricity of the ellipse; (II) The line y=k1x intersects the ellipse in two points C (x1, y1), D (x2, y2) (y2〉0); line y = k2x intersect the ellipse in two points G (x3, y3), H (x4, y4) (y4〉0) Proof: k1x1x2/(x1 + x2) = k2x3x4/(x3 + x4) (III) for (II) C, D, G, H, let CH intersect the X-axis at the point P, GD intersect the X-axis at the point Q Proof: |OP|=|OQ| (The proof process does not consider the case where CH or GD is perpendicular to the X-axis) 2. Answer: The reference answers given by experts from the Admissions Office of the Beijing Education Examination Authority in the published "Compilation of Answers to the 2003 National Unified Examination for the Admissions to Colleges and Universities" are as follows: (18) This subtopic mainly examines the basic knowledge of the straight line and the ellipse, and tests the analytical problem and problem-solving skills full 15 points (Ⅰ) solution: the ellipse equation is x2/a2 + (y - r)2/b2 = 1 focal point coordinates x substituted into the ellipse equation, get b2x2 + a2 (k1x - r)2 = a2b2, ? (II) Proof: The equation of the line CD y=k is rectified to give (b2+a2k12)x2-2k1a2rx+(a2r2-a2b2)=0 According to Vedas theorem, we get x1+x2=2k1a2r/(b2+a2k12), x1x2=(a2r2-a2b2)/(b2+a2k12), so x1x2/(x1+x2)= (r2-b2)/2k1r① Substitute the equation of the line GH y=k2x into the equation of the ellipse and similarly we get x3x4/(x3+x4)=(r2-b2)/2k2r② From ①, ② we get k1x1x2/(x1+x2)=(r2-b2/2r=k2x3x4/(x3+x4) so the conclusion holds (III) Proof: Let the point P(p, o) and the point Q (q, o) by C, P, H common line, we get (x1-p)/(x4-p) = k1x1/k2x4 solve P = (k1-k2)x1x4/(k1x1-k2x4) by D, Q, G common line, the same can be obtained q = (k1-k2)x2x3/(k1x2-k2x3) by k1x1x2/(x1+x2) = k2x3x4/(x3+ x4), deformation gives: x2x3/(k1x2-k2x3)=x1x4/(k1x1-k2x4) that is: (k1-k2)x2x3/(k1x2-k2x3)=(k1-k2)x1x4/(k1x1-k2x4) so |p|=|q|, that is, |OP|=|OQ| 3. Brief comments This subtopic mainly examines the line and the ellipse and other basic knowledge, test the ability to analyze and solve problems test questions easy to start, the first (Ⅰ) question test the elliptic equation, the method of coefficients to be determined, coordinate translation and elliptic properties: focal point coordinates, eccentricity, look at the diagram to speak to solve the problem, but the test is the focus of the content of the first (Ⅱ) question is a typical line and the position of the ellipse problem to be proved in the equation contains x1x2, x1 + x2, x3x4, x3 + x4 The symmetrical structure of the equation is symmetrical and beautiful, harmonious and balanced, which makes it easy to recall Vedders theorem on the relationship between the roots and coefficients of quadratic equations, and reveals the idea of proving the problem. The most fundamental idea of analytic geometry and the most fundamental method are used here to solve two systems of quadratic equations, using the substitution elimination method to obtain a quadratic equation, and separating the coefficients using Vedders theorem to give the relationship between x1x2, x1 + x2, x3x4, x3 + x4 expressions, and then substitute the two sides of the formula to be proved that the purpose of the proof of the process of proof, the similarity of the structure of the two systems of conjunctive equations using the same reasoning can be obtained, the entire proof process is also pleasing to the eye, feel the charm of the smooth, simple beauty of the logical proof and expression The proof of (Ⅲ) used the sufficient condition of the common line of three points The slope formula of the line through two points is used, and after solving p, q, respectively, |OP|=|OQ| is equivalently transformed into p=-q (or p+q=0) At this point, analyzing the precondition (II) and the conclusion to be proved p=-q, the