Ceva va ung dung NHOM zero thang 1 nam 2015

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Ceva va ung dung  NHOM zero thang 1 nam 2015

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Nguyễn Thanh Dũng, Chu Van An high school for gifted students, Lang Son province Nguyễn Việt Hà, Lao Cai high school for gifted students Trần Ngọc Thắng, Vinh Phuc high school for gifted students Lê Anh Chung , Bac Kan high school for gifted students Trương Thanh Tùng, Thai Nguyen high school for gifted students Phạm Thị Thu Trang, Le Quy Don high school for gifted students, Dien Bien province Đinh Ngoc Diệp, Ha Giang high school for gifted students Nguyễn Dũng, Nguyen Trai high school for gifted students, Hai Duong province Do Son, 01/23/2015 The lesson of zerogroup Ceva’theorem and its applications A Learning outcomes for Ceva Theorem -Learners are able to explain and express Ceva’theorem and use it in solving geometric problems in VMO, IMO -To understand the content of this theorem and apply it to solve the specific mathematical problems -To recognize that what signs in the mathematical problems can help learners to think of Ceva’theorem - To creat some similar problems B Vocabulary needed for the lecture * Content-obligatory language Keywords which teachers and learners need in order to understand the topic (need written in Boll Font) Word-phrase Triangle Meaning Tam giác Angle Segment Ratio Concurrent (adj) Concurrency (n) Internal bisector, external bisector Altitude tangent The content of lesson Inductive method 1.Phát biểu nội dung định lý Góc Đoạn thẳng Tỉ số đồng quy Đồng quy Phân giác trong, phân giác Đường cao Tiếp xúc Activating prior knowledge + Let M, N, P be the midpoints of the sides BC, CA, AB of a triangle ABC Calculate the expression MB NC PA MC NA PB + Given a triangle ABC Let M, N, P be the feet of the internal angle bisectors Calculate the expression MB NC PA MC NA PB What is the relationship of that results and the concurrency of three lines AM, BN, CP? Theorem: Let ABC be a triangle and D, E, F be points on the lines BC, CA, AB respectively If AD, BE, CF are concurrent (i.e meet at a point P), then AF BD CE = +1.The + sign emphasizes directed segments were used FB DC EA here We can see the content of the theorem in following example: a) Given the triangle ABC, then its medians concur b) Given the triangle ABC, then its internal bisectors concur 2.Áp dụng vào vài ví dụ dễ hiểu, áp dụng trực tiếp lý thuyết Example 1: Let ∆ABC be a triangle and let L, M, and N, be the points of tangency of the inscribed circle to the sides of the triangle opposite vertices A, B, and C, respectively The line segments AL, BM, and CN are concurrent Proof Let ∆ABC be a triangle Construct the inscribed circle to ∆ABC and let L, M, and N, be the points of tangency of the inscribed circle to the sides of the triangle opposite vertices A, B, and C, respectively See Figure ?? Since AM and AN are common external tangent segments to the inscribed circle, they are congruent Similarly, BL and BN are congruent and CL and CM are congruent Hence, AM CL BN AM CL BN = = MC BL AN AM CM BL By Ceva’s Theorem, the segments AL, BM, and CN are concurrent Note: The concurent point of AL, BM, CN is called Gergonne point - Ask students to rewrite the solution in their own thinking (method) Example Let ∆ABC be a triangle and let L, M, and N, be the points of tangency of the excircle to the sides of the triangle opposite vertices A, B, and C, respectively The line segments AL, BM, and CN are concurrent Proof Similar to Example Example Given triangle ABC Prove that: Three altitudes concur 3.Some problems from VMO, IMO -Một số lưu ý sử dụng định lý sai lầm dễ mắc sử dụng định lý? -Những dấu hiệu nhận biết đề: định lý sử dụng hiệu quả…? -… Let ABCDE be a convex pentagon such that ∠BAC = ∠CAD = ∠DAE and ∠ABC = ∠ACD = ∠ADE The diagonals BD and CE meet