Junior problems ○ J493. In triangle ABC, R = 4r. Prove that ∠A − ∠B = 90 if and only if ¾ ab a b c2 : − = − 2 Proposed by Adrian Andreescu, University of Texas at Austin, USA Solution by Daniel Lasaosa, Pamplona, Spain Note that the second condition is equivalent to 3ab 2 2 2 3 C 1 = a + b − c = 2ab cos C; cos C = ; sin = √ : 2 4 2 2 2 A B C Since it is well known that r = 4R sin 2 sin 2 sin 2 , the second condition in a triangle such that R = 4r also sin A sin B √1 gives 2 2 = 4 2 , hence A − B A + B A B C A B 1 ○ cos = cos + 2 sin sin = sin + 2 sin sin = √ = cos 45 ; 2 2 2 2 2 2 2 2 ○ ○ A B C which implies to ∠A − ∠B = 90 . Conversely, if ∠A − ∠B = 90 , then using R = 4r so 1 = 16 sin 2 sin 2 sin 2 we find 1 A B C A B C 1 cos − sin 2 sin sin sin : √ = 2 = 2 + 2 2 = 2 + C 2 8 sin 2 sin C sin C √1 This quadratic equation in 2 has a double root and hence implies 2 = 2 2 , which we saw above was equivalent to the second equation. The conclusion follows. Also solved by Ioannis D. Sfikas, Athens, Greece; Polyahedra, Polk State College, USA; Arkady Alt, San Jose, CA, USA; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Dumitru Barac, Sibiu, Romania; Marin Chirciu, Colegiul Nat,ional Zinca Golescu, Pites, ti, Romania; Nguyen Viet Hung, Hanoi University of Science, Vietnam; Titu Zvonaru, Comănes, ti, Romania; Taes Padhihary, Disha Delphi Public School, India; Vicente Vicario García, Sevilla. Spain. Mathematical Reflections 5 (2019) 1 J494. Let a; b; c be positive real numbers. Prove that ab + bc + ca + a + b + c 3 1 ≤ 1 + : (a + b)(b + c)(c + a) 8 abc Proposed by Florin Rotaru, Focşani, România Solution by Polyahedra, Polk State College, USA Let p = (a + b + c)~3, q = (ab + bc + ca)~3, and r = abc. Then pq ≥ r, so (a + b)(b + c)(c + a) = 9pq − r ≥ 8pq, and thus it suffices to prove pq(r + 1) ≥ (p + q)r. 2 Note that p ≥ q. If either p ≤ 1 or q ≥ 1, then (p−1)(q−1) ≥ 0 and pq(r+1)−(p+q)r = (p−1)(q−1)r+pq−r ≥ 3~2 0. It remains to consider p > 1 > q. Then r ≤ q < q, and therefore pq(r+1)−(p+q)r = (p−1)qr+p(q−r) > 0, completing the proof. Also solved by Daniel Lasaosa, Pamplona, Spain; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Ioannis D. Sfikas, Athens, Greece; Nguyen Viet Hung, Hanoi University of Science, Vietnam; Pooya Esmaeil Akhondy, Atomic Energy High School, Tehran, Iran; Titu Zvonaru, Comănes, ti, Romania. Mathematical Reflections 5 (2019) 2 J495. Let a; b; c be positive numbers such that abc = 1. Prove that 1 1 1 1 1 1 9 + + + + + ≥ : a b c a2 + b b2 + c c2 + a 2 Proposed by An Zhenping, Xianyang Normal University, China Solution by the author x y z Let a ; b ; c then the equation rewrites as = y = z = x y z x y2z z2x x2y 9 + + + + + ≥ : x y z x2z + y3 y2x + z3 z2y + x3 2 LHS = 2 2 2 2 2 2 2 2 2 2 2 2 y (y + z) z (z + x) x (x + y) y z z x x y + + + + + ≥ xy(y + z)2 zy(z + x)2 xz(x + y)2 x2z2 + y3z y2x2 + z3x z2y2 + x3y 2 [(xy + yz + zx) + x(x + y) + y(y + z) + z(z + x)] = xy(y + z)2 + zy(z + x)2 + xz(x + y)2 + x2y2 + y2z2 + z2x2 + x3y + y3z + z3x 4 4 (x + y + z) (x + y + z) = (xy + yz + zx)(x2 + y2 + z2 + xy + yz + zx) (xy + yz + zx)[(x + y + z)2 − (xy + yz + zx)] 2 With x + y + z = m; xy + yz + zx = k and m ≥ 3k get 4 m 9 4 2 2 2 2 ≥ ⇔ 2m − 9m k + 9k ≥ 0 ⇔ (2m − 3k)(m − 3k) ≥ 0 k(m2 − k) 2 and we are done. Also solved by Ioannis D. Sfikas, Athens, Greece; Michael - Athanasios Peppas, Evangeliki Model High School of Smyrna, Athens, Greece. Mathematical Reflections 5 (2019) 3 J496. Let a1; a2; a3; a4; a5 be positive real numbers. Prove that a1 a2 Q ⋅ Q ≤ 1: cyc 2 (a1 + a2) + a3 cyc 2 (a1 + a2) + a3 Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Polyahedra, Polk State College, USA 2 Note that ∑cyc a3(2a1 + 2a2 + a3) = ∑cyc a3 . By the weighted AM-HM inequality, 2 a3 ⎛ ⎞ 1 Q ≥ Q a3 ⋅ = 1: cyc 2(a1 + a2) + a3 ⎝cyc ⎠ ∑cyc a3(2a1 + 2a2 + a3) Hence, by the AM-GM inequality, 2 a1 a2 1 ⎛ a1 + a2 ⎞ Q ⋅ Q ≤ Q cyc 2(a1 + a2) + a3 cyc 2(a1 + a2) + a3 4 ⎝cyc 2(a1 + a2) + a3 ⎠ 2 2 1 ⎛5 1 a3 ⎞ 1 5 1 = − Q ≤ − = 1: 4 ⎝2 2 cyc 2(a1 + a2) + a3 ⎠ 4 2 2 Also solved by Daniel Lasaosa, Pamplona, Spain; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Ioannis D. Sfikas, Athens, Greece; Titu Zvonaru, Comănes, ti, Romania. Mathematical Reflections 5 (2019) 4 J497. Prove that for any positive real numbers a; b; c a2 b2 c2 √ √ √ bc ca ab 2 a b c : b + c + a + + + ≥ ( + + ) Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by Daniel Lasaosa, Pamplona, Spain 1 11 1 By the weighted AM-GM inequality, and since 14 + 14 + 7 + 1 = 2, we have 2 2 2 ¼ 1 a 11 b 1 c √ 1 − 1 − 1 + 11 + 1 − 11 + 2 + 1 bc 2 a 7 7 b 14 7 2 c 14 7 2 2b; 14 b + 14 c + 7 a + ≥ ⋅ ⋅ = a2 b2 c2 √ with equality iff b = c = a = bc, ie iff a = b = c. Adding the cyclic permutations of this inequality produces the proposed result, where equality holds iff a = b = c. Also solved by Albert Stadler, Herrliberg, Switzerland; Polyahedra, Polk State College, USA; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Dumitru Barac, Sibiu, Romania; Taes Padhihary, Disha Delphi Public School, India; Ioannis D. Sfikas, Athens, Greece; Pooya Esmaeil Akhondy, Atomic Energy High School, Tehran, Iran; Titu Zvonaru, Comănes, ti, Romania; Arkady Alt, San Jose, CA, USA. Mathematical Reflections 5 (2019) 5 ○ J498. Let ABC be a triangle with ∠A ≠ ∠B and ∠C = 30 . On the internal angle bisector