Geometric Inequalities Marathon - The First 100 Problems and Solutions

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Geometric Inequalities Marathon The First 100 Problems and Solutions

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Titu Andreescu Old and new

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Secrets in Inequalities Vol II

Algebraic Inequalities

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Geometric Inequalities Marathon



1

The First 100 Problems and Solutions

1

Con Contrib tribut utor orss

Typese ypesett ttin ing g and and Edit Editin ing g

Memb Member erss of Math Mathli link nkss

Same Samerr Seraj Seraj (BigS (BigSam ams) s)

Preface

On Wednesday, April 20, 2011, at 8:00 PM, I was inspired by the existing Mathlinks marathons to c a marathon on Geometric Inequalities - the fusion of the beautiful worlds of Geometry and Multivar Inequalities. It was the result of the need for expository material on GI techniques, such as the crucial which were well-explored by only a small fraction of the community. Four months later, the thread has 100 problems problems with full solutions, solutions, and not a single pending pending problem. On Friday Friday,, August August 26, 2011 PM, I locked the thread indefinitely with the following post:

The reason reason is that most of the known techniques techniques have been been displaye displayed, d, which was my goal. goal. Recent ecent pr are tending to to be similar to old ones or they require methods that few are capable of utilizing at this Until the community is ready for a new wave of more diffcult GI, and until more of these new generatio have been distributed to the public (through journals, articles, books, internet, etc.), this topic will re locked.

This collection is a tribute to our hard work over the last few months, but, more importantly, it is a so of creative problems for future students of GI. My own abilities have increased at least several fold sinc exposure to the ideas behind these problems, and all those who strive to find proofs independently wi themselv themselves es ready to tackle tackle nearly any geometric geometric inequalit inequality y on an olympiad olympiad or competition. competition.

The following f ollowing document is dedicated to my friends Constantin Mateescu and R´ eda eda Afare (Thalesma and the pioneers Panagiote Ligouras and Virgil Nicula, all four of whom have contributed  much t Read Free Foron 30this Days Sign up to vote title evolution of GI through the collection and creation of GI on Mathlinks.  Not useful  Useful  anytime. The file ma may y be distri distribut buted ed physi physical cally ly or electr electroni onical cally ly,, in whole whole orCancel in part, par t, but for and only for Special offer for students: Only $4.99/month. commercia comm erciall purposes. purposes. References References to problems problems or solutions solutions should credit credit the corresponding corresponding authors.

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Notation

For a

ABC :

• Let AB = c, BC = a, CA = b be the sides of ABC . • Let A = m (∠BAC ), B = m (∠ABC ), C = m (∠BC A) be measures of the angles of ABC . • Let ∆ be the area of ABC . • Let P be any point inside ABC , and let Q be an arbitr arbitrary ary point point in the plane. plane. Let the ce through P and A, B , C intersect a, b, c at P a , P b , P c respectively.

• Let the distance from P to a, b, c, extended if necessary, be d , d , d respectively. • Let arbitrary cevians issued from A, B, C be d, e, f respectively. • Let the semiperimet semiperimeter, er, inradius, and circumradi circumradius us be s, r, R respectively. • Let the heights issued from A, B , C be h , h , h respectively, which meet at the orthocenter • Let the feet of the perpendiculars from H to BC , CA , AB be H , H , H respectively. • Let the medians issued from A, B, C be m , m , m respectively, which meet at the centroid • Let the midpoints of A, B , C be M , M , M respectively. • Let the internal angle bisectors issued from A, B, C be l , l , l respectively, which meet at the inc a

a

b

b

c

c

a

a

a

b

b

b

c

c

c

a

b

c

I , and intersect their corresponding opposite sides at La , Lb , Lc respectively.

• Let the feet of the perpendiculars from I to BC , CA , AB be Γ , Γ , Γ respectively. • Let the centers centers of the excircles tangent tangent to BC , CA , AB be I , I , I respectively, and the excirc a

a

b

b

c

c

tangent to BC , CA , AB at E a , E b , E c .

• Let the radii of the excircles tangent to BC , CA , AB be r , r , r respectively. • Let semester the symmedians issued fromScribd A, B , C be s , s , s respectively, which meet at the Lemoine Master your with  a

a

b

b

c

c

respectively. S , and intersect their corresponding opposite sides at S a ,Read S b , S cFree Foron 30this Days Sign up to vote title

& The New Times Not  UsefulA, B, C • LetYork Γ be the Gergonne Point, and the Gergonne cevians through beuseful g ,g ,g Cancel anytime.

a

b

c



respectiv

Special offer for students: Only $4.99/month. Let N be the Nagel Point, and the Nagel cevians through A, B , C be na , nb , nc respectively.

•

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Prob Probllems 1. For

ABC , prove that R ≥ 2r. (Euler’s Inequality) Inequality) 2. For ABC , prove that AB > P A.





bc + ca ABC , prove that ab +4∆ ≥ s(s 1− a) . 2 √ 4. For ABC , prove that r(4R + r) ≥ 3∆. B − C ≥ 2r . 5. For ABC , prove that cos

3. For



2

6. For

ABC , prove that



12(R2

 R

− Rr + r2) ≥



AI

≥ 6r .

7. A circle with center center I is inscribed inside quadrilateral ABCD . Prove that 8. For

ABC , prove that 9R2 ≥



Inequality) a2 . (Leibniz’s Inequality)



AB

≥

9. Prove Prove that for any non-degener non-degenerate ate quadrilateral quadrilateral with sides a,b,c,d, it is true that 10. For

ABC , prove that 3

11. For acute 12. For

      ≥ · ≥   ≥ ·   ·   ≥ √   ·

a sin A

ABC , prove that

ABC , prove that

cos

sin A

a

cot3 A + 6

A

csc

2

·



a2 + b2 + c d2

cot A.

6 3+

2

2

3(a sin C + b sin B + c sin A).

cot A

A

√

cot

A

2

.

13. A 2-dimensional 2-dimensional plane is partitione partitioned d into x regions by three families of lines. All lines in a fami parallel to each other. What is the least number of lines to ensure that x 2010. (Toronto (Toronto 20

√ ABC , prove that 3 3R ≥ 2s. 1Scribd Master your ABC , prove thatwith ≥ 2≥3· 15. For semester 2 − cos A

≥

14. For



& The New York Times 1  sin A . 16. For ABC , prove that ≥ Special offer for students: Only $4.99/month.

8

2



5

− √

1 . Free Foron 30this Days Sign to vote title cosRead A up

3 3



 Not useful Cancel anytime.

Useful

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24. Of all triangles triangles with a fixed perimeter, perimeter, dtermine dtermine the triangle with the greatest area. area.

25. Let ABCD be a parallelogram such that ∠A 90. Altitudes from A meet BC,CD at E, F respect Let r be the inradius of CE F . Prove that AC 4r . Determine when equality holds.

≤



≥

26. For ABC , the feet of the altitudes from B, C to AC, AC, AB respectively, are E, D respectively. L feet of the altitudes from D, E to BC be G, H respectively. respectively. Prove that DG + EH BC . Dete when equality holds.



≤

27. For ABC , a line l intersects AB,CA at M, N respectively. K is a point inside lies on l. Prove that ∆ 8 [BM K ] + [ CN K ]. ].



28. For

ABC , prove that

 ≥ ·   15 + 4

cos(A

− B) ≥



ABC such t

sin A.

29. Let pI be the perimeter of the Intouch/Contact Triangle of

ABC . Prove that p ≥ 6r I

ABC , let A B C  be an arbitrary triangle. Prove that 1 + Rr ≥ B − C ≥ 24 · sin A . 31. For ABC , prove that cos2

30. In addition to

32. For

ABC , prove that

33. For

ABC , prove that

   



2

≥ 9r. A−B ≥ cos ha

2



sin





s 4R

sin A . sin A

2

3A . 2

ABC , prove that sin2 A2 + cos B −2 C ≥ 1. 35. For ABC , AO,BO,CO are extended to meet the circumcircles of BOC, COA, AOB 9 AK BL CM ≥ tively, at K,L,N respectively. Prove that + + . 2 OK OL OM √ 9abc ≥ 36. For ABC , prove that 4 3∆. a+b+c 34. For



Master your semester with  a b(aScribd − b) ≥ 0. 37. For ABC , prove that & The New York Times π sin a

Read Free Foron 30this Days Sign up to vote title

2

3

38. ShowOnly that$4.99/month. for all 0 < a, a, b < Special offer for students:

2

we have

sin b

+

cos3 a cos b




Useful

≥ sec(a − b)

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

a2 +

44. For

ABC , prove that

45. For

ABC , prove that 6R ≥


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√abc ≥ 4(abc) 3R




2 3

.

a2 + b2 . mc2 BC

DE

46. For a convex convex hexagon ABCDEF with AB = BC,CD = DE,EF = F A, prove that + BE DA 3 . Determine when equality holds. 2 47. For

√ ABC , prove that s 3 ≥



la .

    − ≥ · · −   ≥√ · 

48. For

ABC , prove that R

49. For

ABC , prove that

1 12

2r

a2

2

ab . R

ma

4 3∆ max

ma mb mc . , , ha hb hc

50. A1 A2 B1 B2 C 1 C 2 is a hexagon with A1 B2 C 1 A2 = A, B1 C 2 A1 B2 = B , C 1 A2 B1 C 2 = C and AA2 = BC , BB 1 = BB 2 = CA , CC 1 = CC 2 = AB. Prove that [A1 A2 B1 B2 C 1 C 2 ] 13 [ABC

∩

∩

∩

1

≥ ·

1

51. For

ABC , let r1, r2 denote the inradii of ABM , ACM . Prove that r1 + r2 ≥ 2

52. For

ABC , prove that

a



csc2

A



a

− B) + 9 ≥ 8 · 2 2s4 − a4 53. For ABC , find the minimum of the expression . ∆2 √3 B − C A · ≥ 54. For ABC , prove that cos cos . 2 4 2 ≥

cos(A





·

1 r

cos A.

      · ·

2

A ABC , prove that 3 a2 > ∆ cot . 2 ABC , c ≤ b ≤ awith 56. For semester . Through Through interior interior point point P and the vertices A , B , C , lines are drawn me Master your Scribd 

55. For

Free For 30 Days the opposite sides at X , Y , Z respectively. Prove that AX Read + BY + 2athis + .
& The New York Times

3

s 57. For ABC , prove that Special offer for students: Only $4.99/month. 2abc

58. F

 ABC , let P A

x,PB

 ≥

cos4

y,PC

A

2

 .

. Prove that


Useful

+ bzx +



≥ abc, with equality hold

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ABC , prove the following and determine which is stronger: ( Samer Seraj) Seraj)

  · ·  

(a) ∆

≥r

(b) ∆

≥r·

1 3 2 3

·

ma mb +

1 2

·



ab.

ma mb + r(r + 4R).

