D. M. Bˇatinetu-Giurgiu and Neculai Stanciu In Memory of Ion Ionescu
Abstract. The aim of this note is to prove that the Weitzenb¨oc’s inequality must be named the Ionescu-Weitzenb¨oc’s inequality.
1. Introduction
In Romanian Mathematical Gazette, Vol. III (15 September 1897 – 15 August 1898), No. 2 , 15 October 1897, on page 52, Ion Ionescu, the founder of Mathemat-ical Gazette, published the following problem:
*273. Prove that there is no triangle for which the inequality 4S√
3 > a2+ b2+ c2 is satisfied.
The solution of the problem 273, appeared in Gazeta Matematicˇa, Vol. III (15 September 1897–15 August 1898), No. 12, 15 August 1898, on pages 281-283, as follows:
Problema 273. Prove that there is no triangle for which the inequality 4S√
3 > a2+ b2+ c2 is satisfied.
Solution by N.G. Muzicescu (Student, Ia¸si). Let ABC be a triangle, and we construct around the side BC the equilateral triangles BCA0 and BCA00; so M = AA0∩ BC will be the middle of these two lines. Let AA0 = d and AA00= d0. In triangle AA0A00, and applying well–known theorems we deduce that:
d2+ d02= 2 · AM2+ 2 · A00M2 (1) d2− d02= 4 · A00M2· M H; (2) and from the triangle ABC, we have
b2+ c2= 2 · AM2+a2
2 . (3)
On the other hand, because A00M is the height of the equilateral triangle with length of the side a, we have:
A00M = 1 2a√
3. (4)
By the relations (1), (2), (3), and (4) we de-duce:
d2+ d02= a2+ b2+ c2, (5) d2− d02= 2a · M H√
3, (6)
but 2a · M H represents 4 times the area of the triangle ABC, because M H is evidently equal to the height of this triangle; making the substitutions given by (5) and (6) we get
2d02= a2+ b2+ c2− 4S√ 3, which in turn gives
a2+ b2+ c2≥ 4S√ 3, so the given inequality is impossible.
A
B
C H
M
A
A
Solution by I. Moscuna (Bucharest) and I. Penescu (Bucharest). We have S = 1
2bc sin A, a2= b2+ c2− 2bc cos A,√
3 = cot 30◦= cos 30◦ sin 30◦
Replacing, simplifying with 2 and letting the second member by only b2+ c2 we have
bc
sin A ·cos 30◦
sin 30◦ + cos A
> b2+ c2, which yields
2bc sin(A + 30◦) > b2+ c2. (7) On the other hand we have (b − c)2≥ 0, so
2bc ≤ b2+ c2. (8)
The inequality (7) is satisfied for A < 150◦, because otherwise the first member would be negative and could not be greater like the second member which is positive.
Assuming this condition is satisfied we can divide (7) by (8). Therefore
sin(A + 30◦) > 1, (9)
which is impossible, since the sine function cannot be greater than the unity. So the inequality (7) and therefore the given inequality are absurd for any triangle. If (8) and (9) become equalities then so does (7). So we must have A + 30◦= 90◦, i.e.
A = 60◦and b = c, i.e. the triangle must be equilateral. Hence the given inequality becomes equality for any equilateral triangle.
Solution by Miss. Maria Rugesu (Student Ia¸si) and by Mrs. Th. M.
Vladimirescu (Rˆımnicul Vˆılcei); G. G. Urechilˇa and I. Sichitiu (Sc. Sp.
De Art. ¸si Geniu) and Corneliu P. Ionescu (Student Cl, VI Lic. Galat¸i).
