Junior problems J601 Let a, b, c, d be real numbers such that 7 > a ≥ b + 1 ≥ c + 3 ≥ d + 4 Prove that 1 7 − a + 4 6 − b + 9 4 − c + 16 3 − d ≥ a + b + c + d Proposed by Adrian Andreescu, University o.
Junior problems J601 Let a, b, c, d be real numbers such that > a ≥ b + ≥ c + ≥ d + Prove that 16 + + + ≥ a + b + c + d 7−a 6−b 4−c 3−d Proposed by Adrian Andreescu, University of Texas at Dallas, USA Solution by Arkady Alt, San Jose, CA, USA Let s ∶= a + b + c + d Since a < 7, b < 6, c < 4, d < then s < 20 By Cauchy-Schwarz Inequality (7 − a + − b + − c + − d) ( Hence, 12 22 32 42 + + + ) ≥ (1 + + + 4)2 = 100 7−a 6−b 4−c 3−d 16 100 100 (s − 10)2 + + + ≥ and −s= ≥ − a − b − c − d 20 − s 20 − s 20 − s Equality iff (7 − a, − b, − c, − d) = (1, 2, 3, 4) ⇐⇒ (a, b, c, d) = (6, 4, 1, −1) Also solved by Polyahedra, Polk State College, USA; Ivan Hadinata, Jember, Indonesia; Alex Grigoryan, Quantum College, Yerevan, Armenia; Corneliu Mănescu-Avram, Ploieşti, Romania; Daniel Văcaru, Pites, ti, Romania; Hyunbin Yoo, South Korea; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Henry Ricardo, Westchester Area Math Circle, USA; Sundaresh H R, Shivamogga, India; Theo Koupelis, Cape Coral, FL, USA; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Muhammad Thoriq, Yogyakarta, Indonesia; Israel Castillo Pilco, Huaral, Peru; Adam John Frederickson, Utah Valley University, UT, USA; Dao Van Nam, MAY High School, Hoáng Mai, Ha Noi, Vietnam Mathematical Reflections (2022) J602 Prove that for any positive real number x, √ x+ √ ≥ x √ 1+ 1 +√ x+1 x+1 When does equality hold? Proposed by An Zhenping, Xianyang Normal University, China Solution by Polyahedra, Polk State College, USA Rationalizing a numerator, we have √ √ x+1 x+1 1 √ = √ −√ x+ √ − 1+ −√ x+1 x x x+1 x + 1( x + − 1) By the Cauchy-Schwarz inequality, √ √ (x + 1)(x + 2) = √ √ √ √ (( x) + 12 ) (12 + ( x + 1) ) ≥ x + x + 1, √ √ √ with equality if and only if 1/ x = x + 1, that is, if and only if x = ( − 1)/2 Also solved by Ivan Hadinata, Jember, Indonesia; Alex Grigoryan, Quantum College, Yerevan, Armenia; Muhammad Thoriq, Yogyakarta, Indonesia; Israel Castillo Pilco, Huaral, Peru; Adam John Frederickson, Utah Valley University, UT, USA; Corneliu Mănescu-Avram, Ploieşti, Romania; Daniel Văcaru, Pites, ti, Romania; G C Greubel, Newport News, VA, USA; Marin Chirciu,Colegiul Nat, ional Zinca Golescu, Pites, ti, Romania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Soham Dutta, DPS, Ruby Park, West Bengal, India; Sundaresh H R, Shivamogga, India; Theo Koupelis, Cape Coral, FL, USA; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Prajnanaswaroopa S, Bangalore, India; Daniel Pascuas, Barcelona, Spain; Arkady Alt, San Jose, CA, USA Mathematical Reflections (2022) J603 Let ABC be a triangle with centroid G and M, N, P, Q be the midpoints of the segments AB, BC, CA, AG, respectively Prove that