65. For any convex convex pentagon A1 A2 A3 A4 A5 , prove that

5

5





(Ai Ai+2 + Ai+1 Ai+4 ) >

i=1

66. For


ABC , prove that s2 ≥



Ai A2i . Ai 2

i=1

la2 .

67. ABCD is a quadrilateral inscribed in a circle with center O. P is the intersection of its diagona R is the intersection of the segments joining the midpoints of the opposite sides. Prove that OP

ABC , prove that 54 · bc > m m . 69. For ABC √ , let M ∈ [AC ],], N ∈ [BC ] and L ∈ [M N ], where [XY ] denotes the line segment X Y that: ∆ ≥ S 1 + S 2 , where S 1 = [AML AM L] and S 2 = [BN L]. 70. For ABC , prove that (b + c)P A ≥ 8∆. 71. Right ABC has hypotenuse AB . The arbitrary arbitrary point point P is on segment CA , but different fro AB − BP AB − BC vertices A, C . Prove that . >

 

68. For

3

b

c

   3

3

AP

72. For

ABC , prove that max

73. For

ABC , prove that





BP CP , AC AB

CA

≥√ − 2

1.

√ ≥ 6 3R. s−a a2

74. Let P be a point inside a regular n-gon, with side length s, situated at the distances x1 , x n 1 2π from the sides, sides, which which are extended if necessary necessary.. Prove Prove that . > xi s i=1  Read Free Foron 30this Days Sign up to vote title β 75. A point A is taken inside an acute angle with vertex O. The line OA forms angles α and Useful Not useful   sides sides of the angle. angle. Angle Angle φ is given such that α + β + φ < π . On the sides of the former former angle Cancel anytime. Special offer for students: Only $4.99/month. points M and N such that ∠M AN = φ, and the area of the quadrilateral OMAN is maximal.

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

ABC , prove that 9R2 ≥ a2 ≥ 18Rr. 83. For ABC , prove that (P A · P B · c) ≥ abc. 84. For ABC , prove that 8R3 ≥ IE . 82. For



85. For

ABC , prove that

86. For

ABC , prove that



a

  ·   ≥ √  ≥   ·    sin

A

tan

2

a3 + 6abc

3 3 . 2

A

2

a

ab >

a3 + 5abc.

87. D and E are points on congruent sides AB and AC, respectively, of isosceles CE . Prove that 2 EF BC. Determine when the equality holds.

≥

ABC such that

√ ABC , prove that AH ≥ 3 3. a 89. Let M, A1 , A2 , · · · , A (n ≥ 3), be distinct points in the plane such that A1 A2 = A2 A3 = · · · A −1 1 1 ≥ . A A1 . Prove that M A · M A +1 M A1 · M A 88. For

n



n

n



n

i=1

90. For

i

i

QA2 .

ABC , determine min

91. For

n

   ·  √  ≥   · ≥ ≥  ≥  − 8

a2 + 4 3∆

a2 ab

1+

ABC , prove that GA. 3 92. For ABC , prove that a2 + b2 + R2 c2 , and determine when equality holds. 93. For ABC , prove that ab (s2 + r2 ) 4abcs + 36R2 r2 . 94. For

ABC , prove that

1

2r R

.

A

4cos 2 sin B sin C ABC , prove that sin ≥ + . 2 2 − sin 1 sin 2 2 2 Master your semester with Scribd  96. In ABC , the internal angle bisectors of angles A,B,C intersect circumcircle Read Free For 30this Days Sign up tothe vote on titleof ABC  95. For

C

B

A

1982) , Y , Z respectively. Prove that AX + BY + CZ > a + b + c. (Australia 1982) & The NewXYork Times  Useful  Not useful 97. An arbitrary line  through through the incenter incenter I of Special offer for students: Only $4.99/month. a2 BM CN

≥

anytime. ABC cuts ABCancel and AC at M and N. Show

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Sol Solutions ons 1. Euler’s Original Proof R(R 2r) = OI 2 0

≥ ⇐⇒ R ≥ 2r. 

−

1. Author: Author: tonypr tonypr 3 r r Rewrite the inequality as 1 + . Then note the identity 1 + = cos A + cos B + cos C . 2 R R So it is sufficient sufficient to prove prove that 2 cos A + 2 cos cos B + 2 cos cos C 3. It’s easy to verify that this inequality is equivalent to (1 (cos B + cos C )) ))2 + (sin B sin C which is true by the Trivial Inequality. 

≤

≤ −

−

1. Author: Author: BigSams BigSams x+y For positive reals x , y , z it is true that ( x + y)(y + z )(z + x) 8xyz by AM-GM: 2 xyz . By Ravi Substitution, let a,b,c be side lengths of a triangle such that a = x + y, b = y + z, c The inequalit inequality y becomes becomes abc 8(s Heron’s ’s Theore Theorem, m, the the ineq ineq a)(s b)(s c). By Heron 2∆ abc abc . Using the fact that ∆ = = sr, R 2r.  sabc 8S 2 4∆ 4R s



≥

≥

⇐⇒

≥

≥

−

−

−

≥

≥

Author: BigSams BigSams 1. Author: Note that By CS,



4R + r r

=

1

   ·   ≥

ra = 4R + r and ra

ra

1

ra

=

1

.

r

9

⇐⇒ R ≥ 2r. 

2. Author: Author: 1=2 Lemma. AB + AC > PB + BC Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title  the extension of BP intersect AC at N . Then the triangle inequality gives us & The NewProof. YorkLetTimes  Useful  Not useful Special offer for students: Only $4.99/month.

P N + NC > PC

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3. Author: Author: Goutham Goutham Let x = s We have But

a, y = s b, z = s x2 xy =

c all greater than 0, and s = x + y + z , ∆2 = xyzs. (x2 + 3xy ) 4 xy .

− ≥ − ⇒ −    ≥

And so,

(x2 + 3xy ) = ab 4xyzs

(x + y)(x + z ) =

≥





ab. ab

 ≥

xy . Therefore, we have 4∆ xyzs

1

− a) . 

s(s

4. Author: Author: Mateescu Mateescu Constantin Constantin

√

Using Using the wellwell-kno known wn formula formula for area i.e. ∆ = sr, the inequality rewrites as: s 3 4R 2 2 2 Of course, this is weaker than Gerretsen’s Inequality i.e. s 4R + 4Rr + 3 r , since the inequ 2 (4 + ) R r 4R2 + 4Rr + 3r2 reduces to Euler’s inequality i.e. R 2r . Howeve However, r, there is also a 3 method to obtain directly the inequality ( ). In the well known inequality: x = (s b)(s c)

≤

≤

≤

≥

∗  

  

− − y = ( s − c)(s − a) 3(xy + yz + zx ) ≤ (x + y + z )2 we take: and thus we obtain: z = (s − a)(s − b) √ 2 3s(s − a)(s − b)(s − c) ≤ [r(4R + r )] , whence 3∆ ≤ r(4R + r) ⇐⇒ (∗). 

5. Author: Author: Thalesmaste Thalesmasterr

 −    − −    − Master your semester with  Scribd & The New York Times  cos

Note the identities

cos

B

C

2

A

2

=

= cos (s

B

2

cos

b)(s bc

s(s a) sin = 2 bc a=y+z

C

2

+ sin

c)

b =z+x

2

sin

C

2 and

B

Using Ravi’s substitution: Special offer for students: Only $4.99/month.

B

  

r=

R=

∆ s abc

4∆




Useful  Not useful  equiv Cancel , the inequality inequality is equivalen alentanytime. t to: (2x + y + z )2

c=x+y

which is true according to AM-GM Inequality.





≥ 8x

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3

 6

abc 64(abc)2

·

≥6

of 40

 

√

abc, as desired.





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

Left Side. Lemma. AI + BI + CI

≤ 2(R + r) (Author: Author: Mateescu Mateescu Constantin Constantin)) 1 √ bc A · Proof. Show easily that AI = cos = · bc · s(s − a) a.s.o. Thus, Thus, we have: have: 2 s s 2 √ 1 1 · · − ≤ · (ab + bc + ca) · s(s − a) = ab + bc + ca ≤ 4(R + ( ) bc s s a 2 s s2 The last inequality is due to Gerretsen i.e. s2 ≤ 4R2 + 4Rr + 3r2 . Therefore, Therefore, we have sho AI + BI + CI ≤ 2(R + r ) . 

  



 

C.B.S.

As a direct direct consequ consequenc encee of the lemma, lemma, it suffices suffices to prove prove 2( R + r ) 2R2 5Rr + 2r2 0. However, ths is equivalent to (2 R r)(R 2r) 0, which is indeed true. For both inequalities, equality holds for ABC equilateral.

−

≤

≥

−

− 

≥



2 3(R2



− Rr + r

Author: Thalesmaste Thalesmasterr 6. Author: Left Side.

 

AI 2 = bc

− 4Rr Note that: BI 2 = ca − 4Rr According to C.S Inequality: CI 2 = ab − 4Rr 3(AI 2 + BI 2 + CI 2 ) ≥ (AI + BI + CI )2 ⇐⇒ 3(s2 + r2 − 8Rr ) ≥ AI + BI + CI So it suffices to show that 3(s2 + r 2 − 8Rr) ≤ 12(R2 − Rr + r 2 ) ⇔ s2 + r 2 + 8Rr ≤ 4R2 − 4Rr + 4r2 ⇔ s2 ≤ 4R2 + 4Rr + 3r2, which is the Gerretsen Inequality. 







6. Author: Author: tonypr tonypr

Right Side. r with Scribd Master your semester  Note that AI = . Applying this cyclically to BI andRead hand side is equivalent t CI , the Free For 30this Days Sign up toleft vote on title sin  1 1 1 Useful  Not useful & The New York Times  + + A

2

r r r Special offer for students: 6r Only A$4.99/month. + + B sin 2 sin 2 sin C 2

≤

⇐⇒ 2 ≤

sin A 2

sin B 2 3

sin C 2

Cancel anytime.