key is to communicate k1x1x2/(x1+x2)=k2x3x4/(x3+x4) and x1x4/(k1x1-k2x4)=- If the two equations are deformed as follows, it is obvious that the idea is smooth and natural Let: k1x1x2/(x1+x2)=k2x3x4/(x3+x4) be ① The two sides of the same inverse, 1/k1x2+1/k1x1=1/k2x4+1/k2x3①’ Set: x1x4/(k1x1-k2x4)=-x2x3/(k1x2-k2x3) as ②, the two sides of the same inverse, k1/x4-k2/x1=k2/x2-k1/x3, shift the term to get k2/x1+k2/x2 = k1/x3+k1/x4 ②’multiply ①’both sides together by k1k2, that is, to get k2/x1+k2/x2=k1/x3+k1/x4 It is exactly the same as ②’here using the method of simultaneous deformation of the two equations can be easier to achieve the purpose, there is analysis, synthesis, thinking, the choice of arithmetic ideas depends on the characteristics of the equation Observation and association Looking at the characteristics of this question and the answer process, we see the role and power of using algebraic equations but the method of dealing with geometric problems 4. What is its background? What does it tell us about mathematics learning and teaching, senior review and exam preparation? There is an interesting theorem about circles: Butterfly Theorem Let AB be the chord of circle O, M be the midpoint of AB, and M be the two chords CD and EF of circle O. CF and DE intersect AB at H and G respectively, then MH = MG. The geometric diagram drawn by this theorem resembles a fluttering butterfly, so it is called Butterfly Theorem (Figure 2).　　Like, and like the proof process and the results of the test question tell us that the butterfly theorem in the ellipse still holds, and is proved by the analytical method if the long axis and short axis of the 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, and the above proof applies as the ellipse can also be regarded as a circle by compression transformation, so the butterfly theorem on the circle Theorem by compression transformation can also become ellipse butterfly theorem fluttering butterfly dance ellipse, fly down the college entrance examination mathematics flower readers appreciate to this, whether you appreciate the mathematical proposition geometry experts to design the college entrance examination questions and good intentions?　　The butterfly on the ellipse, Zhang Jingzhong academician in his book "A Mathematicians Vision" gift to secondary school students has a subtle discussion, interested readers refer to the book P54-59] 5. Background, however, here to prove it, but only to the textbook repeatedly mentioned in the three-point common line problem and the slope formula, using the most basic methods of analytic geometry senior high school textbook "plane analytic geometry" all a book (compulsory) several references to the three-point common line problem, such as P13 exercise 1, problem 14: known three points A (1, -1), B (3, 3), C (4, 5) to prove: three points in a straight line P17 exercise 4: prove: three points A, B and C are known, if the slopes of the lines AB and AC are equal, then these three points are on the same line; P27 exercise II, question 9: prove that the three points A (1, 3), B (5, 7), C (10, 12) are on the same line; P47 review reference question I, question 3: prove in two ways: three points A (-2, 12), B (1, 3 ), C(4, -6) on the same line You see, the exercises, problems, and review reference questions in the textbook repeatedly mention the proof that the three points are co-linear, and emphasize the use of different methods to prove why? Do you (teachers, students) pay attention to it?　　