Geometric inequalities marathon the first 100 problems and solutions

Geometric Inequalities Marathon 1 The First 100 Problems and Solutions Contributors Typesetting and Editing Members of Mathlinks Samer Seraj (BigSams) 1 Preface On Wednesday, April 20, 2011, at 8:00 PM, I was inspired by the existing Mathlinks marathons to create a marathon on Geometric Inequalities - the fusion of the beautiful worlds of Geometry and Multivariable Inequalities. It was the result of the need for expository material on GI techniques, such as the crucial Rrs, which were well-explored by only a small fraction of the community. Four months later, the thread has over 100 problems with full solutions, and not a single pending problem. On Friday, August 26, 2011, at 5:30 PM, I locked the thread indefinitely with the following post: The reason is that most of the known techniques have been displayed, which was my goal. Recent problems are tending to to be similar to old ones or they require methods that few are capable of utilizing at this time. Until the community is ready for a new wave of more diffcult GI, and until more of these new generation GI have been distributed to the public (through journals, articles, books, internet, etc.), this topic will remain locked. This collection is a tribute to our hard work over the last few months, but, more importantly, it is a source of creative problems for future students of GI. My own abilities have increased at least several fold since the exposure to the ideas behind these problems, and all those who strive to find proofs independently will find themselves ready to tackle nearly any geometric inequality on an olympiad or competition. The following document is dedicated to my friends Constantin Mateescu and R´eda Afare (Thalesmaster), and the pioneers Panagiote Ligouras and Virgil Nicula, all four of whom have contributed much to the evolution of GI through the collection and creation of GI on Mathlinks. The file may be distributed physically or electronically, in whole or in part, but for and only for non- commercial purposes. References to problems or solutions should credit the corresponding authors. To report errors, a Mathlinks PM can be sent BigSams, or an email to samer_seraj@hotmail.com. Samer Seraj September 4, 2011 1 The original thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=403006/ 1 2 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 (∠BCA) 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 arbitrary point in the plane. Let the cevians 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 a , d b , d c respectively. • Let arbitrary cevians issued from A, B, C be d, e, f respectively. • Let the semiperimeter, inradius, and circumradius be s, r, R respectively. • Let the heights issued from A, B, C be h a , h b , h c respectively, which meet at the orthocenter H. • Let the feet of the perpendiculars from H to BC, CA, AB be H a , H b , H c respectively. • Let the medians issued from A, B, C be m a , m b , m c respectively, which meet at the centroid G. • Let the midpoints of A, B, C be M a , M b , M c respectively. • Let the internal angle bisectors issued from A, B, C be l a , l b , l c respectively, which meet at the incenter I, and intersect their