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·

·

AI DI r r BI CI = = = sin ∠AID sin ∠BI C AD BC AI DI AD = . BI CI BC

Combining, =


· ·

AI 2 + DI 2 AD2 BI 2 + CI 2 BC 2 and and 2 cos cos ∠BI C = . AI DI BI CI AI 2 + DI 2 AD2 BI 2 + CI 2 BC 2 Combining, = 2 cos cos ∠AID = 2cos ∠BI C = AI DI BI CI AI 2 + DI 2 AD2 BC 2 BI 2 CI 2 = = AI DI BI CI AI 2 DI 2 BI 2 CI 2 AD2 BC 2 = + + + = + AI DI AI DI BI CI BI CI AI DI BI CI

By the Cosine Cosine Law, 2 cos ∠AID =

⇒ ⇒

−

·

·

·

−

·

−

·

−

−

−

·

·

−

·

·

·

−

·

−

·

It is well-known that for a tangential quadrilateral, the sum of two opposite sides is equal t semiperimeter. So AB + BC + CD + DA = 2(AD + BC ) = 2(AD + BC )2 = 4(AD2 + AD BC + BC AD + BC 2 ) = =

    

·

·

AD2 BI CI BC 2 AI DI 4 AD2 + + + BC 2 AI DI BI CI

4

·

·

·

AD2 BC 2 + AI DI BI CI

·

·

·

· · ·



(AI DI + BI CI )

·

·

AI 2 DI 2 BI 2 CI 2 + + + AI DI AI DI BI CI BI CI (AI + BI + CI + DI )2 2(AI DI + BI CI )

By Cauchy-Schwarz,

≥



⇐⇒ ⇐⇒

 · 4

·

·

·

AD2 BC 2 + AI DI BI CI

·

·

AB + BC + CD + DA

·

≥

·

·

(AI DI + BI CI )

√

·

·

≥

√

2(AI + BI + CI + DI )

2(AI + BI + CI + DI )



8. Author: Author: RSM

Master your semester with Scribd a +b +c − = OG ≥ 0 ⇐⇒ 9R ≥ a . 9 & The NewRYork Times 2

2

2

2

Special offer for students: Only $4.99/month. 9. Author: Author: RSM

2

2

2







Useful

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Let

      

 

of 40

p =

cot A

q=

cot B cot C = 1

cot A The inequality is equivalent to: ( p3 3 pq + 3r) + 6 r p p3 Which is Schur’s Inequality. 





 Search document

r=

−

≥ ⇐⇒

− 3 pq + 9r ≥ pq ⇐⇒ p3 − 4 pq + 9r ≥ 0

12. Author: gaussintraining

  −    −       −    · cos

Using the identities

sin

A

2

A

2

=

=

cot

A

2

s(s a) bc

(s

=

b)(s bc

− c)

s r

s(s a) bc

the inequality is equivalent to

−

By Cauchy-Schwarz, LHS =

s(s a) bc

bc a)(s

,

   · −   

(s b) using Heron’s Formula. Thus, we have to prove s ( s a + s b + s c)2 6 3 +

−

−

(s

≥

4

bc b)(s

− c)



≥6

√

3+

s r

  √ − √ 2

s(s a) (s b)(s c)

−

−

−

=

s

r

√ − √ − √ − ≥ √ r √ √ √ =⇒ 2( s − a s − b) ≥ 6r 3 (s − a)(s − b) + (s − b)(s − c) + (s − c)(s − a) ≥ (s − a)(s − b)(s − c). By AM-GM, 3 √2 Master your semester  − − − Using Heron’s Formulawith again, weScribd find that ( )( )( ) = . Days s a s b s c r s30 √2 Free For √ Read Sign up to √ vote on this title  we finally have to show that 3 r s ≥ 3r 3 =⇒ s ≥ 3r 3, which is well-known. & The NewTherefore, York Times  Useful  Not useful







3

3

Special offer for students: Only $4.99/month. 12. Author: Author: Thalesmaste Thalesmasterr





 3

3

Cancel anytime.

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2

Let sa + sb + sc = n. Then the number of lines is 2010 in sa = sb = 26, sc = 25 works, so our answer is 77. 

≤ n3

+ n + 1. Thus, n

≥ 77. Indeed, plu

Author: mcrasher mcrasher 14. Author: Since



s sin A = , it suffices to show that R



sin A

≤

√

3 3 , which is true by Jensen’s Inequalit 2

15. Author: Author: BigSams BigSams Left Side. By Euler’s Inequality, 2r R 8r2 4Rr 4R2 + 4Rr + 3r2 8Rr 5r 2 + 4R2 . By Gerretsen’s Inequality, s2 4R2 + 4Rr + 3r2 . Combining, s2 8Rr 5r2 + 4R2 . r s2 (2R + r)2 s2 + r2 4R2 4 1+ +2 4 + 3 4R2 4R2 R 4 cos A + 2 cos A 4 + 3 cos A cos B 1 (2 cos A) (2 cos B ) 2 (2 cos A) 2 2 cos A

≤ ⇐⇒ ≤ ≤ − −

⇐⇒   ⇐⇒   ⇐⇒  − · − ⇐⇒

≥

Right Side.

≤

⇐⇒

≥   ·  − ≥

≤

−

⇐⇒



−



≥

−

72Rr 9r2 5

− ≤ 16Rr − 5r2. ≤ R ⇐⇒ 72Rr − 9r2 2 2 − ≤ ⇐ ⇒ ≤ By Gerretsen’s Inequality, 16Rr 5r s . Combining, s2 5 r s2 − (2R + r)2 s2 + r2 − 4R2 ⇐⇒ 20 1 + + 2 ≤ 25 + 7 By Euler’s Inequality, 2r

     ⇐⇒  − · − ⇐⇒ R

20

4R2

   ·  ⇐⇒

cos A + 2 cos A 25 + 7 (5 cos A) (5 cos B ) 2 (5 cos A) 3

−

≤

≤

cos A cos B 1 5 cos A

Master your semester with Scribd 16. Author: Author: Mateescu Mateescu Constantin Constantin & The New York Times A r Using the$4.99/month. relation: sin = Special offer for students: Only 2 4R

4R2

−



≤ 23 .  Read Free Foron 30this Days Sign up to vote title



 

Useful  Not useful Cancel anytime. , the inequalit inequality y reduce reducess to 2 r R, which is due to Eule

≤

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18. Author: Author: BigSams BigSams By CS, 9



≤ ( a + b + c)

a+b+c = 2[ABC ] sh = [ABC ]



⇐⇒



1

+

1

+



b c 2[ABC ] 2[ABC ] 2[ABC ] + + a b c

9[ABC ]

a

1



≤ sh

Equality holds if and only if a = b = c, which is derived from the CS equality condition.



19. Author: Author: Goutham Goutham

Lemma. In ABC,M,N,P are points on sides BC,CA,AB respectively such that perime the M N P is minimal. Then M N P is the orthic triangle of ABC . (Author: Author: Farenhajt









Proof. Let M be an arbitrary point on BC , and M  and M  its reflections about AB and AC respect Then, for a given M , the points N, P which minimize the perimeter of M N P are the interse of M  M  with AB and AC . Triangles AMM AM M  and AMM AM M  are isosceles, hence ∠M  AM  = 2∠A = const, thus M  M  , i. required perimeter, is minimal when AM  = AM  = AM is minimal, which is obviously attained is the foot of the perpendicular from A to BC ( ). Now we note that the orthic triangle has the property that, when one of its vertices is reflected a the remaining remaining two two sides of the initial triangle, the two two reflections reflections are collinear collinear with the two rema vertices of the orthic triangle - which is easy to prove: ∠M P N = π 2∠C ∠M P B = ∠C . Therefore Therefore the triangle triangle obtained by the argument argument ( ) is indeed the orthic triangle, as claimed.



∗

∗

−

∧

Using the lemma, the orthic triangle does not have a greater perimeter than the medial triangle, w has a perimeter equal to the semiperimeter of the original triangle. 

Master your semester with Scribd  Read Free Foron 30this Days Sign to constant vote title a. Sinc Let ABC be an arbitrary triangle with a constant area ∆up and base Si ncee th & The NewandYork Times useful a base are constant, then the height h with foot on  also constant since it can be expres a is Useful  Not 20. Author: Author: BigSams BigSams

a

·

a h Special offer for students: $4.99/month. a = ∆ = termsOnly of constants:

Let AB

c,CA

⇒

2X . ha =

Cancel anytime.

2 a (extended if necessary) at P . Let P C b. Let h intersect BC

PB

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21. Author: Author: r1234 Let O be the point of intersection of the two diagonals. Now [ ABCD ] = [ABCD ]

≤ AC · BD .

1 AC BD sin ∠ACD 2

·

·

·

1 1 1 AB BC sin B AB BC similarly we get [ ABCD ] CD 2 2 2 1 1 the other hand we get other two inequalities [ ABCD ] AB CD and [ABCD ] BC 2 2 Adding the last four inequalities we get(AB + CD )(BC + DA) 4. This impli implies es that (AB CD + DA)2 4(AB + CD )(BC + AD) 16 or AB + BC + CD + DA 4. On the other hand we get AC BD 2 or ( AC + BD )2 8 or AC + BD 2 2. Adding we get AB + BC + CD + DA + AC + BD 4 + 2 2.  Now again [ABCD ] =

·

·

·

≤ ·

·

≤ ·

≥

·

≥

≥

≤ · ≤ ·

· ≥

≥ √ ≥

≥√

≥


   −  −    ·  ≥ ·   ⇐⇒          · ≥  a=x+y

Using Ravi’s substitution

b =y+z

c=z+x

We have sin

A

2

=

(s

b)(s bc

c)

yz . (x + y )(x + z )

=

So the inequality is equivalent to

sin

x

2

x(y + z )

sin

2

(x + y + z )3

−

2

sin

A

x

2

y+z

2

x2 (y + z )

x

≥2

3

3

+ z) ≥ (x + y + z)((xxy++y yz + zx ) − 3xyz

(x + y + z )3 4 (x + y + z )(xy + yz + zx ) 3xyz 4(x + y + z )(xy + yz + zx) + 9 xyz + 3xyz

−

Master your semester with Scribd Author: professordad ssordad 23. Author: & The New York profe Times   Special offer for students: Only Using the$4.99/month. half angle identites,

3

2

x(y + z )

It suffices to show that

⇐⇒

C

x

According to Holder’s Inequality,

  ⇐⇒

B

2

sin

A

2

=

≥

≥ 0, which is Schur’s Inequality.




1

−

 

Useful  Not useful Cancel anytime. cos A 3 cos A 3 = . This is equ 2 2 4

−



≥

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Since ∠AEC and ∠AF C are both right, the points AECF are cyclic and AC is a diameter. diameter. The AC is twice the circumradius of CE F . By Euler’ Euler’ss inequa inequalit lity y of a triang triangle le in CE F the circumra circumradiu diuss is at least least twice twice the inradi inradi ◦ 4r1 , with equality iff CE F is equilateral iff ∠C = 60 and A lies on the angle bisec AC ◦ ∠EC F iff ABCD is a rhombus and ∠C = 60 . 