if sin(A − B) sin C = sin(C − A) sin B, then points M, N, P, Q lie on a circle Proposed by Mihaela Berindeanu, Bucharest, România Solution by Kousik Sett, India Draw AD ⊥ BC H is the orthocenter AD cuts the nine-point circle at E which is the midpoint of AH The nine-point circle pass through the points M , N , P , Q, and E We have 1√ 1 bc AN = 2b + 2c2 − a2 , AQ = AN, AE = AH = R cos A, and AD = 2R As sin C = sin(A + B) and sin B = sin(A + C), the given condition sin(A − B) sin C = sin(C − A) sin B, is equivalent to sin(A − B) sin(A + B) = sin(C − A) sin(C + A) Ô⇒ sin2 A − sin2 B = sin2 C − sin2 A Ô⇒ a2 − b2 = c2 − a2 Ô⇒ b2 + c2 = 2a2 By power of point theorem, we get which implies (1) AQ ⋅ AN = AE ⋅ AD, b2 + c2 − a2 1 × (2b + 2c2 − a2 ) = Ô⇒ b2 + c2 = 2a2 , 4 which is true by (1) Also solved by Alex Grigoryan, Quantum College, Yerevan, Armenia; Polyahedra, Polk State College, USA; Corneliu Mănescu-Avram, Ploieşti, Romania; Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Telemachus Baltsavias, Kerameies Junior High School, Kefalonia, Greece; Theo Koupelis, Cape Coral, FL, USA Mathematical Reflections (2022) J604 Let m, n, p be odd positive integers such that (m − n)(n − p)(p − m) = and + + = m n p Find the least possible value of mnp Proposed by Adrian Andreescu, University of Texas at Dallas, USA Solution by the author 15 If m = n, then + = 2, implying m p (2m − 15)(2p − 9) = 135 We get m = n = 9, p = 27 or m = n = 15, p = or m = n = 21, p = or m = n = 75, p = 5, with the least mnp 16 + = 2, implying being 2025 If m = p, then m n mn = 4(m + 2n), a contradiction Finally, if n = p, then 17 + = 2, implying m n (2m − 7)(2n − 17) = 119 We get m = 7, n = p = 17 or m = 63, n = p = 9, with the least mnp being 2023 So, the least possible value of mnp is 2023 Also solved by Theo Koupelis, Cape Coral, FL, USA; Daniel Văcaru, Pites, ti, Romania; Ivan Hadinata, Jember, Indonesia; Alex Grigoryan, Quantum College, Yerevan, Armenia; Muhammad Thoriq, Yogyakarta, Indonesia; Polyahedra, Polk State College, USA; Israel Castillo Pilco, Huaral, Peru; Prajnanaswaroopa S, Bangalore, India; Emon Suin, Ramakrishna Mission Vidyalaya, Narendrapur, India; Sundaresh H R, Shivamogga, India Mathematical Reflections (2022) J605 Prove that (a) for an arbitrary triangle ABC, m2a m2b m2c r + + ≥2+ , bc ca ab 2R (b) if triangle ABC is acute, then m2b m2a m2c r + + ≤1+ 2 2 2 b +c c +a a +b 4R Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by Theo Koupelis, Cape Coral, FL, USA (a) Let s, E be the triangle’s semiperimeter and area, respectively Then we have 4m2a = 2(b2 + c2 ) − a2 ≥ (b + c)2 − a2 = (a + b + c)(b + c − a) = 4s(s − a), and similarly 4m2b ≥ 4s(s − b), and 4m2c ≥ 4s(s − c) Thus, m2a m2b m2c s s + + ≥ [a(s − a) + b(s − b) + c(s − c)] = ⋅ (2s2 − a2 − b2 − c2 ) bc ca ab abc abc 2s 2(s − a)(s − b)(s − c) = ⋅ (ab + bc + ca − s2 ) = + abc abc r E 4E ⋅ =2+ =2+ 2s abc 2R Equality occurs when a = b = c (b) For an acute triangle the quantities cos A, cos B, cos C are positive We have 4m2a = 2(b2 + c2 ) − a2 = b2 +c2 +2bc cos A ≤ (b2 +c2 )(1+cos A), and similarly 4m2b ≤ (a2 +c2 )(1+cos B), and 4m2c ≤ (a2 +b2 )(1+cos C) Thus, m2b m2a m2c 3 r r + + ≤ + ∑ cos A = + (1 + ) = + , 2 2 2 b +c c +a a +b 4 c 4 R 4R where we used the well-known expression b2 + c2 − a2 a2 + c2 − b2 a2 + b2 − c2 + + −1 2bc 2ac 2ac 4(s − a)(s − b)(s − c) r = = abc R cos A + cos B + cos C − = Equality occurs when a = b = c Also solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Alex Grigoryan, Quantum College, Yerevan, Armenia; Polyahedra, Polk State College, USA; Israel Castillo Pilco, Huaral, Peru; Corneliu Mănescu-Avram, Ploieşti, Romania; Daniel Văcaru, Pites, ti, Romania; Marin Chirciu,Colegiul Nat, ional Zinca Golescu, Pites, ti, Romania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Scott H Brown, Auburn University Montgomery, AL, USA; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Dao Van Nam, MAY High School, Hoáng Mai, Ha Noi, Vietnam; Arkady Alt, San Jose, CA, USA; Titu Zvonaru, Comănes, ti, România Mathematical Reflections (2022) J606 Let a, b, c be real numbers in the interval [0, 1] Prove that 2≤ b+c c+a a+b + + + 2(1 − a)(1 − b)(1 − c) ≤ 1+a 1+b 1+c Proposed by Marius Stănean, Zalău, România Solution by Polyahedra, Polk State College, USA The inequalities are equivalent to 2abc ≤ a2 (b + c) b2 (c + a) c2 (a + b) + + ≤ + 2abc 1+a 1+b 1+c By the AM-GM inequality, √ √ √ a2 (b + c) b2 (c + a) c2 (a + b) 2a2 bc 2b2 ca 2c2 ab + + ≥ + + ≥ 3abc ≥ 2abc 1+a 1+b 1+c 2 For the upper bound, we see that a2 (b + c) b2 (c + a) c2 (a + b) a2 (b + c) b2 (c + a) c2 (a + b) + + ≤ + + = ab + bc + ca 1+a 1+b 1+c 2a 2b 2c Since + bc − (b + c) = (1 − b)(1 − c) ≥ 0, + bc ≥ b + c Therefore, + 2abc = a(1 + bc) + b(1 + ca) + c(1 + ab) + (1 − a)(1 − b)(1 − c) − (ab + bc + ca) ≥ a(b + c) + b(c + a) + c(a + b) − (ab + bc + ca) = ab + bc + ca, completing the proof Also solved by Alex Grigoryan, Quantum College, Yerevan, Armenia; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Theo Koupelis, Cape Coral, FL, USA Mathematical Reflections (2022) Senior problems S601 Solve in integers the equation 16x2 y (x2 + 1)(y − 1) + 4(x4 + y + x2 − y ) = 20232 − Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by the author The equation can be rewritten as (4x4 + 4x2 )(4y − 4y ) + (4x4 + 4x2 ) + (4y − 4y ) + = 20232 , which is equivalent to (4x4 + 4x2 + 1)(4y − 4y + 1) = 20232 It follows that (2x2 +1)(2y −1) = ±(7⋅172 ) An easy check shows that there are no solutions for the equation (2x2 + 1)(2y − 1) = −(7 ⋅ 172 ) The equation (2x2 + 1)(2y − 1) = ⋅ 172 implies x2 = 144 and y = The solutions (x, y) are (12, 2), (12, −2), (−12, 2), (−12, −2) Also solved by Ivan Hadinata, Jember, Indonesia; Israel Castillo Pilco, Huaral, Peru; Adam