at P Prove that the line AP bisects the side CD (USA) Solution Let the diagonals AC and BD meet at Q, the diagonals AD and CE meet at R, and let the ray AP meet the side CD at M We want to prove that CM = MD holds The idea is to show that Q and R divide AC and AD in the same ratio, or more precisely AQ AR = QC RD (1) (which is equivalent to QR || CD) The given angle equalities imply that the triangles ABC, ACD and ADE are similar We therefore have: AB AC AD = = AC AD AE Since ∠BAD = ∠BAC + ∠CAD = ∠CAD + ∠DAE = ∠CAE , it follows from AB AD = that the triangles ABD and ACE are also similar Their angle AC AE bisectors in A are AQ and AR, respectively, so that AB AQ = AC AR AB AC AQ AC = = Because , we obtain , which is equivalent to (1) Now AC AD AR AD Ceva’s theorem for the triangle ACD yields AQ CM DR = QC MD RA In view of (1), this reduces to CM = MD , which completes the proof Comment Relation (1) immediately follows from the fact that quadrilaterals ABCD ACDE are similar Home work Problem Connect each vertex of a triangle to the point where its incircle is tangent to the opposite side Prove that the three resulting segments have a common point (It is called Gergonne’s point.) Problem Prove that the altitudes of a triangle have a common point Problem Let ABC be a triangle and let P be a point Lines AP, BP, CP are reflected across the bisectors of angles CAB, ABC, BCA respectively Prove that the resulting lines have a common point (or are parallel) Their common point is called the isogonal conjugate of P Problem Points A1 , B1 , C1 are chosen on sides BC, CA, AB of triangle ABC respectively so that AA1 , BB1 , CC1 are concurrent Let A2 be the reflection of A1 across the midpoint of BC Define B2 , C2 similarly Prove that AA2 , BB2 , CC2 are concurrent (or parallel) Their common point is called the isotomic conjugate of P Problem Points A1 , B1 , C1 are chosen on sides BC, CA, AB of triangle ABC respectively so that AA1 , BB1 , CC1 are concurrent Let A2 , B2 , C2 be the midpoints of sides BC, CA, AB respectively (a) Prove that the lines parallel to AA1 , BB1 , CC1 through A2 , B2 , C2 respectively are concurrent (b) Prove the the lines connecting A2 , B2 , C2 with the midpoints of AA1 , BB1 , CC1 respectively are concurrent Problem Let α, β, γ be angles such that the sum of any two of them does not exceed π Triangles ABC1 , BCA1 , CAB1 are constructed on the interior of triangle ABC such that their angles at A, B, C are α, β, γ respectively Prove that lines AA1 , BB1 , CC1 are concurrent Problem Let the tangents to the circumcircle of triangle ABC at B and C meet at A1 Define points B1 , C1 similarly Prove that AA1 , BB1 , CC1 are concurrent Their common point is called Lemoine’s point Problem Prove that Lemoine’s point of a triangle is the isogonal conjugate of its centroid Problem A circle intersects sides AB, BC, CA of triangle ABC at points C1 and C2 , A1 and A2 , B1 and B2 respectively Prove that lines AA1 , BB1 , CC1 are concurrent iff lines AA2 , BB2 , CC2 are concurrent Problem 10 Let ABCD be a convex quadrilateral inscribed into a circle ω Rays AB and DC meet at P The tangents to ω at A and D meet at M and tangents to ω at B and C meet at N Prove that P, M, N are collinear ... conjugate of P Problem Points A1 , B1 , C1 are chosen on sides BC, CA, AB of triangle ABC respectively so that AA1 , BB1 , CC1 are concurrent Let A2 be the reflection of A1 across the midpoint of BC... triangle ABC at points C1 and C2 , A1 and A2 , B1 and B2 respectively Prove that lines AA1 , BB1 , CC1 are concurrent iff lines AA2 , BB2 , CC2 are concurrent Problem 10 Let ABCD be a convex... called the isotomic conjugate of P Problem Points A1 , B1 , C1 are chosen on sides BC, CA, AB of triangle ABC respectively so that AA1 , BB1 , CC1 are concurrent Let A2 , B2 , C2 be the midpoints

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