≥





Author: truongtansan truongtansang89 g89 26. Author: 1 Note that DG BC = DB DC DG BC = BC 2 cos B sin B DG = BC sin2B . 2 1 Similarly, EH = BC sin2C DG + EH = BC sin A cos(B C ) BC . 2 π π Hence, equality holds when A = and B = C = .  2 4

·

·

⇒

·

·

⇒

·

⇒ − ≤

·

Author: Mateescu Mateescu Constantin Constantin 27. Author:

AM AN M K =q, =r, = t, where q,r,t > 0. MB N C KN [AMN AM N ] AM AN qr Observe that: = = , [ABC ] (q + 1)(r + 1) bc qr From where: where: [AM N ] = [ABC ]( ). Moreov Moreover, er, we can write the following following relatio (q + 1)(r + 1) [BM K ] 1 [AMK AM K ] = = [BM K ] = [AMK AM K ] q q t [AM N ] (∗) rt [ABC = [BM K ] = = (q + 1)(r + 1) q(t + 1) [AMK AM K ] t [AM N ] =t = [AMK AM K ] = [AN K ] t+1

Let us denote:

·

·  ⇒   · ⇒   ⇒    semester⇒with Scribd Master your [CN K ] 1 = [AN K ] r

[AN K ] 1 = [AMK AM K ] t

=

=

[CN K ] =

[AN K ] =

∗

[AN K ] r

[AM N ] t+1

& The NewThus, York Times Th us, the proposed propo sed inequali inequality ty reduces reduces to:

  

  

=

⇒

[ABC ]

·

⇒

[CN K ] =

[AM N ] (∗) q [ABC ] = (q + 1)(r + 1)(t + 1 r (t + 1)




·

·

 

qrt ≥ 8 · Useful  Not2useful 2 · [ABC ] 2 Cancel anytime. (q + 1) (r + 1) (t + 1)

Special offer for students: Only $4.99/month. (q +1)(r +1)(t + 1) 8 qrt , which which is clearly clearly true by AM-GM inequalit inequality y. Equality Equality occurs if AM AN M K

≥ √

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9 + 2

 

of 40

2

sin A


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2

      ⇐⇒ ≥     ⇐⇒ ≤   ⇒ −    ⇒ −  −          −     cos A


2

2

9 ( cos A) + ( sin A) sin A + 4 2 2 2 Note that sin A + cos A = 1 = sin2 A + cos2 A = 3. Note that cos( A B ) = cos A cos B + sin A sin B = 2 cos(A B ) = 2 (cos A cos B ) + 2 (sin A sin B ). Adding Adding these gives 3 + 2 = =

sin2 A + cos A

So 3 + 2

2

cos(A

cos2 A + 2

+

sin A

cos(A

B)

(cos A cos B ) + 2

2

(sin A sin B )

.

B) =

2

cos A

+

2

sin A

.

Applyin Applyingg the above above ident identit ity y, the previo previousl usly y derive derived d becomes

 ⇐⇒

sin A

  ≤ 15 + 4

cos(A



sin A

  ≤ 9 ( + 4

2

cos A) + ( 2



− B), as desired. 

29. Author: Author: BigSams BigSams

Let the sides of ABC be AB = c,BC = a,CA = b, with corresponding sides of the intouc cle being a , b , c respectively.



    

a = 2(s

− a)sin A2

 −  −  −       Master your semester with ≥ · Scribd· Note that

b = 2( s

b)sin

c = 2(s

By AM-GM, s =

c)sin

a


Special offer for students: Only 30. Author: Autho r: $4.99/month. Thalesmaste Thalesmasterr

3

B

2

(s

, and

sin

C

a) = sr2

A

2

=

r 4R

2

a

1 3

=3

2(s

A



1 3

1 3

s − a)sin = 6r .  2 Free For 30 4R Days Read

 

Sign up to vote on this title




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⇐⇒ (2R +2r)(R − 2r) ≥ 0 ⇐⇒ 16Rr − 5r2 ≥ 22Rr − 4R2 − ≥ 16Rr − 5r . 2 2 3+ 1+ + r 2 2 2 ≥ Combining, s ≥ 22Rr − 4R − r ⇐⇒ 24 4 4R

By Euler’s Inequality, R 2r By Gerretsen’s Inequality, s2

≥

r R

     ·      −   ≤ ⇐⇒ ·  − ≥ ·

sin A =

Note the identities:

sin

1 = 4

2

3+

s R

cos A = 1 +

cos A + 2

A

=

r R

2 cos A cos B +

sin2 A + cos2 A = 1

cos(A

Note the identities:

Thus,

sin

cos

2

A

2

A

B

2

24



2

 ·

sin

sin A + 2

x

 

A

2



≤

3+(



sin A sin B

·

2

cos A) + ( 4



sin A)

2



B ) = cos A cos B + sin A sin B

1 + cos x 2 2 1 + cos A cos B + sin A sin B = 2 A 24 sin .  2

cos2

24

⇐⇒

r 4R

·

s R

  

=

·

·



1 + cos(A 2

− B) =



cos2

A

− 2

32. Author: Author: applepi2000 applepi2000

Note that ∆ = rs. Let hi be the altitude to side i. We wish wish to prove prove ha + hb + hc 1 1 1 18∆ 1 1 1 9 2∆ + + + + a b c a+b+c a b c a+b+c 3 a+b+c Take the reciprocal of both sides, then multiply by 3: 1 1 1 . This This is just AM 3 + + a b c so we are done. 



≥

≥

⇐⇒

≥

≤

Master your semester with Scribd 33. Author: Autho r: Thalesmaste Thales masterr & The New York Times

Special offer for students: $4.99/month. AfterOnly expanding it, the inequality is equivalent to:


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We find that the previous inequality is equivalent to: 3

·         ≥ · · 4

cos X

12

+

cos Y cos Z +

cos X

2

sin Y sin Z + 12

cos Y cos Z + 3

·

2

2

cos X

·



cos X

3

− 6r) + 20R r + 13Rr + 2r ≥ 0 20R2 r + 13Rr 2 + 2r3 ≥ s2 Using the ineq If R ≥ 6r, this is it. If R ≤ 6r , then it’s equivalen equivalentt to 6r − R √ 20R2 r + 13Rr 2 + 2r3 (4R + r)2 ≥ ⇐⇒ 4R2 − 7Rr 4R + r ≥ 3s, it suffices to show that: 6r − R 3 0 ⇐⇒ (R − 2r)(4R + r) ≥ 0, which is true by Euler’s Inequality.  ⇐⇒

s (R

  ·

34. Author: Author: r1234 Note sin2

A

2

1

=

− cos A

− C ≥ 8 · 2 B − C Using cos =



cos

B

2



A



cos A = 1 + 4

·



sin

A

2

the inequality redu

. 2 (ra + r) A and r = 4 R sin the inequality reduces to (ra + r) 32Rr A 2 2 4R sin 2 ∆ ∆ abc We know that r = and ra = . So writing rb , rc and putting R = the inequality redu 4∆ s s a (b + c) 8abc which trivially comes from AM-GM inequality. 



sin

and then putting





≥

−

≥


− C = b + c sin A . 2 2 a B − C A ≥ 8 sin ⇐⇒ cos

Note that cos Then



B

2



2



(b + c)

≥ 8abc, which is true according to AM-GM.

35. Author: Author: truongtansan truongtansang89 g89

Master your semester with Scribd  Let R be the radius of (O). Read Free Foron 30this Days Sign up to vote title 9 9 AK BL CM OK + OA OB + OL OC + OM R R ≥ ⇐ ⇒ ⇐ ⇒ + + + +  Useful ≥ Not + & The NewOK York Times useful + 2 2 OL OM OK OL OM OK OL OM Special offer for students: Only $4.99/month. Using Ptolemy’s Theorem on the cyclic quadrilateral BOCK ,

Cancel anytime.

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Now let BD : DC = x : y, CE : EA = y : z and AF : F B = z : x. Now using Menelaus’ Menelaus’ss th we get OD : OA = (x + y + z ) : (y + z ) and similar similar for others. others. Hence Hence the inequa inequalit lity y red 1 9 (x + y + z ) which comes from AM-GM or CS.  2 y+z

·   ≥

36. Author: Author: bzprules bzprules

√

√

√

We have that 2 s 3R 3 = 6s 9R 3 = 2rs2 3 9Rrs = 4(2s)∆ 3 36Rrs. Since 4∆R = 4Rrs = abc, we have 4(2s)∆ 3 9abc. 9abc Dividing yields 4 3 ∆ , as desired.  a+b+c

√

≤ √

≤

⇒

≤

⇒

≤√

· ≤

⇒

≤

√

8rs2 3

≤ 36Rrs

37. Author: Author: applepi2000 applepi2000 Use Ravi Substitution a = x + y, b = x + z, c = y + z . Then it becomes (x2 + y 2 + 2xy )(xy + yz xz z 2 )





≥0 x ≥ 0 ⇐⇒

− −







After expanding and simplifying x3 y 2xyz x3 y 2xyz By Cauchy-Schwarz we have (x3 y + xy 3 + x3 z + xz 3 + y3 z + yz 3 )(xyz 2 + xyz 2 + xy2 z + xy 2 z + x2 yz + x2 yz ) (x2 yz + x2 yz + xy 2 z + x2 yz + xyz 2 + xyz 2 )2 . Dividing by 2xyz x gives the desired result. 

≥

·

−

≥

x



37. Author: Author: Thalesmaste Thalesmasterr Lemma. Let a, b, c be three reals and x, y , z be three nonnegative reals. The inequality holds if x, y, z are the side-lengths of a triangle (sufficient condition). Proof. Use the identity



x(a

− b)(a − c) = 12



(y + z



x(a

− b)(a

− x)(b − c)2 ≥ 0. 

Master your semester Scribd  a b(a − b)with   30this Days ≥ 0 ⇐⇒ c(a + b − c)(a − b)(Read ≥ vote We have ) to 0,For which istitle true according t a − up cFree Sign on  side lengths a triangle. c(a + b − c), b(c + a − b) and a(b + c − a) are & The Newlemma, YorksinceTimes Useful the  Notofuseful 2



Special offer for students: Only $4.99/month.

38. Author: Author: BigSams BigSams

Cancel anytime.

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40. Author: Author: KrazyFK KrazyFK Clearly AC AB + BC and AC CD + DA. We have two similar inequalities for BD and adding them we get the result.

≤

≤

41. Author: Author: xyy Let A1 , B1 , C 1 be the intersection of P A , P B , P B with BC,CA,AB , respectively.

BL CM AN P C 1 P A1 P B1 = . LC M A N B P C P A P B P A1 P B1 P C 1 Let x = . ,y = ,z = AA1 BB 1 CC 1 S P BC S P CA S P AB We know that x + y + z = + + = 1. S ABC S ABC S ABC ABC ABC ABC 1 x z z (x + y )(y + z )(z + x) S = 1 x 1 z 1 z 8

We have S =

·

·

·

·

− · − · − ≤ ⇐⇒

≥ 8xyz , which is true by AM-GM.