John Frederickson, Utah Valley University, UT, USA; Le Hoang Bao, Tien Giang, Vietnam; Sundaresh H R, Shivamogga, India; Theo Koupelis, Cape Coral, FL, USA Mathematical Reflections (2022) S602 For every integer n > let A= 1 1 1 (1 + ) (1 + + ) ⋯ (1 + + + ⋯ + ) n+1 3 5 2n − and B= 1 1 1 1 ( + )( + + )⋯( + + ⋯ + ) 2 4 2n Compare A and B Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by Theo Koupelis, Cape Coral, FL, USA We have A2 = 31 ⋅ (1 + 13 ) = 49 , and B2 = 21 ⋅ ( 12 + 14 ) = 83 , and thus A2 > B2 Also, An+1 n + 1 Bn+1 1 = = ⋅ (1 + + ⋯ + ⋅ (1 + + ⋯ + ) , and ), An n+2 2n + Bn 2 n+1 for all n ≥ When n is odd, the number of terms inside the parentheses for both An+1 /An and Bn+1 /Bn is even For An+1 /An , the average value of the sum of two terms that are symmetrically positioned in the summation is given by 1 n+1 ( + )= , m 2n + − m m(2n + − m) where m is an odd number from to n For Bn+1 /Bn , the average value of the corresponding two terms is 1 2(n + 2) ( + )= (m + 1)/2 n + − (m + 1)/2 (m + 1)(2n + − m) We now have n+1 n+1 2(n + 2) ⋅ ≥ ⋅ ⇐⇒ (2n + 3)(n − m + 1)2 ≥ 0, n + m(2n + − m) (m + 1)(2n + − m) which is obvious When n is even, the number of terms inside the parentheses for both An+1 /An and Bn+1 /Bn is odd In this case, the middle term in the summation for An+1 /An is 1/(n + 1), and the middle term in the summation for Bn+1 /Bn is 2/(n + 2), and thus n+1 1 ⋅ = ⋅ n+2 n+1 n+2 Therefore, for all n ≥ we have An+1 /An > Bn+1 /Bn ; taking into account that A2 > B2 , we get An > Bn for all integer n ≥ Also solved by G C Greubel, Newport News, VA, USA; Dao Van Nam, MAY High School, Hoáng Mai, Ha Noi, Vietnam; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Arkady Alt, San Jose, CA, USA Mathematical Reflections (2022) S603 Let ABC be a triangle with incenter I and centroid G Line AG intersects BC in E and line AI intersects BC in D and the circumcircle in M Point P is the orthogonal projection of E onto AM Prove that if (AB + AC)2 = 16AP ⋅ DM then GI is parallel to BC Proposed by Mihaela Berindeanu, Bucharest, România Solution by Kousik Sett, India M is the midpoint of minor arc BC Join M, E Then ∠M ED = 90○ We have ca ba BD = and DC = b+c b+c By intersecting chord theorem, we get AD ⋅ DM = BD ⋅ DC = Ô⇒ (AP − DP ) ⋅ DM = a2 bc (b + c)2 Ô⇒ AP ⋅ DM = DP ⋅ DM + We have DE = BE − BD = a2 bc (b + c)2 a2 bc (b + c)2 (1) a ca a(b − c) − = b + c 2(b + c) Since ∠M ED = 90○ and EP ⊥ DM , we get M E = M P ⋅ M D Again, since EP ⊥ DM , we get M P − DP = M E − DE Ô⇒ M D ⋅ (M P − DP ) = M E − Ô⇒ M D ⋅ DP = Mathematical Reflections (2022) a2 (b − c)2 4(b + c)2 a2 (b − c)2 4(b + c)2 (2) By (1) and (2), we get AP ⋅ DM = a2 (b − c)2 a2 bc a2 + = 4(b + c)2 (b + c)2 The given condition can be written as (c + b)2 = AP ⋅ DM Ô⇒ (b + c)2 = 4a2 Ô⇒ b+c AI AG = Ô⇒ = a ID GE Hence GI parallel to BC Also solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Telemachus