Đề tài geometric inequality and related issues (bất đẳng thức hình học và các vấn đề liên quan)

Trang 1 TRƯỜNG ĐẠI HỌC VINHTRƯỜNG SƯ PHẠMKHOA TỐN HỌCĐỒ ÁN MƠN HỌCTIẾNG ANH CHUN NGÀNHENGLISH FOR MATHEMATICSTÊN ĐỀ TÀI: GEOMETRIC INEQUALITY AND RELATEDISSUESBẤT ĐẲNG THỨC HÌNH HỌC VÀ C

TRƯỜNG ĐẠI HỌC VINH TRƯỜNG SƯ PHẠM KHOA TOÁN HỌC

ĐỒ ÁN MÔN HỌC TIẾNG ANH CHUYÊN NGÀNH (ENGLISH FOR MATHEMATICS)

TÊN ĐỀ TÀI: GEOMETRIC INEQUALITY AND RELATED

ISSUES (BẤT ĐẲNG THỨC HÌNH HỌC VÀ CÁC VẤN ĐỀ LIÊN QUAN)

2 Dương Thị Xuân Tình 2257140209

3 Nguyễn Thị Huyền Trang 2257140209

4 Đào Thị Thanh Huyền 2257140209

Nghệ An, 2023

5 Trigonal lines Đường phân giác

7 Straight section Đoạn thẳng

9 Median line Đường trung tuyến

22 Isosceles triagle Tam giác cân

23 Right triagle Tam giác vuông

24 Perpendicular lines Đường vuông góc

Geometric Inequality

Group 11 March 12, 2024

Abstract

In this work we investigate about geometric inequalities and related problems (plane geometry) Geometric inequalities are among the difficult problems that make high school students face many difficulties, even good students The problems of ge-ometric inequalities presented in this thesis can be divided into three main contents: Triangle inequality,quadrangular inequality,Circle inequalities.

In the history of world mathematics, geometric inequalities are one of the earliest problems The inequality was first described by Euclid around the 4th century BC, Eu-clid states that in 1 triangle the total length of 2 sides is always greater than the other Since then, the history of mathematics in the world has had many other inequalities related to geometry such as Cauchy’s inequality, AM-GM’s inequality, Ptolemy’s in-equality,

Problems of geometric inequalities are of the type of difficult problems, making

it difficult for high school students to encounter problems of this type It is a very important part of geometry and the knowledge of geometric inequalities also enriches the scope of application of mathematics Compared to algebraic inequalities, geometric inequalities have not received much attention One of the reasons is the way to solve problems about geometric inequalities require both logical thinking and the application

of pure algebraic

This project introduces a number of geometric inequalities from basic to advanced and frequently appears in common problems For example,

• Inequalities in triangle

• Quadrangle inequalities

• Circle inequalities

The project "Some geometric inequalities" includes introduction, new words, 3 content chapters, 1 number of exercises, conclusions and references

This project was completed by a group of 11 class LT 01 at Vinh University with the guidance of Dr Nguyen Huy Chieu The group would like to thank you for your support and guidance

I.Inequalities in triangles and quadrangles

This section presents the inequalities in triangles and quadrangles from fundamen-tal to advanced The content is mainly formed from data [1], [2], [3] and [4]

The symbol △ABC is the triangle ABC with vertices A,B,C We denote the magni-tude of the corresponding angles of vertices A,B,C as A,B,C respectively.

The length of the sides of the triangle: BC=a,CA=b,AB=c.

Half circumference of the triangle: p = a+b+C2

High line with edges: ha, hb, hc

Median line with edges: ma, mb, mc

Trigonal line with edges: la, lb, lc

Radius of outer circle and inline circle: R and r.

Radius of the circle equal to the edges: ra, rb, rc

Triangular area ABC: S, SABCand [ABC].