corresponding opposite sides at L a , L b , L c respectively. • Let the feet of the perpendiculars from I to BC, CA, AB be Γ a , Γ b , Γ c respectively. • Let the centers of the excircles tangent to BC, CA, AB be I a , I b , I c respectively, and the excircles be 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 a , r b , r c respectively. • Let the symmedians issued from A, B, C be s a , s b , s c respectively, which meet at the Lemoine Point S, and intersect their corresponding opposite sides at S a , S b , S c respectively. • Let Γ be the Gergonne Point, and the Gergonne cevians through A, B, C be g a , g b , g c respectively. • Let N be the Nagel Point, and the Nagel cevians through A, B, C be n a , n b , n c respectively. Let [X] denote the area of polygon X. All  and  symbols without indices are cyclic.  denotes the end of a proof, either for a lemma or the original problem. 2 3 Problems 1. For ABC, prove that R ≥ 2r. (Euler’s Inequality) 2. For ABC, prove that  AB >  P A. 3. For ABC, prove that ab + bc + ca 4∆ 2 ≥  1 s(s −a) . 4. For ABC, prove that r(4R + r) ≥ √ 3∆. 5. For ABC, prove that cos B −C 2 ≥  2r R . 6. For ABC, prove that  12(R 2 − Rr + r 2 ) ≥  AI ≥ 6r. 7. A circle with center I is inscribed inside quadrilateral ABCD. Prove that  AB ≥ √ 2 ·  AI. 8. For ABC, prove that 9R 2 ≥  a 2 . (Leibniz’s Inequality) 9. Prove that for any non-degenerate quadrilateral with sides a, b, c, d, it is true that a 2 + b 2 + c 2 d 2 ≥ 1 3 . 10. For ABC, prove that 3 ·  a sin A ≥   a  ·   sin A  ≥ 3(a sin C + b sin B + c sin A). 11. For acute ABC, prove that  cot 3 A + 6 ·  cot A ≥  cot A. 12. For ABC, prove that   cos A 2  ·   csc A 2  ≥ 6 √ 3 +  cot A 2 . 13. A 2-dimensional plane is partitioned into x regions by three families of lines. All lines in a family are parallel to each other. What is the least number of lines to ensure that x ≥ 2010. (Toronto 2010) 14. For ABC, prove that 3 √ 3R ≥ 2s. 15. For ABC, prove that  1 2 −cos A ≥ 2 ≥ 3 ·  1 5 −cos A . 16. For ABC, prove that 1 8 ≥  sin A 2 . 17. In right-angled ABC with ∠A = 90 ◦ , prove that 3 √ 3 4 · a ≥ h a + max{b, c}. 18. For ABC, prove that s ·  h a ≥ 9∆ with equality holding if and only if ABC is equilateral. 19. Prove that the semiperimeter of a triangle is greater than or equal to the perimeter of its orthic triangle. 20. Prove that of all triangles with same base and area, the isosceles triangle has the least perimeter. 21. ABCD is a convex quadrilateral with area 1. Prove that AC + BD +  AB ≥ 4 + 2 √ 2. 22. For ABC, prove that  csc A 2 ≥ 4  R r . 23. For ABC, prove that  sin 2 A 2 ≥ 3 4 . 3 24. Of all triangles with a fixed perimeter, dtermine the triangle with the greatest area. 25. Let ABCD be a parallelogram such that ∠A ≤ 90. Altitudes from A meet BC, CD at E, F respectively. Let r be the inradius of CEF . Prove that AC ≥ 4r. Determine when equality holds. 26. For ABC, the feet of the altitudes from B, C to AC, AB respectively, are E, D respectively. Let the feet of the altitudes from D, E to BC be G, H respectively. Prove that DG + EH ≤ BC. Determine when equality holds. 27. For ABC, a line l intersects AB, CA at M, N respectively. K is a point inside ABC such that it lies on l. Prove that ∆ ≥ 8 ·  [BM K] + [CN K]. 