42. Author: Author: Mateescu Mateescu Constantin Constantin The inequalit inequality y rewrites rewrites as: 2R cause it is well-kno well-known wn that:



·



A

A ≥ ≥ sin A sin s ⇐⇒ 2 2 2 A = π − 2X π B = π − 2Y , where X , Y , Z ∈ 0, 2 C = π − 2Z

sin A sin

s sin A = . Using the substitutio substitutions ns R

 

 



 



sin A

we will tran

the inequality in any triangle ( ) into one restricted to an acute-angled triangle. Indeed, the inequ ( ) is now equivalent to: 2 sin2X cos X sin2X

∗ ≥

∗

 

≥

 − ⇐⇒    Master your semester with − Scribd −  Times & The New York 

⇐⇒





4 sin X 1 sin2 X sin2X 4 sin X 4 sin3 X + sin2X . For convenience, we will denote by s, R, r the semiperimeter, circumradius and inradius respec of the acute triangle XY Z . s sin X =

≥

R

Since:

sin3 X =

2s(s2

Special offer for students: Only $4.99/month. 2rs sin2X = 2 R

6Rr 8R3

3r 2 )


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AR AP = = AC AB

AR = a

⇒

     ≥

BQ BP = = BC BA

⇒

k

· k+1

. Consequently:

1 BQ = a k+1

·

[P QB ] =

2a2 , the conclusion can be restated as: 9 1 k2 k max k>0 = , , 2 2 2(k + 1) 2(k + 1) (k + 1)2 2 k2 = 5k 2 8k 4 0 = k 2 2 2(k + 1) 9

    

a2

2 a2

2

2

· (k +k 1)2 · (k +1 1)2

[PQCR] = a2



⇒

≥

1 2(k + 1)2

≥

⇒

2 9

− − ≥

and since: [

· (k +k 1)2

2 , which follows from the following: 9

⇒ ≥

2

=



 Search document

[ARP ] =

  


⇒ −4k − 8k + 5 ≥ 0

2 ≥ =⇒ −2k 2 + 5k − 2 ≥ 0 2 (k + 1) 9 k

  ⇒ ∈   ⇒ ∈ =

0,

k

=

1 2



1 ,2 2

k

44. Author: Author: fractals fractals 1 By the AM-GM, = 3 (s a)(s b)(s Thus, 3

(s

−a) + (s−b) + (s−c) s

s

s

3

≥

 − 3

(s

a)(s

− b)(s − c) .

s3

− c) ≤ 1 , so s(s − a)(s − b)(s − c) ≤ s4 . Thus rs = 27 27 s √ 1 s2 r s √ , so s ≤ 3√3 , so r ≥ 3 3, which is Mitrinovic’s Inequality.  3 3 −

−



s(s

− a)(s − b)(

45. Author: Author: r1234 Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title Let AD be the median of triangle ABC which intersects the circumcircle at the pointD . & The Newsecant York Times BC a a  Useful  Not useful property, we get AD · DD  = = . So DD  = . 2


Now AD

≤ 2R ⇐⇒

4

AD + DD 

≤ 2R ⇐⇒

2

4

m +

2

Cancel anytime.

a2

4ma

≤ 2R ⇐⇒

4m2a + a2

≤ 2R

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true by Nesbitt’s Inequality. Equality holds if, and only if, all of the following conditions are true: cyclic, CDEA is cyclic, EFAC is cyclic. ACE is equilateral, ABCE is cyclic, From this we easily infer the congruence of ABC , CDE CD E and EF A which tells us the hexagon is lateral. lateral. We can also easily get that it is equiangular, equiangular, and so it is regular, which is therefore therefore the equality case.  47. Author: Author: Mateescu Mateescu Constantin Constantin

 −   − ≤ √ ·  − ≤ − √   − · − ≤    − −   −  −    √− √ − − − − −   − − ≤ −  −   ≤ · − √ − √ −  −  − ≤ √ ·  − ⇒ (1)

(2)

We will prove that: la + lb + mc s(s a) + s(s b) + mc Inequality (1) follows from the well-known fact: la s(s a). 2 bc Indeed, la = s(s a) s(s a). b+c 4mc2 = a + b + 2 (s a)(s b)

2

For inequality (2) let’s note that:

s

≤

Whence we obtain that: 4 mc2

2s

a+b

2 (s

a)(s

2 (s

b)(s

c)

2s

s

a+

b) = 2s

(s

s

a) + ( s

b

s2

a+b

a+

mc2 + mc

2 (s s

b

(3)

≤

a)(s 2

b) = c

2

= 2 s(s a) + s(s b) s2 mc2 . The inequality (3) is clearly true since it follows from Cauchy-Schwarz Inequality, so we are don 48. Author: powerofzet p owerofzeta a 1 It’s known that: ma = 2b2 + 2c2 a2 2 1 1 3 3 By CS 2b2 + 2c2 a2 3 (2b2 + 2c2 a2 ) = 2s2 ma = a2 = 2 2 2 2 By Gerresten’s Inequality, s2 4R2 + 4Rr + 3r2  3 2 vote Free Foron 30this Days Sign up title = 2(4R2 + 4Rr + 3r2 ) 2r2 8Rr = 3 Read 2R2 + ma rto  2 Not useful 2  Useful  R Cancel9anytime. and by Euler’s Inequality R 2r, we get: = R ma 3 2R2 + Special offer for students: Only $4.99/month. 4 2 18 ab + ac + bc s2 + r2 + 4Rr So it suffices to prove that 12( R 2 ) 12(R 2 ) R

 −    − ≤  ·  Master your semester with≤ Scribd    ⇒ ≤ − −  & The New York Times  ≤ ≥ ≥

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49. Author: Author: BigSams BigSams Lemmata. (1) m2a =

2b2 + 2c2 4

− a2 , and the cyclic versions hold as well.

a2 (2) 2 = , and the cyclic versions hold as well. 4S 2 ha

1

2

(1)

× (2) = mh2

a

=

a

a2 (2b2 + 2c2 16S 2

− a2) =⇒

and the cyclic versions hold as well.

a2 (2b2 + 2c2

2

2

− a2) = 16S h2m

a

,

a



By Trivial Inequality, (2a2

− b2 − c2)2 ≥ 0 2 2 ⇐⇒ (a2 + b2 + c2)2 ≥ 3a2(2b2 + 2c2 − a2) = 3 · 16hS 2 m ⇐⇒ a

2

2

a +b +c

a

2

≥

√

4 3Sm a ha

.

Clearly the cyclic versions of the above result can be derived by starting with the cyclic versio (2a2 b2 c2 )2 0 and proceeding by the same manipulations and cyclic versions of identities, ma mb mc inequalit inequality y always always holds for any of . , ,

− − 2

≥

2

2

Thus, a + b + c

≥

ha hb hc ma mb mc 4 3S max . , , ha hb hc

√

·







50. Author: Author: RSM AB2 = AC 1 = b + c, so [AB2 C 1 ] =

[CC 1 C 2 ] =

c2 sin C

2

(b + c)2 sin A and similar for others. 2

. Adding up all these we get the desired result.

(a + b + c)(a2 + b2 + c2 ) [A1 A2 B1 B2 C 1 C 2 ] = + 4[ ABC ] where R is the circumradius of ∆ ABC 4R a,b,c are its sides. Note that, (a + b + c)(a2 + b2 + c2 ) 9abc 9abc So [A1 A2 B1 B2 C 1 C 2 ] + 4[ABC ] = 13[ABC ]   4R Read Free Foron 30this Days Sign up to vote title

≥

Master your semester≥with Scribd & The New York Times 51. Author: Autho r: $4.99/month. RSM Special offer for students: Only ∆

∆




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 

        −



≤

Left Side. 9+

   

cos(A

Since:

− B) ≤

s2 + r2 + 2Rr 2R2

8Rr



s2 + r2 4R2 4R2

−

.

s2 + r 2 + 4Rr sin B sin C = 4R2

1 s2 + r2 8Rr = r2 sin2 A2 cos(A B ) 14Rr r2 , which is true since it is weaker s2 16Rr 5r2 .

≥

1 sin2

−

A

2

− 2R2 (G) ≤

⇐⇒

−

⇐⇒

≥

s2 + r2 + 2Rr 2R2

−

− 2R2 ≤ s2 − 8Rr − 8r2 . r2

R2 + 3Rr + 2r2 R2

− 13r2 (G) s2 − 8Rr − 8r2 ≤ r2 r2 R2 + 3Rr + 2r2 R2

It suffices to show that: true by Euler’s Inequality.



r R

cos B cos C =

Thus, 8 cos A 9+ Gerretsen’s Inequality: s2


 Search document

cos A = 1 +

We make use of the identities:




≤

8R

− 13r ⇐⇒ r

(R

− 2r)(8R2 + 2Rr + r2) ≥ 0, wh


 

a + b + c = 2s

Using the system: system:

ab + bc + ca = s2 + r 2 + 4Rr

abc = 4sRr (a + b4 + c4 ) 38 [ABC ]2

− 16Rr − 32R r − 2r ≥38 2s − Master your semester with≥Scribd ⇐⇒ 12s r + 16s RrRead We have: Foron 30this Days Signsup to vote title rFree  √ R s & The New⇐York ⇒ y (8xTimes − 13) ≥ 16x + 8x + 1, where x = r ≥ 2 andy =Useful ≥ 3  3. Not useful r 4

4

2

2 2

2

3

⇐⇒

≥ 16x, ≥ ⇐⇒ (x

4

2 2

2

Special offer for students: Only $4.99/month. Using Gerretsen’ Gerre tsen’ss Inequalit Inequality: y: y 2 + 5 16x2 + 8x + 1 7x2 16x + 4 0

2 2

Cancel anytime.

we just have have to show show that that (16x 2)(7x 2) 0

≥

− 5)(8x −

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Expres Expressing sing every everythi thing ng in terms terms of x , y , z using well-known well-known formulas formulas and then Ravi Subst x= u+v

   

  ⇐⇒

u + v + 2w

y =w+u

w (u + v )

z =v+w

2

 

 ≥ 

w(u + v + 2w)(u + v )

Which is clearly true according to Hölder’s older’s Inequality. Inequality.

u + v + 2w



3



55. Author: gaussintraining By CS, 3

    · ≥ 2

a

a

2

2

2

= 4s > πs = πr

2

s2 r2

2

·  ·   = Z

cot

A

2

.

56. Author: Author: malcolm malcolm Using AX < max AB,AC for X interior to BC and similarly for the other sides we have AX + BY + CZ < max AB,AC + max BC,BA + max CA,CB = AC + BC + BC = 2 a

{

{

}

}

{

}

{

}

57. Author: Author: Michael Michael Niland s2 cos = 2 a abc

1

A

   ≤     ·            ≤ · · · · Master your semester with Scribd  ≤ ·   ·   ≤ & The New York  Times       ≤ · · · Use the following:

2

A

cos2

2

=2 +

9 4

r 2R

By Chebyshev’s Inequality, cos4

1 3

A

2

a cos2

=

a cos2

1

A

2

A

1

2

a

cos2

a 2 A

cos

2

Again using Chebyshev’s Inequality,

1 A Therefore cos4 Special offer for students: Only $4.99/month. 2 3

2s 9 3 4

A

2

=

1 3

a cos2

a cos2

A

2

2

s abc

=

A

2

1 3

s3 . 2abc

s2 abc

2 2 3 Not useful

A s 230 Read Free For Days Sign to vote this title coson a up

Useful



Cancel anytime.

· 94. 