In order to solve problems of geometric inequalities, we first need to equip our-selves with basic knowledge that is basic algebraic inequalities and equalities, inequali-ties in triangles

1.1 Basic algebraic inequalities

Theorem 1.1 (Inequalities AM-GM) Suppose a1, a2, , an is non-negative real num-bers, then

a1+ a2+ + an

a1a2 an

Equality occurs if and only if a1 = a2= = an

Corollary 1.1 For all positive numbers a1, a2, , an we have

n

√

a1a2 an ≥ 1 n

a1+a1

2+ +a1

n

Equality occurs if and only if a1 = a2= = an

Corollary 1.2 For all positive numbers a1, a2, , an we have

1

a1+

1

a2+ +

1

an

2

a1+ a2+ + an

Equality occurs if and only if a1 = a2= = an

Corollary 1.3 For every non-negative number a1, a2, , an and m = 1, 2, we have

am1 + am2 + + amn

a1+ a2+ + an

n

 m

Equality occurs if and only if a1 = a2= = an

1

Theorem 1.2 (Inequalities Cauchy - Schwarz) For 2 sequences of real numbers a1, a2, , an

and b1, b2, , bn Then

(a1b1+ a2b2+ + anbn)2 ≤a21+ a22+ + a2n b21+ b22+ + b2n

Equality occurs if and only if a1

b1 = a2

b2 = = an

bn

Theorem 1.3 (Inequalities Nesbitt) For a, b, c is a real numbers The following

inequal-ity is always true

a

b+ c+

b

c+ a+

c

a+ b≥ 3

2.

Equality occurs if and only if a = b = c.

1.2 Basic equalities and inequalities in triangles

1.2.1 Basic equalities in triangles

Theorem 1.4 (Law of sine) In triangle ABC, we have

a sin A =

b sin B =

c sinC = 2R

Theorem 1.5 (Law of cosin) In triangle ABC, we have

a2= b2+ c2− 2bc cos A

b2= c2+ a2− 2ca cos B

c2 = a2+ b2− 2ab cosC

1.2.2 Basic inequalities in triangles

Theorem 1.6 (Inequalities in triangles) In triangle ABC we have

|b − c| < a < b + c, |c − a| < b < c + a, |a − b| < c < a + b

Theorem 1.7 (Elementary trigonometric inequalities) For every triangle ABC we

al-ways have the following inequalities

sin A + sin B + sinC ≤ 3

√ 3

2 , cos A + cos B + cosC ≤ 3

2, cos A cos B cosC ≤ 1

8, tan A + tan b + tanC ≥ 3√

3, cot A + cot B + cotC ≥√

3

2

1.3 Geometry inequality and related issues

Triangle is the simplest of polygons, any polygon can be divited into triangles and use it’s properties Therefore, learning inequalities in triangle will be helpful in solving inqualities in polyfons First, we study the following basic inqualities:

1.3.1 Inequality of side lengths

Theorem 1.8 For two circles whose radius are R and R’ (R ≥ R′) , respectively, the distance between their centers is equal to d The necessary and sufficient condition for two circles to intersect is R − R′≤ d ≤ R + R′

Proof. We notice that the two circles are outside each other (Fig A) so we have R +

R′ < d Tf two circles contain each other (Fig B) then we have d < R − R′ If two

circles intersect at a point M, then unconsciously the three sides of the triangle OO’M, with O and O’ being the centers of the circle of radius R and R’ respectively, we have

R− R′≤ d ≤ R + R′

Conversely, if R − R′ ≤ d ≤ R + R′ then the two given circles cannot be apart or contain each other Therefore, they can only intersect

Theorem 1.9 The positive numbers a,b,c are the lengths of the three sides of a triangle

if and only if a + b > c, b + c > a, c + a > b.