Baltsavias, Kerameies Junior High School, Kefalonia, Greece; Theo Koupelis, Cape Coral, FL, USA Mathematical Reflections (2022) 10 U603 Find all functions f ∶ R Ð→ R which satisfy the conditions: (a) f (x + 1)f (y) − f (xy) = (2x + 1)f (y), for all x, y ∈ R, (b) f (x) > 2x for all x > Proposed by Mircea Becheanu, Canada Solution by the author It is easy to see that f = and f (x) = x2 satisfy condition (a) The constant function f = does not satisfy condition (b) and we will prove that only f (x) = x2 is a solution for our problem For the pair (x, 0) we have f (x + 1)f (0) − f (0) = (2x + 1)f (0) If f (0) =/ we get f (x + 1) = 2(x + 1), for all x This means f (x) = 2x for all x But f (x) = 2x does not verify the equation Therefore we have f (0) = For the pair (0, 1) we have f (1)2 − f (0) = f (1) ⇔ f (1)2 = f (1) Hence, f (1) = or f (1) = Assume that f (1) = For that pair (x, 1) we obtain f (x + 1)f (1) − f (x) = (2x + 1)f (1), giving the solution f (x) = 0, for all x So, we are in the case f (0) = and f (1) = Plugging the pair (−1, y) in the condition (a) one obtains f (0)f (y) − f (−y) = −f (y), which shows that f (−y) = f (y) Hence, the function f is even and we will study it for x > For the pair (1, 1) we obtain f (2)f (1) − f (1) = 3f (1), then f (2) = For the pair (2, 1) we obtain f (3)f (1) − f (2) = 5f (1), showing that f (3) = We prove by induction that f (n) = n2 for all positive integers n Assume that f (n) = n2 and plug the pair (n, 1) We obtain f (n + 1)f (1) = f (n + 1) = f (n) + (2n + 1)f (1) = n2 + (2n + 1) = (n + 1)2 For the pair (n, n1 ) we obtain 1 f (n + 1)f ( ) = f (1) + (2n + 1)f ( ) ⇒ ((n + 1)2 − (2n + 1)) f ( ) = 1, n n n giving that 1 f( )= n n (1) Plugging in condition (a) the pair ( n1 , 1) we obtain f( n+1 + 1) f (1) = f ( ) = f ( ) + + 1, n n n n and using (1) one obtains f( n+1 n+1 )=( ) n n (2) Plugging the pair ( n1 , m) and using (1) and (2) one obtains f( n+1 m ) f (m) = f ( ) + ( + 1) f (m) ⇒ n n n f( m n+1 2 n+2 m2 )=( ) m −( ) m2 = n n n n This shows that f (a) = a2 for every rational number a From condition (a) we obtain f (xy) = f (y)[f (x + 1) − (2x + 1)] (3) Taking x > we have xy > y and every positive real number z > y cand be written under the form z = xy for some x > If x > we have x+1 > and by condition (b), f (x+1) > 2(x+1) giving that f (x+1)−(2x+1) > Going back in the equation (3) we obtain f (xy) > f (y) (4) Mathematical Reflections (2022) 17 showing that the function f is monotonic increasing on [0, +∞) We will show now that that f (x) = x2 for all x > Let (an )n≥1 and (bn )n≥1 be two sequences of positive rational numbers converging to x and such that an < an+1 < < x < < bn+1 < bn < Then limn→∞ (bn − an ) = Since f is monotonic increasing we have the inequalities f (an ) < f (an+1 ) < < f (x) < < f (bn+1 ) < f (bn ) < This gives f (bn ) − f (an ) = b2n − a2n = (bn − an )(bn + an ) Let M > be a real number such that an < M and bn < M For every real number ε > there exists a positive integer N such that for n > N we have bn − an < ε/2M Then f (bn ) − f (an ) < (ε/2M )(bn + an ) < ε for all n > N This shows that limn→∞ (f (bn ) − f (an )) = 0,which means that f (bn )n≥1 and f (an )n≥1 have the same limit We obtain that limn→∞ a2n = limn→∞ b2n = x2 = f (x) Since f is even we have f (x) = x2 for all x ∈ R Remark: One can prove that the function is monotonic without using condition (b) Also solved by Theo Koupelis, Cape Coral, FL, USA Mathematical Reflections (2022) 18 U604 If f (m) ∶= ∫ then evaluate, lim ( m→0 (xm − 1) ln(1 + x) dx x ln(x) 2π 8f (m) − ) 9m 3m2 Proposed by Ty Halpen, Florida, USA Solution by the author f (m) = ∫ =∫ =∫ (xm − 1) ln(1 − x) dx x ln(x) m ∫ 0 m ∫ 0 xy−1 ln(1 − x) dy dx (−1)k+1 xk dy dx k k=1 ∞ xy−1 ∑ m (−1)k+1 y+k−1 dx dy ∫ ∫ x k 0 k=1 ∞ =∑ m (−1)k+1 dx ∫ k y+k k=1 ∞ =∑ ∞ =∑ (−1)k+1 ln (1 + m ) k k k=1 ∞ ∞ (−1)j+1 (−1)k+1 mj jk j+1 k=1 j=1 =∑∑ (−1)j+1 mj ∞ (−1)k+1 ∑ j+1 j j=1 k=1 k ∞ =∑ Recognize the inner sum as the Dirichlet eta function, whose representation in terms of the zeta function form will be briefly derived: ζ (j + 1) = Notice that 2j+1 1j+1 + 2j+1 + 3j+1 + 4j+1 + 5j+1 + 6j+1 + can factor out of the even terms: even terms = = 2j+1 ( 1j+1 + 2j+1 + 3j+1 + ) ζ(j + 1) 2j+1 Now, S= 1j+1 − 2j+1 + 3j+1 − 4j+1 + 5j+1 − 6j+1 + 1 1 1 1 + + + + + ) − ( j+1 + j+1 + j+1 + ) 1j+1 2j+1 3j+1 4j+1 5j+1 ζ(j + 1) = ζ(j + 1) − ( j+1 ) =( = ζ(j + 1) (1 − 2−j ) Mathematical Reflections (2022) 19 Thus, ∞ ζ(j + 1) (1 − 2−j ) (−1)j+1 mj j=1 j f (m) = ∑ Now, apply the limit and notice that the terms after j = will vanish in the limit: lim ( m→0 2π 8f (m) 2π mζ(2) 3m2 ζ(3) − ) = lim ( − ( − + o(m2 ))) m→0 9m 9m 3m2 3m2 = lim ( 2π 4ζ(2) − + ζ(3) − o(1)) 9m 3m = lim ( 2π 2π − + ζ(3) − o(1)) 9m 9m m→0 m→0 = ζ(3) Also solved by Daniel Pascuas, Barcelona, Spain; Seán M Stewart, Thuwal, Saudi Arabia; Matthew Too, Brockport, NY, USA; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Brian Bradie, Christopher Newport University, Newport News, VA, USA; G C Greubel, Newport News, VA, USA; Theo Koupelis, Cape Coral, FL, USA; Yunyong Zhang, Chinaunicom Mathematical Reflections (2022) 20 ... Solution by the author 15 If m = n, then + = 2, implying m p (2m − 15) (2p − 9) = 1 35 We get m = n = 9, p = 27 or m = n = 15, p = or m = n = 21, p = or m = n = 75, p = 5, with the least mnp 16... Focşani, Romania Mathematical Reflections (2022) 11 S6 05 Let ABC be an acute triangle with circumcenter O and circumradius R Let Ra ; Rb ; Rc be the circumradii of triangles OBC, OCA, OAB, respectively... Inequality we have 5 ∏ (xk − 1) ≤ ( ∑ (xk − 1)) k=1 k=1 10 − 5 ) Hence, −1 + 10 − (a + b + c + d) ≤ 1, yielding a + b + c + d ≥ The minimum is 8, achieved for P (x) = (x − 2 )5 = x5 − 10x4 + 40x3