The given inequality becomes successively 4p
p(p − a)(p − b)(p − c) ·√
3 > a2+ b2+ c2, 16p(p − a)(p − b)(p − c)3 > (a2+ b2+ c2)2, 6p(2p − 2a)(2p − 2b)(2p − 2c) > (a2+ b2+ c2)2, 3(a + b + c)(b + c − a)(a + c − b)(a + b − c) > (a2+ b2+ c2)2, 3(b + c)2− a2 a2− (b − c)2 > (a2+ b2+ c2)2, 32bc + c2+ b2− a2 2bca2− b2− c2 > (a2+ b2+ c2)2, 34b2c2− (a2− b2− c2)2 > (a2+ b2+ c2)2, 32a2b2+ 2a2c2+ 2b2c2− a4− b4− c4 > a4+ b4+ c4+ 2a2b2+ 2a2c2+ 2b2c2, 4a2b2+ 4a2c2+ 4b2c2− 4a4− 4b4− 4c4> 0, 22a4+ 2b4+ 2c4− 2a2b2− 2a2c2− 2b2c2 < 0, 2(a2− b2)2+ (a2− c2)2+ (b2− c2)2 < 0.
But the last inequality is impossible, because all the terms on the left-hand side are positive. Hence, the given inequality is impossible for all triangles. The last inequality becomes equality if and only if a = b = c, i.e. when the triangle is equilateral.
Also solved, in different ways, by Mrs. A. Iliovici, I. Nicolaescu, Emil G. Nit¸escu, V. V. Cambureanu and C. Vintilˇa.
In the year 1919, Roland Weitzenb¨ock published in Mathematische Zeitschrift, Vol.
5, No. 1-2, pp. 137-146 the article Uber eine Ungleichung in der Dreiecksgeometrie, where he proved that:
In any triangle ABC, with usual notations, holds the inequality a2+ b2+ c2≥ 4√
3S
We observe that the inequality of Ion Ionescu is the same as the inequality of Weitzenb¨ock, and therefore from this moment the inequality of Weitzenb¨ock must be named the inequality Ionescu–Weitzenb¨ock. The inequality Ionescu-Weitzenb¨ock was given as a problem at the third IMO, Vespr´em, Hungary, 8–15, July, 1961.
Eleven proofs of the Ionescu–Weitzenb¨ock ’s inequality were presented by Arthur Engel in his book Problem solving strategies, Springer Verlag, 1998.
We discovered the above while we were working on an article about Weitzenb¨ock ’s inequality. Moreover, we have 23 demonstrations and 10 generalizations other than those published of the Ionescu–Weitzenb¨ock ’s inequality.
Department of Mathematics, Matei Basarab National College, Bucharest, Romania
Department of Mathematics, George Emil Palade School, Buzˇau, Ro-mania
JUNIOR PROBLEMS
Solutions to the problems stated in this issue should arrive before June 15, 2013
Proposals
1. Proposed by D.M. B˘atinet¸u–Giurgiu, “Matei Basarab” National College, Bucharest, Romania, Neculai Stanciu, “George Emil Palade” School, Buz˘au, Romania. Let ABC be a triangle with side lengths a, b, c, area S, circumradius R and inradius r, respectively. Prove that
a3
b · R + c · r + b3
c · R + a · r + c3
a · R + b · r ≥ 4√ 3 R + rS
2. Proposed by D.M B˘atinet¸u–Giurgiu, “Matei Basarab” National College, Bucharest, Romania, Neculai Stanciu, “George Emil Palade” School, Buz˘au, Romania. Prove that if a, b, c are real positive numbers, then
a a2+ bc ≤1
4
1 b +1
c
3. Proposed by Francisco Javier Garc´ıa Capit´an, I.E.S. ´Alvarez Cubero de Priego de C´ordoba, Spain. Let AA0, BB0, CC0 be any three diameters of the same circle.
Let us consider the intersection points P = A0C ∩ BC0, Q = AC0 ∩ B0C and T = P Q ∩ AB. Show that T C is tangent to the circle.
(Dedicated to Juan Bosco Romero M´arquez) 4. Proposed by Paolo Perfetti, Math. Dept. “Tor Vergata” Univ., Rome, Italy.
Without using calculus, find with proofs the coordinates of all the extremum points of
x2
x4+ x3+ 4x + 16
5. Proposed by Ercole Suppa, Teramo, Italy. In ∆ABC let D be the midpoint of BC, let P be an arbitrary point on AD, let E = BP ∩ AC, F = CP ∩ AB and let E1, F1 be the second intersection of BP, CP with the circumcircle, respectively.
Show that E, E1, F1, F are concyclic.