28. For ABC, prove that  15 4 +  cos(A −B) ≥  sin A. 29. Let p I be the perimeter of the Intouch/Contact Triangle of ABC. Prove that p I ≥ 6r  s 4R  1 3 . 30. In addition to ABC, let A  B  C  be an arbitrary triangle. Prove that 1 + R r ≥  sin A sin A  . 31. For ABC, prove that  cos 2 B −C 2 ≥ 24 ·  sin A 2 . 32. For ABC, prove that  h a ≥ 9r. 33. For ABC, prove that  cos A −B 2 ≥  sin 3A 2 . 34. For ABC, prove that  sin 2 A 2 +  cos B −C 2 ≥ 1. 35. For ABC, AO, BO, CO are extended to meet the circumcircles of BOC, COA, AOB respec- tively, at K, L, N respectively. Prove that AK OK + BL OL + CM OM ≥ 9 2 . 36. For ABC, prove that 9abc a + b + c ≥ 4 √ 3∆. 37. For ABC, prove that  a 2 b(a −b) ≥ 0. 38. Show that for all 0 < a, b < π 2 we have sin 3 a sin b + cos 3 a cos b ≥ sec(a −b) 39. For all parallelograms with a given perimeter, explicitly define those with the maximum area. 40. Show that the sum of the lengths of the diagonals of a parallelogram is less than or equal to the perimeter of the parallelogram. 41. For ABC, the parallels through P to AB, BC, CA meet BC, CA, AB respectively, at L, M, N respectively. Prove that 1 8 ≥ AN NB · BL LC · CM MA . 42. For ABC, prove that  a sin A 2 ≥ s 43. For ABC, it is true that BC = CA and BC ⊥ CA. P is a point on AB, and Q, R are the feet of the perpendiculars from P to BC, CA respectively. Prove that regardless of the location of P , max{[AP R], [BP Q], [P QCR]} ≥ 4 9 ∆. (Generalization of Canada 1969) 4 44. For ABC, prove that  a 2 + abc √ 3R ≥ 4(abc) 2 3 . 45. For ABC, prove that 6R ≥  a 2 + b 2 m 2 c . 46. For a convex hexagon ABCDEF with AB = BC, CD = DE, EF = F A, prove that BC BE + DE DA + F A F C ≥ 3 2 . Determine when equality holds. 47. For ABC, prove that s √ 3 ≥  l a . 48. For ABC, prove that R −2r ≥ 1 12 ·  2 ·  m a −  ab R  . 49. For ABC, prove that  a 2 ≥ 4 √ 3∆ ·max  m a h a , m b h b , m c h c  . 50. A 1 A 2 B 1 B 2 C 1 C 2 is a hexagon with A 1 B 2 ∩C 1 A 2 = A, B 1 C 2 ∩A 1 B 2 = B, C 1 A 2 ∩B 1 C 2 = C and AA 1 = AA 2 = BC, BB 1 = BB 2 = CA, CC 1 = CC 2 = AB. Prove that [A 1 A 2 B 1 B 2 C 1 C 2 ] ≥ 13 ·[ABC]. 51. For ABC, let r 1 , r 2 denote the inradii of ABM a , ACM a . Prove that 1 r 1 + 1 r 2 ≥ 2 ·  1 r + 2 a  . 52. For ABC, prove that  csc 2 A 2 ≥  cos(A −B) + 9 ≥ 8 ·  cos A. 53. For ABC, find the minimum of the expression 2s 4 −  a 4 ∆ 2 . 54. For ABC, prove that √ 3 2 ·  cos B −C 4 ≥  cos A 2 . 55. For ABC, prove that 3 ·  a 2 > ∆ ·   cot A 2  2 . 56. For ABC, c ≤ b ≤ a. Through interior point P and the vertices A, B, C, lines are drawn meeting the opposite sides at X, Y, Z respectively. Prove that AX + BY + CZ < 2a + b. 57. For ABC, prove that s 3 2abc ≥  cos 4 A 2 . 58. For ABC, let P A = x, P B = y, P C = z. Prove that ayz + bzx + cxy ≥ abc, with equality holding if and only if P ≡ O. (China 1998) 59. For ABC, prove that 3 ·  d 2 a ≥  P A 2 sin 2 A. 60. For ABC, if CA + AB > 2 ·BC, then prove that ∠ABC + ∠ACB > ∠BAC. (Euclid Contest 2010) 61. For ABC, prove that  7 ·  a 2 + 2 ·  ab 2 ≥  m a . (Dorin Andrica) 62. For ABC, prove that  cos A 2 ≥ √ 2 2 +  1 2 + (3 √ 3 −2 √ 2) · s 2R . 63. For ABC, prove that   a 2 b 2 2∆ ≥ max  a b + b a , b c + c b , c a + a c  . 5 64. For ABC, prove the following and determine which is stronger: (Samer Seraj) (a) ∆ ≥ r ·  1 3 ·  m a m b + 1 2 ·  ab. (b) ∆ ≥ r ·  2 3 ·  m a m b + r(r + 4R). 