In fact, the different proofs of the three-point common line can fully mobilize and organize the key basic knowledge of the first chapter of analytic geometry You can use the basic formula for the distance between two points on the plane to prove that |AC|=|AB∣+∣BC∣; you can also apply the fixed-point formula x=(x1+λx2)/(1+λ), y=(y1+& lambda;y2)/(1+λ) to prove λ=(x1-x)/(x-x2)=(y1-y)/(y-y2); you can use the formula for the slope of a line past two points, Kp1p2=(y2-y1)/(x2-x1), to prove that KAB=KAC; you can also first establish the equation of the line AB f(x, y) = 0, and then verify that the coordinates of the point C fits the equation of the line AB that f(x, y) = 0; you can also use the distance formula from the point to the line after establishing the equation of the line AB to prove that dc-AB = 0; you can also calculate the area of △ABC, to prove that S△ABC = 0 you see, there are five or six ways to solve the same problem, of course, there are high and low difficulty a problem in the choice of methods, multiple solutions. Optimization method is also a reflection of the ability (insight, observation), from the comparison can identify the advantages and disadvantages of the method It is said that the test down, some key secondary school students of the top of their own failed to answer the (Ⅲ) question is very remorseful, some teachers also said that the topic is too large arithmetic difficult to complete! I wonder if the readers can find out where the crux of the above problem lies. Beijing has many key secondary school teachers and students, high school mathematics textbook exercises do not care, rarely to study the textbook examples, exercises, to seek and find the inner connection between knowledge, to summarize the principles of problem solving, ideas and laws of a variety of review materials, dozens of sets of dozens of sets of mock test papers around the world, so that senior students jump into the sea of questions to do dizzy and difficult to extricate themselves, which where to talk about quality education The first thing you need to do is to learn how to use your own skills. We should appreciate the fluttering elliptical butterflies in the heart to appreciate the selected topics to make full use of the nutritional value of the topic in mathematics teaching and review of the important role, thus liberating the mind, bravely and boldly abandon the sea of questions tactics and to make students jump out of the sea of questions, teachers must first jump into the sea of questions, the sea of questions to explore the pearl, to understand the true meaning of mathematics education reform focus on the foundation, focus on understanding, focus on the connection, focus on ability Supplementary. Butterfly theorem in chaos theory A branch of mathematics is chaos theory Chaos theory has a very famous theorem butterfly theorem It is said that some of the slightest factors, can be in a complex environment, causing monstrous waves, as if a butterfly in the southern hemisphere of the Earth gently flapping beautiful wings, the tiny airflow, has been enough to cause hurricanes and tsunamis in the northern hemisphere and how can we track the tip of the leaf a slight tremor? So the economy and weather are unpredictable, just as life can not predict the extension of the butterfly theorem as in Figure I, is the butterfly theorem, there is a conclusion EP = PF; as in Figure II, is the evolution of the butterfly theorem, the point P, Q, R, S whether there is also a certain relationship? So over the center of the circle O of the two concentric circles in the chord of the midpoint M made two lines intersecting the circle in A, B, C, D, E, F, G, H, even AF, BE, CH, DG, respectively, intersecting the chord in the point P, Q, R, S, then there is an equation: established proof: the introduction, as shown on the right, there are conclusions by and sine theorem can be obtained: the original conclusion for OM1AD in M1, OM2EH in M2, so that MA-MD = MB-MC = 2MM1 = 2Msin; MH-ME = MG-MF = 2MM2 = 2Msin and