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

59. Author: Author: RSM

Suppose, P A , P B  , P C  are the perpendiculars perpendiculars from P to the sides BC,CA,AB and P A = p , P  q , P C = r. Note that B  C  = dA sin A and similar for others. So the inequality is equivalent to A B 2 + B  C 2 + C  A2 3(P A2 + P B 2 + P C 2 ) A B 2 + B  C 2 + C  A2 Which Which is true since (P A2 + P B 2 + P C 2 ) = + 3P G2 where is G is the cen 3 of A B  C  . Equality holds when P and G coincides, i.e. when P is the symmedian point of ABC . 

≤

60. Author: Author: Thalesmaste Thalesmasterr Using the condition , we have ( b

≥ c or c > b) =⇒

(b > a or c > a ).

∠ABC + In the two cases, a is not the greatest side, so A < We want to show that: ∠BAC < 2 2 π b+c a b c We have: a < + 2sin A < sin B A < < 3 2 2R 2R R 3 3 3 s π 3sin A < sin A = So: So: sin sin A = sin The function sin is increasi 2 2 3 R π π π the interval ]0; [. Hence A since we proved that A < .  2 3 2

π

⇔ ⇐⇒



≤

√ ⇐⇒

≤

⇐⇒

√

≤


By squaring both sides of this inequality and taking into account the identity: m2a + mb2 3(a2 + b2 + c2 ) 1 1 , we are left to prove that: mb mc a2 + bc, which follows by 4 2 4 2

2a2 + bc

2



≤





2

a bc ≤ a2 + bc4 a.s.o. ⇐⇒ 16m a.s.o. Indeed Indeed,, m m ≤ + 2 4 2 2 2 2 2 2 ⇐⇒ 16 · 2(c + 4a ) − b · 2(a + 4b ) − c ≤ 2a2 + bc 2 ⇐⇒ (b − c)2(a

ming up the inequalities: mb mc







b

c



Master your semester with Scribd  Read Free For 30this Days Sign up to vote on titlebi-monthly jou (BigSams used the same method in his submission to the Mathematical Reflections  & The Newwhere York Times the problem proble m was originally originally from)  Useful  Not useful (a

− b − c) ≤ 0, which is true from the Trivial and Triangle Inequslities. 


Autho

Thales

Cancel anytime.

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√ √ ≥ 2R2 + (6 6 − 8)Rr2 + (352 − 12 6)r2 2 Using Walker’s Inequality: s ≥ 2R + 8Rr + 3r (since XY Z is acute-angled), we just have to that: √ √ 2R2 + 8Rr√+ 3r2 ≥ 2R2 + (6√ 6 − 8)Rr + (35 − 12 6)r2 ⇔ (16 − 6 6)Rr ≥ 2(16 − 6 6)r2 ⇔ R ≥ 2r, which is Euler’s Inequality.  ⇐⇒

s2

63. Author: Author: Mateescu Mateescu Constantin Constantin Lemma. Let ABC be a triangle and let D be a point on the side [ BC ] so that: BD c2 + kb2 = k , k > 0. Then: 2R. DC (1 + k )(c2 + kb2 ) ka2



≤

−

c2 + kb2 ka2 Proof. Using the dot product, one can show the distance: AD = ( ). 1+k (1 + k )2 Let w be the circumcircle of ABC and let X = AD w. AD DX = BD CD ka 2 ka2 ka a Thus, = AD DX = = = DX ; CD = BD = (1 + k )2 (1 + k)2 AD 1+k 1+k Moreover, since AX is a chord in the circle w, it follows that: AX 2R AD + DX 2 ka 2R AD + (1 + k )2 AD

 

2

 ·

·

 

∗

∩

⇒

≤

·

⇐⇒

≤

⇐⇒ (∗) 2 ⇐⇒ (1 + k)2 · AD + ka2 ≤ 2R · AD ·(1 + k)2 ⇐⇒ c2 + k · b2 ≤ 2R · AD · (1 + k ) c2 + kb2 ⇐⇒ (1 + k)(c2 + kb2) − ka2 ≤ 2R, which is exactly what we wanted to prove.  ·

≤

{ } ⇒ ·

−



a2 b(c2 + a2 ) in the previous lemma we obtain: b2 a2 b2 + b2 c2 + c2 a2 abc c a of the well-known relation R = , our last inequality simplifies to: + 4∆ a c

√

Particularly, Particularly, for k =

≤

≤ 2R and makin √a2b2 + b2c2

2∆ In a similar manner we can prove the analogous inequalities, therefore solving the problem.

Master your semester withtin Scribd 64. Author: Autho r: Mateescu Mateescu Constantin Constan  1  & The NewItYork Times · mm will be shown that: ∆ ≥ r · (1)


3

b

1 c + 2


     · ≥ · ·



 

(2) Useful 2 Not useful Cancel anytime. m m + r(4R + bc r b c 3

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65. Author: Author: BigSams BigSams Problem Rewording. In pentagon ABCDE , prove that: (AC + BE )AB + (BD + CA )BC + (CE + DB )CD + (DA + EC )DE + (EB + AD)EA > AC 2 + BD 2 + CE 2 + DA2 + EB 2

Solution. By Triangle Inequality, AB + BC > CA = (AB + BC )AC > AC 2 . Repeating with BCD, CDE, DEA, EAB and summing all five yields the result.







⇒





66. Author: gaussintraining

√

2bc 2 bc A Since la = cos = s(s a) s(s a) by AM-GM, it follows that la2 2 b+c b+c The analogous relationships also hold, yielding la2 3s2 (a + b + c)s = s2 . 



− ≤

 −  ≤

≤ s(s −

−

67. Author: Author: jatin Let E and F be the midpoints of AC and BD respectivel respectively y. We know R is the midpoint o Note that E and F lie on the circle with diameter OP . And henc hencee OP OE as well as OP Now, OR is a median of OEF . Therefore, OR OF or OR OE . Hence, OP OR. 



≤

≥

≤

≥


1 1 Problem 61 from this marathon was equivalent to: Thus w mb mc a2 + bc. Thus 2 4 1 left to prove that: a2 < bc which is obviously true, since it rewrites as: 2( s2 r2 2 2(s2 + r2 + 4Rr) 0 < r 2 + 4Rr. 



 

≤



⇐⇒



− −

BigSams commented afterwards that a more elementary final step is by Triangle Inequ Master your with Scribd Note.asemester   Read Free Foron 30this Days Sign up to vote title (b + c − a) > 0 ⇐⇒ 2 · ab > a .  & The New York Times  Useful  Not useful 2

Special offer for students: Only $4.99/month. 69. Author: Author: Mateescu Mateescu Constantin Constantin

Cancel anytime.

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

ML = r, where k,q,r > 0. Therefore, LN

     

AM =k M C

=

[AML AM L] =k [CM L]

=

S 1 = k [CM L]

ML =r LN

=

[CM L] =r [CN L]

=

[CM L] =

BN =q N C

=

[BM N ] =q [M N C ]

=

[M N C ] =

1 [BM C ] q+1

AM =k M C

=

[BM A] =k [BM C ]

=

[BM C ] =

1 S k+1

S 1 =

     

BN =q N C

=

⇒ ⇒ ⇒

kr

=

⇒

⇒

(k + 1)(q + 1)(r + 1)

⇒ ⇒ ⇒ ⇒

·

r r+1

· [M N C ] ·

·

· S

=

[BN L] =q [CN L]

=

S 2 = q [CN L]

ML =r LN

=

[CM L] =r [CN L]

=

[CN L] =

BN =q N C

=

[BM N ] =q [M N C ]

=

[M N C ] =

1 [BM C ] q+1

AM =k M C

=

[BM A] =k [BM C ]

=

[BM C ] =

1 S k+1

⇒

S 2 =

⇒ ⇒ ⇒ ⇒

q

(k + 1)(q + 1)(r + 1)

⇒ ⇒ ⇒ ⇒

·

1 [M N C ] r+1

·

·

·

· S

           

Consequently, the proposed inequality reduces to:

√ semesterkrwith Scribd q Master your · ·Free (k + 1)(q + 1)(r S ≥ S + S ⇐⇒ (k + 1)(q + 1)(r + 1) (k + 1)(q + 1)(rRead + 1) up Foron 30this Days Sign to vote title  √ √



3

3

kr + q.Times & The New York 3

3



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3



Useful

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Cancel anytime.

Special offer for students: Only $4.99/month. Taking k = x3 , r = y3 and q = z 3 , where x, y,z > 0 it suffices to show that:

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Let P 1 be the symmetric of point P w.r.t. w.r.t. the midpoin midpointt of side [BC ]. Define Define P 2 and P 3 in a s manner. By Ptolemy’s Theorem, for a convex quadrilater M N P Q, M N P Q + N P M Q 2[M N P Q equality if and only if M N P Q is cyclic and M P N Q. Applying this to convex quadrilaterals ABP 1 C,BCP 2 A,CAP 3 B , we get:

·

⊥

 ·   · ·

b P C + c P B

·

≥

≥ 2(∆ + [P 1BC ])

a P C + c P A

·

·

≥ 2(∆ + [P 2CA])

a PB + b PA

≥ 2(∆ + [P 3AB]) Adding them gives that LHS ≥ 2(3∆+[P 1 BC ] + [P 2 AC ] + [P 3 AB]) for which we use [ P 1 BC ] = and so on to get that LHS ≥ 8∆ = RHS .  ·


  

AP =

AP Let us denote = k, where k > 0. Thus, P C

P C =

By Pythagoras’ theorem, applied in c

to show that:

 

−

a2 + k

k

⇐⇒ ⇐⇒

c + ak > k+1

b2

(k+1)2

+1 · b

a2 +

b2

(k + 1)2

>

k k+1

1 b k+1

·

 P BC one obtains: c

⇐⇒

− a ⇐⇒ c − b

·b

PB =



a2 +



b2

(k + 1)2

(c + ak)2 > a2 (k + 1)2 + b2

c2 + 2ack + a2 k2 > a2 k 2 + 2a2 k + a2 + b2

Master your semester with Scribd 72. Author: Autho r: Mateescu Matee scu Constantin Constantin & The New York Times

2

c

=a2 +b2

⇐⇒

a2 +

>

b2

(k + 1)2 k

k+1

. Hence, Hence, we a

· (c − a) ⇐⇒

⇐⇒

c > a , which is true.







