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Junior Problems J301. Let a and B Be Nonzero Real Numbers Such That Ab ≥ + 1 + 3. Prove That Ab

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Junior Problems J301. Let a and B Be Nonzero Real Numbers Such That Ab ≥ + 1 + 3. Prove That Ab Junior problems 1 1 J301. Let a and b be nonzero real numbers such that ab ≥ a + b + 3. Prove that 1 1 3 ab ≥ p + p : 3 a 3 b Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Arkady Alt, San Jose, California, USA p 1 1 1 1 Let x := 3 ab; y := − p ; z := − p ; then, xyz = 1 and ab ≥ + + 3 becomes equivalent with 3 a 3 b a b x3 + y3 + z3 ≥ 3: However, (x + y + z) x2 + y2 + z2 − xy − yz − zx = x3 + y3 + z3 − 3xyz = x3 + y3 + z3 − 3 ≥ 0 and x2 + y2 + z2 ≥ xy + yz + zx ; therefore, x + y + z ≥ 0; which implies 1 1 3 x ≥ − (y + z) () x3 ≥ (−y − z)3 () ab ≥ p + p : 3 a 3 b Also solved by Polyahedra, Polk State College, USA, Himansu Mookherjee, Kolkata, India; Radouan Bo- ukharfane, Morocco; Daniel Lasaosa, Pamplona, Spain; David Yang, Bergen County Academies, NJ, USA; Debojyoti Biswas, Kolkata, India; Eliott S. Kim, The Lawrenceville School, NJ, USA; Jaesung Son, Rid- gewood, NJ, USA; Jishnu Bose,Indian Statistical Institute, Kolkata, India; Kevin Ren; Corneliu M«nescu- Avram, Transportation High School, Ploie¸sti,Romania; Nicu¸sorZlota, “Traian Vuia” Technical College, Foc¸sani,Romania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Peter C.Shim, Pingry School, Basking Ridge, NJ, USA; Woosung Jung, Korea International School, South Korea; Hyun- seo Yang, Daecheong Middle School, Seoul, South Korea; Joshua An, Washington University in St. Louis, MO, USA; Yeonjune Kang, Peddie School, Hightstown, NJ, USA; Seung Hwan An, Taft School, Watertown, CT, USA; Joe Hong, Seoul International School, South Korea; Seong Kweon Hong, The Hotchkiss School, Lakeville, CT, USA; Jin Hyup Hong, Great Neck South High School, New Hyde Park, NY, USA. Mathematical Reflections 3 (2014) 1 J302. Given that the real numbers x; y; z satisfy x + y + z = 0 and x4 y4 z4 + + = 1; 2x2 + yz 2y2 + zx 2z2 + xy determine, with proof, all possible values of x4 + y4 + z4. Proposed by Razvan Gelca, Texas Tech University, USA Solution by Chakib Belgani, Youssoufia, Morocco and Mahmoud Ezzaki, Oujda, Morocco 2 2 4 1 2 2 1 2 Notice that 2x + yz = x + yz − x(y + z) = (x − y)(x − z) and x = 2 x (2x + yz) − 2 x yz, so 1 xyz x y z (x2 + y2 + z2) − + + = 1: 2 2 (x − y)(x − z) (y − x)(y − z) (z − x)(z − y) It is clear that x y z + + = 0; (x − y)(x − z) (y − x)(y − z) (z − x)(z − y) so x2 + y2 + z2 = 2 and 2 (x + y + z)2 − (x2 + y2 + z2) = 4(xy + yz + zx)2 = 4(x2y2 + y2z2 + z2x2 + 2xyz(x + y + z)); which it is equal to 4(x2y2 + y2z2 + z2x2) = 2 (x2 + y2 + z2)2 − (x4 + y4 + z4) Thus 4 = 2(4 − (x4 + y4 + z4)) and the result follows. Also solved by Polyahedra, Polk State College, USA, Himansu Mookherjee, Kolkata, India; Daniel La- saosa, Pamplona, Spain; Radouan Boukharfane, Morocco; Hyunseo