65. For any convex pentagon A 1 A 2 A 3 A 4 A 5 , prove that 5  i=1 (A i A i+2 + A i+1 A i+4 ) > 5  i=1 A i A 2 i 2 . A i+5 ≡ A i . 66. For ABC, prove that s 2 ≥  l 2 a . 67. ABCD is a quadrilateral inscribed in a circle with center O. P is the intersection of its diagonals and R is the intersection of the segments joining the midpoints of the opposite sides. Prove that OP ≥ OR. 68. For ABC, prove that 5 4 ·  bc >  m b m c . 69. For ABC, let M ∈ [AC], N ∈ [BC] and L ∈ [MN], where [XY ] denotes the line segment XY . Prove that: 3 √ ∆ ≥ 3  S 1 + 3  S 2 , where S 1 = [AML] and S 2 = [BN L]. 70. For ABC, prove that  (b + c)P A ≥ 8∆. 71. Right ABC has hypotenuse AB. The arbitrary point P is on segment CA, but different from the vertices A, C. Prove that AB −BP AP > AB −BC CA . 72. For ABC, prove that max  BP AC , CP AB  ≥ √ 2 −1. 73. For ABC, prove that  a 2 s −a ≥ 6 √ 3R. 74. Let P be a point inside a regular n-gon, with side length s, situated at the distances x 1 , x 2 , . . . , x n from the sides, which are extended if necessary. Prove that n  i=1 1 x i > 2π s . 75. A point A is taken inside an acute angle with vertex O. The line OA forms angles α and β with the sides of the angle. Angle φ is given such that α + β + φ < π. On the sides of the former angle, find points M and N such that ∠MAN = φ, and the area of the quadrilateral OM AN is maximal. 76. For ABC, find the smallest constant k such that it always holds that k ·  ab >  a 2 . 77. For ABC, prove that  abd a d b ≤ 4 3 ∆ 2 , and determine when equality holds. 78. For ABC, let AI, BI, CI extended intersect the circumcircle of ABC again at X, Y, Z respectively. Prove that  IX ≥  AI. 79. Let {a, b, c} ⊂ R + such that  a 2 + b 2 − c 2 ab > 2. Prove that a, b, c are sides of triangle. 80. Let AP be the internal angle bisector of ∠BAC and suppose Q is the point on segment BC such that BQ = P C. Prove that AQ ≥ AP . 81. For ABC, prove that ∆ 2 ≥ r ·  l a . 6 82. For ABC, prove that 9R 2 ≥  a 2 ≥ 18Rr. 83. For ABC, prove that  (P A · P B ·c) ≥ abc. 84. For ABC, prove that 8R 3 ≥  IE a . 85. For ABC, prove that   sin A 2  ·   tan A 2  ≥ 3 √ 3 2 . 86. For ABC, prove that  a 3 + 6abc ≥   a  ·   ab  >  a 3 + 5abc. 87. D and E are points on congruent sides AB and AC, respectively, of isosceles ABC such that AD = CE. Prove that 2EF ≥ BC. Determine when the equality holds. 88. For ABC, prove that  AH a ≥ 3 √ 3. 89. Let M, A 1 , A 2 , ··· , A n (n ≥ 3), be distinct points in the plane such that A 1 A 2 = A 2 A 3 = ···A n−1 A n = A n A 1 . Prove that n−1  i=1 1 MA i · MA i+1 ≥ 1 MA 1 · MA n . 90. For ABC, determine min   QA 2  . 91. For ABC, prove that  8 ·  a 2 + 4 √ 3∆ 3 ≥  GA. 92. For ABC, prove that a 2 + b 2 + R 2 ≥ c 2 , and determine when equality holds. 93. For ABC, prove that   ab  · (s 2 + r 2 ) ≥ 4abcs + 36R 2 r 2 . 94. For ABC, prove that  a 2  ab ≥ 1 +  1 − 2r R . 95. For ABC, prove that sin B sin 2 C 2 + sin C sin 2 B 2 ≥ 4 cos A 2 1 −sin A 2 . 96. In ABC, the internal angle bisectors of angles A, B, C intersect the circumcircle of ABC again at X, Y, Z respectively. Prove that AX + BY + CZ > a + b + c. (Australia 1982) 97. An arbitrary line  through the incenter I of ABC cuts AB and AC at M and N. Show that a 2 4bc ≥ BM AM · CN AN . 