Proof If a,b,c is the length of the 3 sides of the triangle, then according to the inequality

of the 3 sides of the triangle we have a + b > c, b + c > a, c + a > b

Conversely, if there are a,b and c are 3 positive real numbers satisfied a + b >

c, b + c > a, c + a > b, then we can choose two points A and B on the plane separated

by some c Take A and B as the center, the corresponding radius is A and B From the

unconscious a + b > c, b + c > a, c + a > b we have |a − b| < c < a + b According to

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theorem 1.8, two circles of the center A and B must intersect at a point C So a,b,c is the length of the sides of the triangle ABC in the above way.

Theorem 1.10 For triangle ABC and the ponit M is in triangle Then we have MB +

MC< AB + AC

Proof Extending BM towards M cuts AC at point N According to the theorem 1.9, we

have

MB+ MC < MB + MN + NC

= BN + NC < AB + AN + NC

= AB + AC

Problem 1.1 Let M be a point in the triangle ABC Prove that:

p< MA + MB + MC < 2p

with p is the half-circumference of the triangle ABC.

Solve Apply theorem 1.9 to triangles MAB,MAC and MCA we have AB < MA +

MB, BC < MB + MC,CA < MC + MA Add the three inequalities above and divide both sides by 2 we have p < MA + MB + MC

Conversely, according to Theorem 1.10 we have MA+MB < CA+CB, MB+MC <

AB+ AC, MC + MA < BC + BA Add the three inequalities above and divide both sides

by 2 we have AM + BM +CM < 2p

1.4 Inequalities born from geometric formulations

Theorem 1.11 (Euler’s formula) Let R and r be the radius of the circumcircle and the

incircle of triangle ABC, respectively, d is the distance between the centers of those two circles We have

d2= R2− 2Rr

4

Proof Let O and I be the centers of the circumcircle and incircle circles of triangle ABC, respectively Know that the circumcircle of triangle BCI has center D as the midpoint of arc BC Let M be the midpoint of BC and Q be the projection of I on OD Then

OB2− OI2= OB2− DB2+ DI2− OI2

= OM2− MD2+ DQ2− QO2

= (MO + DM)(MO − DM) + (DQ + QO)(DQQO)

= DO(MO − DM + DQ + OQ) = R(2MQ) = 2Rr

So OI2 = R2− 2Rr, mean d2 = R2− 2Rr

Corollary 1.4 (Euler’s inequality) Symbols R,r are the radius of the circumcircle and

the radius of the inscribed circle of triangle ABC, respectively Then

R≥ 2r

Equality occurs if and only if triangle ABC is equilateral.

Problem 1.2 Let the triangle ABC where R is the radius of the circumcircle, r is the

radius of the inscribed circle and p is the half- circumference of the triangle ABC Prove that r ≤ p

3 √

3 ≤ R

2

5

Solve. We have [ABC] = abc4R = pr, inferred 2p = a + b + c ≥ 3√3 abc= 3√3

4Rr p Therefore 8p3≥ 27(4Rr p) ≥ 27(8r2p), because R ≥ 2r So p ≥ 3√3r

Second inequality, p

3 √

3 ≥ R

2 is equivalent to a + b + c ≤ 3√3R Using the law of sine, this inequality is equivalent to sin A + sin B + sinC ≤ 3

√ 3

2 This inequality is true because funtion f (x) = sin x is a convex function above (0, π), therefore sin A+sin B+sinC3 ≤ sinA+B+C3 = sin 60 =

√ 3

2

Theorem 1.12 (Leibniz’s inequality) Let the triangle ABC with the lengths of sides

a,b,c (O,R) is the circumcircle of the triangle Suppose G is the focus and (O,R) is the triangular extrinsic circle Then

OG2= R2−1

9



a2+ b2+ c2

Corollary 1.5 (Leibniz’s inequality) Let the triangle ABC with the lengths of sides a,b,c.

(O,R) is the circumcircle of the triangle We have the following inequality

9R2≥ a2+ b2+ c2

Equality occuss if and only if O is the centroid of triangle ABC.

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2 Bibliography

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