Yang, Daecheong Middle School, Seoul, South Korea; Joshua An, Washington University in St. Louis, MO, USA; Seung Hwan An, Taft School, Watertown, CT, USA; Joe Hong, Seoul International School, South Korea; Seong Kweon Hong, The Hot- chkiss School, Lakeville, CT, USA; Adnan Ali, Mumbai, India; Arkady Alt, San Jose, California, USA; David Yang, Bergen County Academies, NJ, USA; Debojyoti Biswas, Kolkata, India; Eliott S. Kim, The Lawrenceville School, NJ, USA; Jaesung Son, Ridgewood, NJ, USA; Jishnu Bose,Indian Statistical Institute, Kolkata, India; Corneliu M«nescu-Avram, Transportation High School, Ploie¸sti,Romania; Peter C.Shim, Pingry School, Basking Ridge, NJ, USA; Neculai Stanciu, Buz«u, Romania and Titu Zvonaru, Com«nes, ti, Romania; Woosung Jung, Korea International School, South Korea; Jin Hyup Hong, Great Neck South High School, New Hyde Park, NY, USA. Mathematical Reflections 3 (2014) 2 J303. Let ABC be an equilateral triangle. Consider a diameter XY of the circle centered at C which passes through A and B such that lines AB and XY as well as lines AX and BY meet outside this circle. Let Z be the point of intersection of AX and BY . Prove that AX · XZ + BY · YZ + 2CZ2 = XZ · YZ + 6AB2: Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Nikolaos Evgenidis, Gymnasium of Agia, Thessalia, Greece Triangles ABZ and XYZ are similar, so it will be YZ = 2AZ , XZ = 2BZ and XY = 2AB. Now, since ◦ \YXZ + \XYZ + \BAC = \YXZ + \XYZ + \YZX, it will be \YZX = 60 . It is also XY 2 2CZ2 = ZX2 + ZY 2 − (1) 2 and XY 2 = XZ2 + YZ2 − 2XZ · YZ · cos 60◦ , XZ · ZY = XZ2 + YZ2 − XY 2 (2) By (1) and (2) the given is written AX · XZ + BY · YZ = XY 2: But it is AX · XZ = XZ(XZ − AZ) = XZ2 − AZ · XZ = 4BZ2 − AZ · XZ and BY · YZ = YZ(YZ − BZ) = YZ2 − YZ · BZ = 4AZ2 − YZ · BZ: By (2), it is XY 2 = 4BZ2 + 4AZ2 − 4BZ · AZ: So, it now suffices to prove 4BZ · AZ = AZ · XZ + BZ · YZ(3): It is known that AZ · XZ = BZ · YZ, which gives that (3) holds if 4BZ · AZ = 2BZ · YZ that is true since YZ = 2AZ. Also solved by Polyahedra, Polk State College, USA, Neculai Stanciu, Buz«u, Romania and Titu Zvonaru, Com«nes, ti, Romania; Daniel Lasaosa, Pamplona, Spain; Hyunseo Yang, Daecheong Middle School, Seoul, South Korea; Joshua An, Washington University in St. Louis, MO, USA; Seung Hwan An, Taft School, Wa- tertown, CT, USA; Joe Hong, Seoul International School, South Korea; Seong Kweon Hong, The Hotchkiss School, Lakeville, CT, USA; Chakib Belgani, Youssoufia, Morocco and Mahmoud Ezzaki, Oujda, Morocco; Adnan Ali, Mumbai, India; Arkady Alt, San Jose, California, USA; David Yang, Bergen County Academies, NJ, USA; Himansu Mookherjee, Kolkata, India; Eliott S. Kim, The Lawrenceville School, NJ, USA; Kevin Ren; Peter C.Shim, Pingry School, Basking Ridge, NJ, USA; Woosung Jung, Korea International School, South Korea; Jin Hyup Hong, Great Neck South High School, New Hyde Park, NY, USA. Mathematical Reflections 3 (2014) 3 p p J304. Let a; b; c be real