of ∠BCA consider ○ the points D and E such that ∠CAD = ∠CBE = 30 and on the perpendicular bisector of AB, on the ○ same side as C related to AB, consider the point F such that ∠AF B = 90 . Prove that DEF is an equilateral triangle. Proposed by Titu Andreescu, USA, and Marius Stănean, România Solution by Polyahedra, Polk State College, USA By the property of an angle bisector, CD intersects the circumcircle of △ABC again at P , the midpoint of ○ ○ the arc of AB not containing C. Hence, ∠P BA = ∠P AB = 15 , and thus ∠P BF = ∠P AF = 60 . Since FP ○ ○ is the perpendicular bisector of AB, ∠BF P = 45 = ∠BEP and ∠AF P = 45 = ∠ADP . Therefore, P BEF ○ ○ and P AF D are cyclic quadrilaterals. Consequently, ∠P EF = ∠P BF = 60 and ∠CDF = ∠P AF = 60 , completing the proof. Also solved by Daniel Lasaosa, Pamplona, Spain; Taes Padhihary, Disha Delphi Public School, India; Titu Zvonaru, Comănes, ti, Romania. Mathematical Reflections 5 (2019) 6 Senior problems S493. In triangle ABC; R = 4r. Prove that 19 1 1 1 25 a b c 2 ≤ ( + + ) a + b + c ≤ 2 Proposed by Adrian Andreescu, University of Texas at Austin, USA Solution by Arkady Alt, San Jose, CA, USA 2 2 1 1 1 2s ab bc ca 2s s + 4Rr + r a b c ( + + ) ( + + ) a + b + c = abc = 4Rrs = 2 2 2 2 2 2 s + 4Rr + r s + 4 ⋅ 4r ⋅ r + r s + 17r = = 2Rr 2 ⋅ 4rr 8r2 Since 2 2 2 2 2 2 s ≤ 4R + 4Rr + 3r = 4 ⋅ (4r) + 4 ⋅ 4r ⋅ r + 3r = 83r and 2 2 2 2 s ≥ 16Rr − 5r = 16 ⋅ 4r ⋅ r − 5r = 59r (Gerretsen’s Inequalities) then 2 2 2 2 s + 17r 83r + 17r 25 8r2 ≤ 8r2 = 2 and s2 17r2 59r2 17r2 19 + + : 8r2 ≥ 8r2 = 2 Also solved by Titu Zvonaru, Comănes, ti, Romania; Nicusor Zlota, Traian Vuia Technical College, Foc- sani, Romania; Corneliu Mănescu-Avram, Ploies, ti, Romania; Ioannis D. Sfikas, Athens, Greece; Marin Chirciu, Colegiul Nat,ional Zinca Golescu, Pites, ti, Romania; Telemachus Baltsavias, Keramies Junior High School Kefalonia, Greece. Mathematical Reflections 5 (2019) 7 S494. Let n > 1 be an integer. Solve the equation n x − ⌊x⌋ = n: Proposed by Alessandro Ventullo, Milan, Italy Solution by Daniel Lasaosa, Pamplona, Spain n n If SxS < 1 we have Sx − ⌊x⌋S < S1S + S1S = 2, or no solution exists in this case. If x = 1 we have x − ⌊x⌋ = 0 for n n any n ≥ 2, whereas if x = −1, we have x − ⌊x⌋ = 0 when n is odd and x − ⌊x⌋ = 2 when n is even, or the only possible solution with SxS ≤ 1 is x = −1 and only for n = 2. n If x > 1, let m = ⌊x⌋, or m ≥ 1 and x = m+δ, where 0 ≤ δ < 1, or the equation rewrites as m+n = (m+δ) ≥ mn. Or, n n−1 n−2 n−3 n ≥ m − m = m m − 1 = m(m − 1) m + m + ⋯ + 1 ≥ m(n − 1)(m − 1): 6 If m ≥ 2, we have n ≥ 2(n − 1), or 2 ≥ n, whereas if m ≥ 3, we have n ≥ 6(n − 1), or n ≤ 5 < 2.