98. For ABC, prove that  GA ≥  2 ·  a 2 + 4 √ 3∆ 3 . (A sequel to Problem 91) 99. For ABC, prove that 3 ≥  SA GA . 100. Let m ∈ R + and φ ∈ (0, π). For ABC, prove that (1 −m cos φ) · a 2 + m (m −cos φ) · b 2 + m cos φ · c 2 ≥ 4m sin φ ·∆ Equality holds if and only if m = a b and φ = C. For m = 1 and φ = 60 ◦ obtain Weitzenb¨ock’s Inequality. (Virgil Nicula) 7 4 Solutions 1. Euler’s Original Proof R(R − 2r) = OI 2 ≥ 0 ⇐⇒ R ≥ 2r.  1. Author: tonypr Rewrite the inequality as 1 + r R ≤ 3 2 . Then note the identity 1 + r R = cos A + cos B + cos C. So it is sufficient to prove that 2 cos A + 2 cos B + 2 cos C ≤ 3. It’s easy to verify that this inequality is equivalent to (1 − (cos B + cos C)) 2 + (sin B − sin C) 2 ≥ 0, which is true by the Trivial Inequality.  1. Author: BigSams For positive reals x, y, z it is true that (x + y)(y + z)(z +x) ≥ 8xyz by AM-GM:  x + y 2 ≥  √ xy = xyz. By Ravi Substitution, let a, b, c be side lengths of a triangle such that a = x+y, b = y+z, c = z+x. The inequality becomes abc ≥ 8(s − a)(s − b)(s − c). By Heron’s Theorem, the inequality is sabc ≥ 8S 2 ⇐⇒ abc 4∆ ≥ 2∆ s . Using the fact that ∆ = abc 4R = sr, R ≥ 2r.  1. Author: BigSams Note that  r a = 4R + r and  1 r a = 1 r . By CS, 4R + r r =   r a  ·   1 r a  ≥ 9 ⇐⇒ R ≥ 2r.  2. Author: 1=2 Lemma. AB + AC > P B + BC Proof. Let the extension of BP intersect AC at N. Then the triangle inequality gives us P N + NC > P C AB + AN > BN = BP + PN Adding NC to both sides of the second inequality gives AB +AN +N C > BP +P N +N C > P B +P C. Note that AN + NC = AC, since N is on AC. Therefore AB + AC > P B + P C.  This lemma implies that                AB + AC > P B + PC BA + BC > P A + P C CA + CB > P A + P B If we add all three inequalities together, we get 2(AB +BC + AC) > 2(P A + P B +P C), which implies the desired result.  8 3. Author: Goutham Let x = s −a, y = s −b, z = s −c all greater than 0, and s = x + y + z, ∆ 2 = xyzs. We have  x 2 ≥  xy =⇒  (x 2 + 3xy) ≥ 4  xy. But  (x 2 + 3xy) =  (x + y)(x + z) =  ab. And so,  ab 4xyzs ≥  xy xyzs . Therefore, we have  ab 4∆ ≥  1 s(s −a) .  4. Author: Mateescu Constantin Using the well-known formula for area i.e. ∆ = sr, the inequality rewrites as: s √ 3 ≤ 4R + r (∗). Of course, this is weaker than Gerretsen’s Inequality i.e. s 2 ≤ 4R 2 + 4Rr + 3r 2 , since the inequality: 4R 2 + 4Rr + 3r 2 ≤ (4R + r) 2 3 reduces to Euler’s inequality i.e. R ≥ 2r. However, there is also a simple method to obtain directly the inequality (∗). In the well known inequality: 3(xy + yz + zx) ≤ (x + y + z) 2 we take:           x = (s −b)(s −c) y = (s −c)(s − a) z = (s −a)(s −b)           and thus we obtain: 3s(s −a)(s −b)(s − c) ≤ [r(4R + r)] 2 , whence √ 3∆ ≤ r(4R + r) ⇐⇒ (∗).  5. Author: Thalesmaster Note the identities                        cos B −C 2 = cos B 2 cos C 2 + sin B 2 sin C 2 cos A 2 =  (s −b)(s −c) bc sin B 2 =  s(s −a) bc and          r = ∆ s R = abc 4∆ Using Ravi’s substitution:      a = y + z b = z + x c = x + y , the inequality is equivalent to: (2x + y + z) 2 ≥ 8x(y + z), which is true according to AM-GM Inequality.  