ISSN: 2217-446X, url: http://www.mathproblems-ks.com Volume 3, Issue 2 (2013), Pages 140–170
Editors: Valmir Krasniqi, Jos´e Luis D´ıaz-Barrero, Armend Sh. Shabani, Paolo Perfetti, Valmir Bucaj, Mih´aly Bencze, Ovidiu Furdui, Enkel Hysnelaj, Anastasios Kotronis, Jozsef S´andor, David R. Stone, Roberto Tauraso, Francisco Javier Garca Capit´an, Emanuele Callegari, Ercole Suppa, Cristinel Mortici, Mo-hammed Aassila.
PROBLEMS AND SOLUTIONS
Proposals and solutions must be legible and should appear on separate sheets, each indicating the name of the sender. Drawings must be suitable for reproduction.
Proposals should be accompanied by solutions. An asterisk (*) indicates that nei-ther the proposer nor the editors have supplied a solution. The editors encourage undergraduate and pre-college students to submit solutions. Teachers can help by assisting their students in submitting solutions. Student solutions should include the class and school name. Solutions will be evaluated for publication by a com-mittee of professors according to a combination of criteria. Questions concerning proposals and/or solutions can be sent by e-mail to: [email protected]
Solutions to the problems stated in this issue should arrive before September 15, 2013
Problems
66. Li Yin, Department of Mathematics and Information Science, Binzhou Uni-versity, Binzhou City, Shandong Province, 256603, China. If xk > 0, 1 ≤ k ≤ n, then show that
n
X
k=1
x2k
n
X
k=1
x4k 6
n
X
k=1
x3k v u utn
n
X
k=1
x6k.
67. Proposed by D.M. B˘atinet¸u-Giurgiu, Matei Basarab National Colege, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzˇau, Ro-mania (Jointly). Let x ∈ R and (Ln(x))n≥2 be the sequence defined by
Ln(x) = ncos2x
n+1p
(n + 1)!sin2x
−√n
n!sin2x Calculate lim
n→∞Ln(x).
c
2010 Mathproblems, Universiteti i Prishtin¨es, Prishtin¨e, Kosov¨e.
140
68. Proposed by Vasile Cirtoaje, Ploiesti, Romania. Let a and b be positive real numbers such that a + b = 2. If k ≥ 1
2, then prove that aakbbbka≥ 1.
69. Proposed by Mih´aly Bencze, Bra¸sov, Romania. Solve the following equation:
m
X
k=1
log2xk+ X
1≤i<j≤m
log xilog xj+m(m + 1)
2 log2a = (m + 1) log a log
m
Y
k=1
xk, where a > 0.
70. Proposed by Anastasios Kotronis, Athens, Greece. Let an =Pn k=0
n kk!n−k. Show that
an=r π
2n−1/2+2
3n−1+ 1 12
r π
2n−3/2+ O(n−2).
71. Proposed by Enkel Hysnelaj, University of Technology, Sydney, Australia. Solve the following system of equations
1
x+y+z1 = 1a
1
y +z+x1 = 1b
1
z+x+y1 = 1c where a, b, c ∈ < − {0}.
72∗. Proposed by Li Yin, Department of Mathematics and Information Science, Binzhou University, Binzhou City, Shandong Province, 256603, China.
Let 1 < p < ∞, we can generalize the inverse of arcsin as follows:
arcsinp(x) = Z x
0
1
(1 − tp)1/pdt 0 ≤ x ≤ 1 and
πp
2 = arcsinp(1) = Z 1
0
1 (1 − tp)1/pdt.
The inverse of arcsinp(x) on 0,π2p
is called the generalized sine function and denoted by sinp. The generalized hyperbolic cosine function is defined as cosp(x) =
d
dxsinp(x). Calculate
Ip= Z πp2
0
ln cosp(x) cosp(x)
ln sinp(x) sinp(x) dx.
Solutions
No problem is ever permanently closed. We will be very pleased considering for publication new solutions or comments on the past problems.
58. Proposed by Ovidiu Furdui, Technical University of Cluj-Napoca, Cluj-Napoca, Romania.
2) Show that
∞ where ζ(z) represents the Riemann zeta function.
Solution by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy.