Useful

 

Special offer for students: Only $4.99/month. Construct the lines passing through the vertices of triangle ABC so that they are parallel t

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CP  P C  = B  P A = 135◦, so 2 1. Equali Equality ty occurs occurs when x = y and A AB A = 45◦ , B = C = 67.5◦ and P is the orthocenter of triangle ABC . 

which implies

≥

−

Author: Mateescu Mateescu Constantin Constantin 73. Author:

  −

a2 (s

Using the identities:

(s

− b)(s − c) = 4s2r(R − r)

the given inequality is equivalent to

2

− b)(s − c) = sr √ √ 4s2 r(R − r) a2 (s b)(s − c) ≥ ≥ 6R 3 ⇐⇒ 6 3 ⇐⇒ s ≥ R (s a) sr2

 − −

a)(s

√ −

3Rr 3 2(R r) We will now show that this inequality is weaker than the known Gerretsen s2 16Rr 5r2 . Indeed, by squaring both sides of our previous inequality, it suffices to prove that: 27R2 r2 16Rr 5r2 4(R r)2 (16Rr 5r2 ) 27R2 r2 r (R 2r) 64R2 47Rr 2 4(R r) 0, which is obviously true since R 2r (Euler). Equality is attained if and only if ABC is equilateral. 

−

≥

⇐⇒

−

− ≥ 

−

≥

⇐⇒

−

≥

−



−

Remark. Here is a sketch of obtaining the first mentioned identity. Since ( s b)(s c) = bc we get: a2 (s b)(s c) = a2 [bc s (s a)] = abc a s2 a2 + s a3 , and furth



−



−

has to use the well known identities:

 − 

−

2

2

2

3

3

3



2

a + b + c = 2(s

a + b + c = 2s(s



− − r − 4Rr)

2

2

−

−

−

. 2

− 6Rr − 3r )

74. Author: Author: BigSams BigSams In an arbitrary regular polygon X , let the inradius be r and the sidelength be s. Note that the perimeter of X is always always greater greater than the circumfere circumference nce of the incircle. 2π n = sn > 2πr . >

⇒

⇐⇒

r s x Master your semester · swith x Scribd sr Also note that [ X ] = = n· =⇒ 2 2 & The New YorkTimes 1 1 n n n n

n

i

i=1

n

2

i

2

By CS, = = . Thus, n Special offer for students: Only $4.99/month. xi nr r x i=1 i i=1

≥



n

i=1

i=1

xi

>

= nr . up Read Free Foron 30this Days Sign to vote title

 2π s

.


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Thus, [M  AM ] < [N  AN ]



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⇒ [OM  AN ] < [OMAN ].]. 

So we have to find out on what conditions we can find on the sides on the sides of the angle p Circumscribe a circle about the triangle triangle M and N such that ∠M AN = φ and M A = AN . Circumscribe Since α + β + φ < π , the point A is located outside outside the circle. If L is the point of intersection o and the circle, then: π

∠AM N

=

π

− φ > ∠LM N = ∠LON and ∠AN M = π − φ > ∠LOM

2 − φ , then it is possible 2to find points M and N such that M A = AN and ∠M AN

if α, β < 2 If the conditions conditions are not fulfilled then such such points cannot be found. found. In this case, the quadrilate quadrilate maximal area degenerates into a triangle (either M or N coincides with O).  76. Author: Author: dr Civot Take a = b = c to get that k > 1. Let a = x + y, b = y + z , c = z + x by Ravi Transformation. The inequality inequality becomes becomes 3k xy + k x2 > 2 x2 + 2 xy .





 

k = 2 works because by Triangle Inequality



a(b + c



− a) > 0 ⇐⇒ 2 ·

Suppose that there exists a 1 < k < 2 which works. Take x =



A

  ab >

,y = z =

2 k 2 The inequality becomes LHS = (3 k 2) xy > (2 k ) x = RHS RH S . It will be shown that there is value of A for each 1 < k < 2 such that RHS mean that 1 < k < 2 does not exist work. RHS > (2 k )x2 = A A(6k 4) + (2 k )(3k 2) LHS =

−

−

RHS

−

A

−

A2

− LHS > 0 ⇐⇒

−



−

1

x

a2 , so k

.

− LHS > 0, whic

−

− A(6k − 4 ) + (k − 2)(3k − 2) > 0, which is true for sufficiently large

77. Author: Author: applepi2000 applepi2000 Let ad = x,bd = y,cd = z . Master your semester with Scribd 1  Free For 30 Days ⇒ Then from triangles MAB,MAC,MBC we have (x + yRead + z )up = = 2∆ =title S vote x + y + z. Sign to on this  2 & The NewWeYork Times Useful  Not useful 4∆  need to show xy + yz + zx ≤ . But this is true by Cauchy-Schwarz: a

b

c

2

Cancel anytime.

3 Special offer for students: Only $4.99/month. 1 4 2 2 (x + y + z ) = ∆ and we are done. Equality holds iff x = y = z , i.e. M = xy + yz + zx

≤

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((x + y )(y + z )(z + x))3 8(x2 (x + y + z ) + xyz )(y 2 (x + y + z ) + xyz )(z 2 (x + y + z ) + xyz ). But we have x2 (x + y + z )+ xyz = x(x + y )(x + z ), so our inequality is equivalent to ( x + y )(y + z )( 8xyz , which is true by AM-GM. 

≥

79. Author: Author: applepi2000 applepi2000 Say without loss of generality a b Multiplying the given by abc gives

≥ ≥ c > 0, since the inequality is symmetric. c(a2 + b2 − c2 ) > 2abc ⇐⇒ a2 b + a2 c > a3 Now, use the identity ( a + b − c)(a − b + c)(−a + b + c) = a2 b − a3 − 2abc. Then the given is ( a + b − c)(a − b + c)(−a + b + c) > 0. Now note that a + (b − c) ≥ a > 0 and (a − b) + c ≥ c > 0, this becomes −a + b + c > 0 ⇐⇒



    

Also, rearranging the two strict inequalities above gives a + b > c and a + c > b . Thus, a,b,c of a triangle. 

80. Author: Author: dr Civot If ∠B = ∠C then it’s clear that AP = AQ. Now assume that ∠B < ∠C . Then ∠APB > 90. Let M be midpoint of BC , then is B M CP = BQ and CM = BM = M P = M Q, but that is possible just if Q M P [ ]. [ ], [ ], Q [BC ] = B Q P . = In triangle AQP ∠QPA > 90 > ∠P QA so AQ > AP

∗ ∗∗

∈

⇒ ⇒ − −

− − − − ∗∗

⇒

80. Author: Author: Mateescu Mateescu Constantin Constantin If D is a point belonging to the segment [ BC ] and

(this can be easily proved by using the dot product i.e. a.s.o.)

2

2

∈ R+ then: AD2 = c 1++kbk − (1 + −−→ − − → − − → − − → AB + 2 AD = AD · AD, where AD =

BD =k DC

1+

BP b BQ = (by Angle Bisector Theorem) and = P C c QC a2 bc  2 = AP bc Read Free Foron 30this Days Sign up to vote title (b + c)2 c  Not useful , whence, by using the previous relation for D oneUseful has: P, Q  Cancel anytime. b b 3 + c3 a2 bc Special offer for students: Only $4.99/month. AQ2 = ( b + c) b+c

Returning to our problem, let’s observe that:

Master your semester with Scribd & The New York Times ∈{ }

  

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Indeed 16Rr

 

of 40

− 5r2 ≥ 14Rr − r2 ⇐⇒



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2Rr

≥ 4r2 ⇐⇒ R ≥ 2r ⇐⇒

Euler’s Inequality. Inequality.



Remark. The first mentioned identity can be proved like this: 8a2 b2 c2 cos A2 2bc 16Rr2 s2 A cos = = 2 la = 2 (a + b + c)(ab + bc + ca) abc b+c s + r2 + 2Rr



 

−

82. Author: gaussintraining Left Side. Since 2r 2 8Rr, the the ineq inequa ualit lity y is equi equiv valen alentt to 2s2 2r2 + 8Rr + 9R a2 = 2s2 compariso comparison n to Gerretsen’s Gerretsen’s Inequalit Inequality y i.e. s2 4R2 + 4Rr + 3r2 , we see that it is weak weak 2 2 2 2 2 9R + 8Rr + 2r 8R + 8Rr + 6r = R 4r2 , which follows from Euler’s Inequality. 



−

−

≥

≤

≤ ≥

⇒

Right Side. Again, since 2r 2 8Rr, the inequality is equivalent to s2 a2 = 2s2 r2 + 13Rr. Ag comparison to Gerretsen’s Inequality i.e s2 16Rr 5r2 , we see that it is weaker since 16 Rr r2 + 13Rr = 3Rr 6r2 , which again follows from Euler’s Inequality. 



−

⇒

−

≥

≥

≥

−

Author: r1234 83. Author: We prove it using complex numbers. Let z1 ,z2 ,z3 be the three vertices of the triangle ABC . (z z1 )(z z2 ) Now we consider the function g (z ) = . (z3 z1 )(z3 z2 ) We see that g(z1 ) = g (z2 ) = g (z3 ) = 1. Since Since this a two two degree degree polynomia polynomiall so we conclu conclu g(z ) = 1.



|−

|| − | | − || − |

≤

z1 z z2 = z1 z3 z2

− −

· ·

DA DB and hence the result follows. BC CA It can be checked that the equality holds when D is the orthocenter. 

So 1 = g (z )

z z3

− −



84. Author: Autho r: Mateescu Mateescu Constantin Constan Master your semester withtin Scribd Read Free Foron 30this Days Sign up to vote title Note that  I I I is acute-angled and I is its orthocenter. orthocenter. Thus, Thus, II = 2R & The New York Times  Useful  Not useful A A a b c

Special offer for students: sinceOnly R I $4.99/month. I I = 2R and r 3 a

b

c

a

∠I a

= 90◦ 3

−2

Cancel anytime.

we obtain: II a = 4R sin

2

I a I b I c



cos   I b I a



. The proposed proposed inequalit

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The function function f (x) =

x √cos sin x

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is concave upward for 0 < x <

Cauchy-Schwarz and Jensen inequality.


π

2

and therefore we are done



Author: KingSmasher3 KingSmasher3 86. Author: Left Side. For the left hand side of the problem, we have ( a + b + c)(ab + ac + bc) = a2 b + a2 c + ab2 + b2 c bc2 + 3abc. By Schur’s Inequality, RHS a3 + b3 + c3 + 3abc + 3abc = a3 + b3 + c3 + 6abc. 

≤

Right Side. For the right hand side of the problem, we use the fact that a,b,c are the sides of a triangle, so w a = x + y, b = x + z, c = y + z . Thus the inequality becomes (3 , 0, 0)+8(2, 1, 0)+18xyz > (3, 0, 0)+8(2, 1, 0)+10xyz , which is c true since x,y, z > 0.  87. Author: Author: applepi2000 applepi2000 Assuming F = D. Then it is equivalent with 4( ED )2 (BC )2 . Let AD = a,AE = b. Then by Law of Cosines, ( ED )2 = a2 + b2 2ab cos A. (BC )2 = 2(a + b)2 2(a + b)2 cos A Now note that we need 4( ED )2 (BC )2 0. Or, in other words 2 a2 + 2b2 4ab + (2a2 + 2b2 4ab)cos A 0. 2(a b)2 (1 + cos A) 0. This is true since cos A > 1. For equality equality to hold, we must must have D, E are the midpoints of AB,AC respectively. 