6. Author: FantasyLover Right Side. Let (I) meet sides AB, BC, CA at P, Q, R, respectively. Furthermore, denote by a, b, c the lengths of AR, BP, CQ. The given inequality is equivalent to  a 2 + r 2 +  b 2 + r 2 +  c 2 + r 2 ≥ 6r. On the other hand, r(a + b + c) =  abc(a + b + c) ⇐⇒ r =  abc a + b + c from Heron’s Formula. Hence, it suffices to prove   a 2 + abc a + b + c ≥ 6  abc a + b + c ⇐⇒   a(a + b)(a + c) ≥ 6 √ abc. However, using AM-GM Inequality twice gives   a(a + b)(a + c) ≥ 3 6  abc(a + b) 2 (b + c) 2 (c + a) 2 ≥ 9 3 6  abc ·64(abc) 2 ≥ 6 √ abc, as desired.  Left Side. Lemma. AI + BI + CI ≤ 2(R + r) (Author: Mateescu Constantin) Proof. Show easily that AI = bc s · cos A 2 = 1 s · √ bc ·  s(s −a) a.s.o. Thus, we have:   AI  2 = 1 s 2 ·   √ bc ·  s(s −a)  2 C.B.S. ≤ 1 s 2 · (ab + bc + ca) ·  s(s −a) = ab + bc + ca ≤ 4(R + r) 2 . The last inequality is due to Gerretsen i.e. s 2 ≤ 4R 2 + 4Rr + 3r 2 . Therefore, we have shown that: AI + BI + CI ≤ 2(R + r) .  As a direct consequence of the lemma, it suffices to prove 2(R + r) ≤ 2  3(R 2 − Rr + r 2 ) ⇐⇒ 2R 2 − 5Rr + 2r 2 ≥ 0. However, ths is equivalent to (2R −r)(R − 2r) ≥ 0, which is indeed true.  For both inequalities, equality holds for ABC equilateral. 6. Author: Thalesmaster Left Side. Note that:      AI 2 = bc −4Rr BI 2 = ca −4Rr CI 2 = ab −4Rr According to C.S Inequality: 3(AI 2 + BI 2 + CI 2 ) ≥ (AI + BI + CI) 2 ⇐⇒  3(s 2 + r 2 − 8Rr) ≥ AI + BI + CI So it suffices to show that  3(s 2 + r 2 − 8Rr) ≤  12(R 2 − Rr + r 2 ) ⇔ s 2 + r 2 + 8Rr ≤ 4R 2 − 4Rr + 4r 2 ⇔ s 2 ≤ 4R 2 + 4Rr + 3r 2 , which is the Gerretsen Inequality.  6. Author: tonypr Right Side. Note that AI = r sin A 2 . Applying this cyclically to BI and CI, the left hand side is equivalent to 6r ≤ r sin A 2 + r sin B 2 + r sin C 2 ⇐⇒ 2 ≤ 1 sin A 2 + 1 sin B 2 + 1 sin C 2 3 2 ≤ csc A 2 + csc B 2 + csc C 2 3 ⇐⇒ csc  A + B + C 6  ≤ csc A 2 + csc B 2 + csc C 2 3 which follows from Jensen’s Inequality since csc x 2 is convex for x ∈ (0, π).  7. Author: BigSams It is well-known that ∠AID + ∠BIC = 180 ◦ . There are two implications: sin ∠AID = sin ∠BIC and cos ∠AID = −cos ∠BIC. Let r be the inradius. [AID] = sin ∠AID ·AI ·DI 2 = AD · r 2 =⇒ AI · DI AD = r sin ∠AID . Similarly, BI ·CI BC = r sin ∠BIC . 10 [...]... AM = ∠N AN and hence M A < N A 32 Thus, [M AM ] < [N AN ] ⇒ [OM AN ] < [OM AN ] So we have to find out on what conditions we can find on the sides on the sides of the angle points M and N such that ∠M AN = φ and M A = AN Circumscribe a circle about the triangle M ON Since α + β + φ < π, the point A is located outside the circle If L is the point of intersection of OA π−φ π−φ and the circle, then: ∠AM... y = z = 1, in other words, when the points M , N and L are the midpoints of the segments [AC], [BC] and [M N ] respectively 70 Author: Goutham 30 Let P1 be the symmetric of point P w.r.t the midpoint of side [BC] Define P2 and P3 in a similar 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], with equality if and only if M N P Q is cyclic and M P ⊥ N Q... sides of the initial triangle, the two reflections are collinear with the two remaining vertices of the orthic triangle - which is easy to prove: ∠M P N = π − 2∠C ∧ ∠M P B = ∠C Therefore the triangle obtained by the argument (∗) is indeed the orthic triangle, as claimed Using the lemma, the orthic triangle does not have a greater