It is well-known that (i − 1)!(j − 1)! The geometric series yields
Z 1
Z a Thus we have
∞ and keeping in mind the minus sign, we get
F0(a) = −2
−ζ(2)/2. Integrating by parts G(1) = −2 On the other hand
F (1) = 2Li2(1) − 2Li2(−1) − 2Li2(1) = ζ(2)
Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Moti Levy, Israel ; Hun Min Park,Advanced Institute of Science and Technology, Dajeon, South Korea; and the pro-poser
59. Proposed by D.M. B˘atinet¸u-Giurgiu, Matei Basarab National Colege, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzˇau, Ro-mania (Jointly). Let n be a positive integer. Prove that
LnL2n+2
Solution by Angel Plaza, University of Las Palmas de Gran Canaria, Spain. By the AM-GM inequality,
1 2
LnL2n+2 Ln+3
+ Ln+1L2n+3 Ln+ Ln+2
≥ s
LnL2n+2 Ln+3
· Ln+1L2n+3 Ln+ Ln+2
= s
Ln+3
Ln+ Ln+2 ·p
LnLn+1· Ln+2
> p
LnLn+1· Ln+2. So it is enough to prove that (Ln+ Ln+2)2
2 ≥√
6 − 1
·p
LnLn+1· Ln+2. Again, by the AM-GM inequality,pLnLn+1< LnL2n+1 = Ln+22 . So, it is sufficient to prove that
(Ln+ Ln+2)2 ≥ √
6 − 1 L2n+2
2Ln+ Ln+1 Ln+ Ln+1
2
≥ √
6 − 1
1 + Ln Ln+ Ln+1
2
≥ √
6 − 1.
Since Ln Ln+ Ln+1
≥ 1
4, and 1 +142
= 2516 >52− 1 >√
6 − 1, the conclusion follows.
Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Moti Levy, Israel ; and the proposer.
60. Proposed by Enkel Hysnelaj, University of Technology, Sydney, Australia. De-termine all functions f : R \ {−2, −1, 0, 1, 2} → R, which satisfy the relation
f (x) + f 7 + x 2 − x
= ax + b where a, b ∈ R.
Solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongo-lia. Consider the function g(x) = 7+x2−x . Then
(g ◦ g)(x) = g(g(x)) = 7 + g(x)
2 − g(x) =7 + 7+x2−x
2 − 7+x2−x =2x − 7 x + 1 , (g ◦ g ◦ g)(x) = g(g(g(x))) = 2g(x) − 7
g(x) + 1 = 2 ·7+x2−x− 7
7+x
2−x+ 1 = x f (x) + f 7 + x
2 − x
= ax + b (1)
In (1) if we do the change of variables x → 7 + x
2 − x, x → 2x − 7 x + 1 ,
then
Now adding up (1) and (3) and subtracting (2), yields 2f (x) = a · from which follows
f (x) =ax3+ (2a + b)x2− (5a + b)x + 21a − 2b 2(x + 1)(x − 2)
Also solved by Moti Levy, Israel ; Adrian Naco, Polytechnic University, Tirana, Albania; Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria; Fozi M. Dannan, Damascus, Syria;
Moubinool OMARJEE, Paris, France; Dorlir Ahmeti, University of Pr-ishtina, Republic of Kosova; and the proposer.
61. Proposed by D.M. B˘atinet¸u-Giurgiu, Matei Basarab National Colege, Bucharest, Romania and Neculai Stanciu, George Emil Palade Secondary School, Buzˇau, Ro-mania (Jointly). If a ∈ R+, b, c, d, xk∈ R∗+, Xn=Pn
Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Let f be the function defined on the interval [0, cn/d) by
f (x) =an + bx cn − dx
Clearly, for x ∈ [0, cn/d) we have f00(x) = 2dn(bc+ad)(cn−dx)3 > 0. Thus f is convex. Now, let tk = nxk/Xn for k = 1, 2, . . . , n. By assumption tk ∈ [0, cn/d) for every k, therefore, using the convexity of f , we find that
n which is the desired inequality.
Solution 2 by Moti Levy, Israel . Let ρk =Xxk
n then 0 < ρk < 1 andPn k=1ρk= 1.