numbers such that a + b + c = 1. Let M be the maximum value of a + b + 3 c and p 1 p 3 let M2 be the maximum value of a + b + c. Prove that M1 = M2 and find this value. Proposed by Aaron Doman, University of California, Berkeley, USA Solution by Daniel Lasaosa, Pamplona, Spain p p By Lagrange's multiplier method, the maximum of f(a; b; c) = a + b + 3 c when a + b + c = 1 occurs for a 1 1 1 1 3 1 real constant λ such that 1 = λ, p = λ, p = λ, ie for b = , c = p , and consequently a = − p , 2 b 3 3 c2 4 3 3 4 3 3 for 3 1 1 1 5 2 M1 = − p + + p = + p 4 3 3 2 3 4 3 3 p p 1 1 Similarly, the maximum of g(a; b; c) = a + b + 3 c occurs when 1 = λ, p p = λ, and p p p = λ. 3 3 3 2 p 2 b+ c 6 b+ c c From the second relation we find b + 3 c = 1 , which inserted in the third yields c = p1 , for b = 1 − p1 , 4 3 3 4 3 and consequently a = 3 + p2 , yielding 4 3 3 3 2 1 5 2 M2 = + p + = + p = M1: 4 3 3 2 4 3 3 The conclusion follows. Also solved by Polyahedra, Polk State College, USA, Radouan Boukharfane, Morocco; Hyunseo Yang, Daecheong Middle School, Seoul, South Korea; Joshua An, Washington University in St. Louis, MO, USA; Seung Hwan An, Taft School, Watertown, CT, USA; Joe Hong, Seoul International School, South Korea; Seong Kweon Hong, The Hotchkiss School, Lakeville, CT, USA; Arkady Alt, San Jose, California, USA; David Yang, Bergen County Academies, NJ, USA; Eliott S. Kim, The Lawrenceville School, NJ, USA; Jishnu Bose,Indian Statistical Institute, Kolkata, India; Kevin Ren; Nicu¸sorZlota, “Traian Vuia” Technical College, Foc¸sani,Romania; Peter C.Shim, Pingry School, Basking Ridge, NJ, USA; Woosung Jung, Korea International School, South Korea; Jin Hyup Hong, Great Neck South High School, New Hyde Park, NY, USA. Mathematical Reflections 3 (2014) 4 ◦ J305. Consider a triangle ABC with \ABC = 30 . Suppose the length of the angle bisector from vertex B is twice the length of the angle bisector from vertex A. Find the measure of \BAC. Proposed by Mircea Lascu and Marius Stanean, Zalau, Romania Solution by Daniel Lasaosa, Pamplona, Spain Let `a; `b be the respective lengths of the angle bisectors from A; B. It is relatively well known (or can easily be found using Stewart's theorem and the bisector theorem) that 2bc A 2ca B ` = cos ; ` = cos : a b + c 2 b c + a 2 ◦ ◦ It follows, after using the Sine Law and setting C = 180 − A − B and B = 30 , that `b = 2`a is equivalent to A B sin (b + c) = 2 sin (c + a); 2 2 A B − C C − A sin A cos 60◦ − = sin A cos = 2 sin B cos = cos (75◦ − A) : 2 2 2 In turn this can be further expressed as A cos (75◦) cos A = sin A cos 60◦ − − cos (15◦) : 2 ◦ ◦ A ◦ ◦ A ◦ ◦ Now, if A > 90 , cos A < 0, hence cos 60 − 2 < cos (15 ), yielding either 60 − 2 > 15 , for A < 90 , ◦ A ◦ ◦ ◦ contradiction, or 60 − 2 < −15 , for A > 150 = 180 − B, absurd. On the other hand, if A < 90 , then ◦ A ◦ ◦ ◦ cos A > 0, and 60 − 2 < 15 , for A > 90 , with contradiction again.
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