perimeter than the medial triangle, which has a perimeter equal to the semiperimeter... BC · sin B ≤ · AB · BC similarly we get [ABCD] ≤ · CD · DA on 2 2 2 1 1 the other hand we get other two inequalities [ABCD] ≤ · AB · CD and [ABCD] ≤ · BC · AD 2 2 Adding the last four inequalities we get(AB + CD)(BC + DA) ≥ 4 This implies that (AB + BC + 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 ≥ √or AC + BD ≥ 2 2 8 Adding we get... x < and therefore we are done using 2 sin x Cauchy-Schwarz and Jensen inequality 86 Author: KingSmasher3 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 + ac2 + 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... that the parallelogram is not a rectangle Then putting it on its base and straightening its slanted side will increase the height, and keep the base constant Thus, the greatest area must be a rectangle Now, we must maximize ab given 2(a + b) By AM-GM we know this is maximized when a = b Thus, the figure is a square 20 40 Author: KrazyFK Clearly AC ≤ AB + BC and AC ≤ CD + DA We have two similar inequalities. .. Thalesmaster We use the well-known inequality ma + mb + mc ≤ 4R + r and the identity ab + bc + ca = s2 + r2 + 4Rr s2 + r2 + 4Rr Then, we just have to show that: 2(4R + r) − ≤ 12(R − 2r) ⇐⇒ s2 + r2 + 4R2 ≥ 22Rr R Which immediately follows by summing up the knows results s2 + r2 ≥ 14Rr and 4R2 ≥ 8Rr 23 49 Author: BigSams Lemmata 2b2 + 2c2 − a2 , and the cyclic versions hold as well 4 1 a2 (2) 2 = , and the cyclic... from the Trivial and Triangle Inequslities (BigSams used the same method in his submission to the Mathematical Reflections bi-monthly journal, where the problem was originally from) 62 Author: Thalesmaster √ √ √ 2 1 A A + + 2(3 3 − 2 2) cos The inequality is equivalent to cos ≥ 2 2 2 2  X = π − A    2  π−B Use the substitution: Y =  2    Z = π − C 2 Denote s, R, r the semi-perimeter, the circumradius... holds if and only if a = b = c, which is derived from the CS equality condition = 19 Author: Goutham Lemma In ABC, M, N, P are points on sides BC, CA, AB respectively such that perimeter of the M N P is minimal Then M N P is the orthic triangle of ABC (Author: Farenhajt) Proof Let M be an arbitrary point on BC, and M and M its reflections about AB and AC respectively Then, for a given M , the points... Construct the lines passing through the vertices of triangle ABC so that they are parallel to the sides BC, CA and AB respectively The intersection of these three lines determines a new triangle A B C , where A is the midpoint of segment B C Thus, AP = BC = AB = AC , so B P C = 90◦ Now it follows that: A P C + B P A = 270◦ , wherefrom one has either B P A ≤ 135◦ or A P C ≤ 135◦ Let us consider the first . inspired by the existing Mathlinks marathons to create a marathon on Geometric Inequalities - the fusion of the beautiful worlds of Geometry and Multivariable Inequalities. It was the result of the need. Geometric Inequalities Marathon 1 The First 100 Problems and Solutions Contributors Typesetting and Editing Members of Mathlinks Samer Seraj (BigSams) 1. O. The line OA forms angles α and β with the sides of the angle. Angle φ is given such that α + β + φ < π. On the sides of the former angle, find points M and N such that ∠MAN = φ, and the