We re-write the inequality using ρk, b
then the inequality in terms of the function f is 1
n
n
X
k=1
f (ρk) ≥ f 1 n
n
X
k=1
ρk
! . The function f (x) is convex in the interval 0,dc
since its derivative f0(x) =
ad+bc
(c−dx)2 is monotonically non-decreasing in the interval. The inequality follows from Jensen’s inequality for convex functions.
Also solved by Ioan Viorel Codreanu, Satulung, Maramure¸s, Romania;
Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; AN-anduud Problem Solving Group, Ulaanbaatar, Mongo-lia and the proposer.
62. Proposed by Mih´aly Bencze, Bra¸sov, Romania. Show that if ak, bk, ck ∈ R∗(k = 1, 2), such that a21+ b21+ c21= a22+ b22+ c22, then at least one of the equations
a1x2+ 2c2x + b1= 0; b1x2+ 2a2x + c1= 0; c1x2+ 2b2x + a1= 0 has real roots.
Solution by Juan Gabriel Alonso (E.S.O. First Course, student, Garo´e Atlantic School), and ´Angel Plaza, University of Las Palmas de Gran Canaria, Spain (Jointly). We argue by contradiction. Let us suppose that none of the given equations has any real root. Then, all the discriminants are negative.
So,
4c22− 4a1b1< 0; 4a22− 4b1c1< 0; 4b22− 4c1a1< 0 from where
2 a22+ b22+ c22 − 2 (a1b1+ b1c1+ c1a1) < 0, 2 a21+ b21+ c21 − 2 (a1b1+ b1c1+ c1a1) < 0.
(a1− b1)2+ (b1− c1)2+ (c1− a1)2< 0, which indeed it is a contradiction.
Also solved by Moti Levy, Israel ; Dorlir Ahmeti, University of Prishtina, Republic of Kosova; Ioan Viorel Codreanu, Satulung, Maramure¸s, Ro-mania; Omran Kouba, Higher Institute for Applied Sciences and Tech-nology, Damascus, Syria; Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy; Muhammad Thoriq, Department of Mathematics, Gadjah Mada University, Yogyakarta, Indonesia; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Roberto De la Cruz Moreno, Mathematics Research Center, Bellaterra Campus, Barcelona, Spain; and the proposer.
63. Proposed by Anastasios Kotronis, Athens, Greece. Let k ≥ 3 be an integer.
Show that
X
n≥0
1
(2n + 1)(3n + 2) · · · (kn + k − 1)
= 1 k!
k
X
m=2
(−1)m−1 k m
(m−1)mk−2 π 2 cot π
m− ln m +
m−1
X
`=1
cos2`π m · ln
sin`π
m
!
Solution by Moti Levy, Israel . or otherwise the series will be greater than harmonic series and thus diverges.
Hence 0 = limn→∞nun= limn→∞Pk The Digamma function ψ (z) can be used to evaluate the infinite sumX
n≥0
un. It is known that
X
The Digamma function for rational arguments is given by Gauss’s Digamma The-orem,
and Using the fact thatPk
m=2(γ + ln 2) am= 0, we obtain
and by substituting (4) in (6), the required result follows X
Also solved by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria and the proposer.
64. Proposed by Florin Stanescu, Serban School Cioculescu, Gaesti, Romania. Let f : [0, 1] → R be a continuous function. If
Solution 1 by Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria. Note that
−2
So, using the Cauch-Schwarz inequality, we obtain Thus, the minimum value that this integral may take is 8494 and it corresponds to λ = −260283. Choosing this value for λ we conclude that
which is much stronger than the proposed inequality. Moreover, this is the best possible inequality since we have equality if f (x) = 1 −260283g(x).
Solution 2 by Paolo Perfetti, Department of Mathematics, Tor Vergata University, Rome, Italy. First we note that,
Z 1
We get by Cauchy–Schwarz and the fact that the integrands in the second line are positive
which implies our inequality.
Also solved by Moti Levy, Israel ; Moubinool Omarjee, Paris, France and the proposer.