≥

−

−

−

≥

−

≥

−

−

≥

−

88. Author: chronondecay chronondecay

First assume that the triangle has an obtuse angle at A. It is well-known that A is also the orthoc of HB C , which is an acute triangle. triangle. Thus we we have BH BA,CA CH since ∠HAB, ∠HA obtuse. Thus we may swap H and A, and the LHS of the inequality decreases.

≥

≤

Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title Now assume that ABC is non-obtuse.   the feetTimes of altitudes from A, B be A , B respectively. Then & The NewLetYork Useful  Not useful Cancel anytime. 2[ABC ] AB AC sin A    Special offer for students: Only $4.99/month. = , AB = AC cos A, AH AA = AB AB  = AA = BC BC π

·

·

·

·

 

⇒

 

AH = cot A BC

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90. Author: Author: Mateescu Mateescu Constantin Constantin It is well-known well-known that: 3 QA2 + QB 2 + QC 2 = 9 QG2 + a2 + b2 + c2 . Therefore, Therefore, QA2 + 1 a2 + b2 + c2 , which is attained for Q = G.  QC 2 a2 + b2 + c2 , so the minimum is 3 3

 



·

≥ ·





·



91. Author: Author: BigSams BigSams Let m be the median with side lengths equal to the medians of Applying the reverse Hadwiger-Finsler Inequality to m , (ma mb )2 = 4 3S m + 6 m2a 4 3S m + 3 m2a 6





≤ ⇐⇒ 6 ·

√

  ·  ·



·√

√

−

. − ·

 ·





ma mb

 ≤ ·  · ·  ≤√ √     ≤ · · · ⇐⇒ ·   √  · ≤ · ·

4 3S m + 5 m2a 3 3 Note the identities S m = S and m2a = a2 . 4 4 3 3 6 8 ma mb 4 3 S + 5 a2 ma mb 4 3S + 5 4 4 2 1 2 8 ma a2 + 4 3S . Note that GA = ma .  3 3 3

⇐⇒ ⇐⇒

ma mb

·



a2

92. Author: Author: creatorvn creatorvn a2 + b2

− c2 + R2 ≥ 0 ⇐⇒ 2ab

R2 cos C + 2ab

The inequality is equivalent to 0 If cos C > 0 the problem has been solved. If not, then the ineq is equivalent to R2 2ab

≥

≥ − cos C = cos(A + B ) ⇐⇒ 2ab cos(A + B ) ≤ R2 1 π − sin A sin B sin A − B ≤ , which is true because 2 8 3 sin A + sin B + sin 2 − A − B ≤ sin A + B + 2 − A − B LHS ≤







3



π



π



3

3



=

1 . 8



Master your semester with Scribd  92. Author: Autho r: Virgil Nicula Read Free Foron 30this Days Sign up to vote title  & The NewLetYork Times Useful Not useful   the reflection reflection w.r.t. the midpoin midpointt M of [BC ], i.e. i.e. ABPC is a para parall llelo elogr gram am  P of A w.r.t. 2 4 OB M B 2 = 4 OM 2 = Special offer for students: Only $4.99/month. 2 OA2 + OP 2 AP 2 = 4R2



−

−

·

⇒

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− a2 = 2



R2 + OP 2

−

4 m2

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We will rewrite the whole inequality in terms of R,r,s by using the identities: ab + bc +ca = s2 + and abc = 4 Rrs 16Rrs2 + 36R2 r2 s2 + r 2 + 4Rr s2 + r2 36R2 r 2 4Rr3 r4 s4 + s2 2r2 12Rr



⇐⇒ ⇐⇒ ⇐⇒

 ≥ −  ≥ ≥

−

2 2

−

2 2

− 6Rr + r 6Rr − r + 36R r − 4Rr3 − r4 2 s2 − 6Rr + r2 ≥ 72R2 r 2 − 16Rr3 . 2 By Gerretsen’s Inequality i.e. s2 ≥ 16Rr − 5r2 , one gets: s2 − 6Rr + r2 ≥ 10Rr − 4r2 2 Thus it suffices to prove the following inequality 10Rr − 4r2 ≥ 72R2 r2 − 16Rr 3 which redu the obvious one: (R − 2r) (7R − 2r) ≥ 0. Equality holds iff  ABC is equilateral. 



2

 

s

2 2





 



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94. Author: Author: creatorvn creatorvn 2

 − ≤  − ⇐⇒  − −  ≤ −      − − − ≤ − − −  ≤ − a2 ab

2r s2 3r2 12Rr 1 1 1 2 2 R s + r + 4Rr R 2 2 2 2 2 2 2 4R + 3r + 4Rr 3r 12Rr 2Rr 1 2t R r = = where t = LHS 2 2 2 2 16Rr 5r + r + 4Rr 5Rr r 5t t R 2 1 2t 1 We need to prove (1 2 ), which is true, since Euler’s Inequality states . t t 5t t2 2 2r



− −

≤

−

95. Author: Author: creatorvn creatorvn 2∆

2∆

ac

ab

ab

ac

s(s

+ (s−c)(s−a) (s−b)(s−a)

 − ⇐⇒ −  ⇐⇒ − (s

b) (s s a

− c)

  −

≥4

1

b

b)

+

−b)(s−c) bc

c b (s

 ⇐⇒ − −  ≥ √  −  ≥ √  − −  − −  2∆

(s

  −  −− c (s

−a)

bc

− c) b c + s (s b) (s − c) c (s b) b (s − c) √ By AM-GM, s ≥ bc + (s b) (s − c) and Master your semester with Scribd

Multiplying them yields the necessary result.


Special offer for students: Only 96. Author: Autho r: $4.99/month. luisgeometra luisgeometra

s

2

2

(s



b

+

a

c (s

b)

bc +

(s s (s

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bc +

b) (s

(s

c b (s

− c)

b) (s

c)

− c)

b

− c) +



 ≥ √   − 4

c

c (s b) b (s c) Read Free Foron 30this Days Sign up to vote title



−

−


Useful

s

(

bc

≥

2  

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For convenience, let us denote

 

  

=



−→  −→  −→

−−→ −→

bBA + cCA IA = 2s

, where k, q

CN =q AN

∈ R+. Note:

  −−→ −−→ −−→




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BM =k AM

−−→ + aAB −−→ −→ + k · IA −→ cCB −IM −→ = IB = 1+k −−→ −−→ (a − kb − kc) AB + (−c − kc ) BC


+ k bBA + c CB + BA

−−→ −−→

cCB + aAB IB = 2s

−→ −−→

aAC + bBC IC = 2s

2s (1 + k)

2s (1 + k )

−−→ + BC −−→ + bBC −−→ + q bBA −−→ + c CB −−→ + BA −−→ −→ + q · IA −→ a AB −IN → = IC = 1+q 2s (1 + q) − − → − − → (a − qb − qc) AB + (a + b − qc) BC = 2s (1 + q) −−→ −→ Therefore, the colinearity of vectors IM and IN implies: (a − kb − kc ) (a + b − qc) = (−c − kc) a − bk a2 a − bk ≥ which after expanding is equivalent to q = . The inequality becomes: k· 4bc c c ⇐⇒ a2 ≥ 4bk (a − bk) ⇐⇒ a2 + 4k2b2 ≥ 4abk ⇐⇒ (a − 2kb)2 ≥ 0, which is clearly true. MB a N C a Equality Equality is attained attained iff a = 2 k · b i.e. = and = .  2b 2c AM AN





 





97. Author: Author: Virgil Nicula

Lemma. Let d be a line, three points A , B , C anotherr line d and a point P d. For anothe note intersections K , L, M of δ with the lines P A, P B , P C respective respectively ly.. Prove Prove that there

{

relation

}⊂

∈

LA

MB N C · · · AB = 0. BC + CA + LP M P N P

LA  , d  d. Deno Master your semester with · BC Proof. Let d for which Denote te X ∈ d ∩ δ , Y ∈ d ∩ δ. Thus Thus,, P ∈ dScribd Read Free Foron 30this Days Sign up to vote title LP  N C Times AX & The NewCAYork useful ·AB = 0 ⇐⇒ ·BC + BX ·CA + CX ·AB =0 Useful ⇐⇒ AX  ·BC Not ·CA +CX ·AB + +BX N P Special offer for students: Only $4.99/month.

Denote D

AI

P Y

P Y

∩ BC and apply the lem

P Y

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MB

DC +

N C

BD

ID

BC

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 



·

· √ ≥ · ⇐⇒ · √ 2 (a2 + b2 + c2 ) + 4 3S

    · ≥

⇐⇒

2 3

ma

 



a2 . ma mb

. Note that

3


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3 3 Note the identities S m = S and m2a = 4 4 3 4 2 2 ma mb a + 3 3S 4 3

⇐⇒ ·


≥ 12 ·





2 3

GA =

√

a2 + 2 3S

·



ma .



Author: Virgil Nicula 99. Author: Denote the midpoint M of [BC ] and N

∈ AS ∩ BC .

Is wellwell-kno known wn that

AS SN AN = 2 = 2 . 2 +c a a + b2 + c2 Denote AM = ma , AN = sa , m  BAN = m(  CAM = φ.

Apply van Aubel’s relation to S

b2

            ⇒     M AC

Apply the Sine Law to : (

+c2 ) a2 +b2 +c2 2ma 3 2

=

⇒

=

⇒

M C ma = sin φ sin C

=

N AB

sa b

sa NB = sin B sin φ

3 b2 + c2 AS = = 2 (a2 + b2 + c2 ) AG 3(ab + bc + ca) AS = 3.  AG a2 + b2 + c2



NB N C = 2 = 2 c b

·

2bc sa = 2 ma b + c2

3 b 2 + c2 sa = 2 (a2 + b2 + c2 ) ma

· b22+bcc2 = a2 +3bbc2 + c2 a.s.o.

≤

100. Author: Author: Mateescu Mateescu Constantin Constantin Note that the inequality can be written as: a2 + m2 b2 m cos φ a2 + b2 c2 + 4m sin φ a2 + b2 c2 = 2 ab cos C And since 2∆ = ab sin C Our inequality becomes: a2 + m2 b2 2ab cos C m cos φ + 2ab sin C m sin φ  2 2 2 2 2 Read Free For2on 30this Days 2abm (cos C cos φ + sin C sin φ) cos(C φ), a +m b a2 +up m ab mtitle Sign tob vote  which which is obviously obviously true b ecause ecause a2 + m2 b2 2abm 2abmUseful C φ).Not useful  cos(  Cancel anytime. Equality occurs if and only if a = m b and φ = C .  Special offer for students: Only $4.99/month.

 

−

≥

≥ · Master your semester with Scribd ⇐⇒ ≥ · ⇐⇒ ≥ ≥ & The New York Times ·

·

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· ≥ −

·

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·

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Geometric Inequalities Marathon - The First 100 Problems and Solutions

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