Ceva va ung dung NHOM zero thang 1 nam 2015

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 gi

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)

Angle Góc

Internal bisector, external bisector Phân giác trong, phân giác ngoài

The content of lesson

Inductive method

1.Phát biểu nội dung định lý

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 .

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

FB DC EA= + .The + sign emphasizes directed segments were used

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 1 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 2 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 1

Example 3 Given triangle ABC Prove that: Three altitudes concur

3.Some problems from VMO, IMO

-Một số lưu ý khi sử dụng định lý này hoặc những sai lầm dễ mắc khi

sử dụng định lý?

-Những dấu hiệu nhận biết trong đề: định lý này sẽ được sử dụng hiệu quả…?

-….

1 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

(which is equivalent to QR || CD) The given angle equalities imply that the triangles ABC, ACD and ADE are similar We therefore have:

AC = AD = AE

Since ∠BAD= ∠BAC+ ∠CAD= ∠CAD+ ∠DAE= ∠CAE, it follows from

AC = AE that the triangles ABD and ACE are also similar Their angle

bisectors in A are AQ and AR, respectively, so that

AC = AR .

Because AC AB = AC AD , we obtain AQ AC

AR = AD , which is equivalent to (1) Now

Ceva’s theorem for the triangle ACD yields

AQ CM DR

In view of (1), this reduces to CM =MD, which completes the proof

Comment Relation (1) immediately follows from the fact that

quadrilaterals ABCD và ACDE are similar

Home work Problem 1 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 2 Prove that the altitudes of a triangle have a common point Problem 3 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 4 Points A B C1 , , 1 1 are chosen on sides BC, CA, AB of triangle ABC respectively so that AA BB CC1 , 1 , 1 are concurrent Let A2 be the

reflection of A1 across the midpoint of BC Define B C2 , 2 similarly Prove that AA BB CC2 , 2 , 2 are concurrent (or parallel) Their common point is called the isotomic conjugate of P

Problem 5 Points A B C1 , , 1 1 are chosen on sides BC, CA, AB of triangle ABC respectively so that AA BB CC1 , 1 , 1 are concurrent Let A B C2 , , 2 2 be the midpoints of sides BC, CA, AB respectively

(a) Prove that the lines parallel to AA BB CC1 , 1 , 1 through A B C2 , , 2 2

respectively are concurrent

(b) Prove the the lines connecting A B C2 , , 2 2with the midpoints of

1 , 1 , 1

AA BB CC respectively are concurrent

Problem 6 Let α, β, γ be angles such that the sum of any two of them

does not exceed π Triangles ABC BCA CAB1 , 1 , 1 are constructed on the interior of triangle ABC such that their angles at A, B, C are α, β, γ

respectively Prove that lines AA BB CC1 , 1 , 1 are concurrent

Problem 7 Let the tangents to the circumcircle of triangle ABC at B

and C meet at A1 Define points B C1 , 1similarly Prove that AA BB CC1 , 1 , 1

are concurrent Their common point is called Lemoine’s point

Problem 8 Prove that Lemoine’s point of a triangle is the isogonal

conjugate of its centroid

Problem 9 A circle intersects sides AB, BC, CA of triangle ABC at

points C1 and C2, A1 and A2, B1 and B2respectively Prove that lines

1 , 1 , 1

AA BB CC are concurrent iff lines AA BB CC2 , 2 , 2are 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