MA*MD = ME*MH, MB*MC = MF*MG, substituted into the above formula, and so the original formula is established proof of the Butterfly Theorem Butterfly Theorem 1815, Western Europe, "Mens Journal" magazine published a difficult problem solved, the topic is as follows: over the chord of the circle ABs midpoint M, cited any two chords CD, EF, connecting ED, CF respectively cross AB in P, Q two points to find PM = QM (see figure) because of the shape of the cool butterfly, the proposition is known as the butterfly theorem A value of four years are unanswered July 1819 quadrilateral butterfly theorem if a quadrilateral diagonal bisects another diagonal, then the two lines through its intersection to the intersection of the four sides (neighboring sides) If a quadrilateral has a diagonal bisecting another diagonal, then the two lines passing through its intersection, with the intersection of the four sides (neighboring sides), are connected at equal distances from the two intersections of the bisected diagonal to the focal point of the diagonal. S△ACD*S△PGQ/S△SGR=AB*BD/BP*BQ*SC*CR/AC*DC*PG*QG/RG*SG=S△ABC*S△BCD/S△BCP*BCQ*S△BCS*S△BCR/S△ABC*S△BCD*S△BCP*S△BCQ/S△BCR*S△BCS=1EG/BG = GF/CGEG = GF, a self-taught high school teacher Horner gave the first answer, but tedious and difficult to understand but from 1819 onwards, people sought new proofs that were concise and easy to understand until 1973, when the high school teacher Stirwin gave very elementary proofs, after which many new proofs were published Stirwins proof: let MQ = x, MA = y, AM = BM = a, ∠E = &alpha ang;C=α, ∠D=∠F=β, ∠BMF=∠AME=δ, ∠DMA=∠CMB=γ with △1, △2, △3, △4 representing △PME, △QMC, △PDM, △QFM area respectively, then △1/△2* △2/△3*△3/△4*△4/△1=(EP*EMsinα/CQ*CMsimα)*(MQ*CMsinγ/PM*MDsinγ)*(PD*MDsinβ/MF*QFsinβ)*(MQ*MFsin& delta;/MP*MEsinδ)=(EP*PD*MQ*MQ)/(CQ*FQ*MP*MP)=1 by the intersecting chord theorem EP*PD=AP*PB=(a-y)(a+y)CQ*FQ=BQ*QA=(a-x)(a+x)(EP*PD*MQ*MQ)=(CQ*FQ*MP*MP) The solution is (aa-yy)xx=(aa-xx)yy, which gives x=y that PM=MQ, which proves Bi Since, the elliptical plane is the oblique section of the positive column surface as shown in the figure the bottom of the cylinder is the projection of the ellipse, so the butterfly theorem holds for the ellipse What is the butterfly theorem? How to prove the Butterfly Theorem? Butterfly Theorem: In circle O, CD, EF for any two chords over the midpoint of the AB chord M, connecting CF, DE respectively cross AB in H, K, then there is MK = MH known: as shown in Figure 8-30 B in circle O, CD, EF for any two chords over the midpoint of the AB chord M, connecting CF, DE respectively cross AB in H, K proof: MK = MH idea 2: according to the symmetry of the circle Make the chord centroid; from the triangle similar and then deduce the triangle similar, from the four points of the common circle, deduce ∠ MOH = ∠ MOK is the key; proof: over O as OS ⊥ FC, OT ⊥ DE, even OH, OK, SM, MT, and then even MO ∵ AM = MB; ∴ OM ⊥ AB, ∠ AMO=∠BMO=90°;In △FCM and △DEM;∠CMF=∠DME;(equal to the top angle);∠MFC=∠MDE;(equal arc to equal circumference angle)∴△FCM∽△DEM;(AA)∴∠FCM=∠ DEM; ∵ FS=SC=½FC;DT=TE=½DE;∴FS/FC=TD/ED;∵ FC/ED=FM/MD∴FS/FM=TD/MD in △FSM and △DTM;∠MFS=∠MDT;(equal arcs to equal circumferential angles); FS/FM=TD/MD;∴△FSM∽△DTM;(SAS)∠FSM=∠DTM;∠MSH=∠MTK; ∵∠AMO=90°,∠HSO=90°;O,S,H,M four points co-circle;∴&ang ;MSH=∠MOH;∵∠BMO=90°,∠KTO=90°;O,T,K,M four points co-circle;∴∠MTK=∠MOK;∴∠MOH=∠MOK;In △MOH and △MOK;∠ MOH=∠MOK;MO=MO;∠AMO=∠BMO=90°;∴△MOHì△MOK;(ASA)∴MH=MK Conclusion: make chordal centroid is the most effective auxiliary line, the starting point of this proof is to prove that △HOK is isosceles triangle, using the trilinear The proposition has many other proofs, so I will not repeat them.

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