65∗. Proposed by Li Yin, Department of Mathematics and Information Science, Binzhou University, Binzhou City, Shandong Province, 256603, China. Let Ωn the volume of the unit ball in Rn, that is Ωn = Γ(n/2+1)πn/2 , n = 1, 2 . . . . where Γ(x) =R∞
0 tx−1e−tdt , x > 0. For all positive integers n, Prove that r n + 2α
n + 3 6 (Ωn+1)n+11
(Ωn)n1 6 r n + 2β n + 3 with the best possible constant factors α = 2 and β = ln(34)
ln(
√π
2 ) = 2.3818 · · · . Solution by Moti Levy, Israel . We will use the following results:
Lemma 1: For x > 0, 1
(x + 1)2 + 1
x +32 > ψ0(x + 1) > 1
(x + 1)2 + 1 x + 1 + π62
(7) Proof. See Theorem 1 in ”Feng Qi and Bai-Ni Guo, Sharp Inequalities for polygamma functions. arXiv:0903.1984v1, May 1, 2009”.
Lemma 2. The function
hv(x) = (Γ (x + 1))x1 (x + 1)v
is strictly increasing for x ≥ 1 and v ≤ vc = ππ22+1+3 ∼= 0.844 60 and is monotonic decreasing for x ≥ 0 and v ≥ 1.
Proof. Let gv(x) = ln hv(x) and fv(x) = x2 dgdxv. gv(x) = 1
xln Γ (x + 1) − v ln (x + 1) , fv(x) = − ln (Γ (x + 1)) + xψ0(x + 1) − vx2
x + 1. dfv(x)
dx = x ψ0(x + 1) −v (x + 2) (x + 1)2
!
(8) We will show that
dfv(x)
dx > 0 for x ≥ 1 and v ≤ vc (9) Substitute the right hand of Lemma 1 in (8) ,
dfv(x)
dx > x 1
(x + 1)2 + 1
x + 1 +π62 −v (x + 2) (x + 1)2
!
= x
(x + 1)2 x + 1 +π62Pv(x) , where
Pv(x) = (x + 1)2+ (1 − v (x + 2))
x + 1 + 6 π2
It is clear that
Pv(x) > Pvc(x) f or v < vc and x ≥ 1 dPvc(x)
dx = 4
π2+ 3x − 6 π2(π2+ 3) Pvc(1) = 0
Since dPvcdx(1) > 0 and Pvc(1) = 0, then Pvc(x) ≥ 0 for x ≥ 1 and consequently
dfv(x)
dx > 0 for x ≥ 1 and v ≤ vc.
The fact fvc(1) = 1 −12ππ22+1+3− γ ∼= 4.868 × 10−4 > 0 and (8) imply that gv(x) is strictly increasing for x ≥ 1 and v ≤ vc. It follows that hv(x) is strictly increasing for x ≥ 1 and v ≤ vc.
We substitute the left hand of Lemma 1 in (8) to show that dfv(x)
dx < 0 for x ≥ 0 and v ≥ 1 (10) dfv(x)
dx < x 1
(x + 1)2 + 1
x + 32 −v (x + 2) (x + 1)2
!
= x
(x + 1)2 x + 32Qv(x) , where
Qv(x) = (x + 1)2+ (1 − v (x + 2))
x +3
2
= (1 − v) x2+
3 − 7
2v
x + 5
2− 3v
The polynomial Qv(x) is negative for x ≥ 0 and v ≥ 1. The fact f1(0) = 0 and (10) imply that gv(x) is monotonic decreasing for x ≥ 0 and v ≥ 1. It follows that hv(x) is also monotonic decreasing for x ≥ 0 and v ≥ 1.
We return now to the original problem.
The ratio (Ωn+1)
1 n+1
(Ω)n1
is equal to (Γ(n2+1))1n
(Γ(n2+32))n+11 , so the inequality in the problem statement is equivalent to
α
sn
2+ 1
n
2+32 ≤ Γ n2 + 11n Γ n2 +32n+11 ≤ β
sn
2 + 1
n
2 +32 (11)
Let us begin with the right hand side of inequality (11), which is equivalent to
0 < Γ n2 + 1n1
n 2 + 1β1
≤ Γ n+12 + 1n+11
n+1 2 + 1β1 Squaring both sides,
Γ n2 + 1n1 2
n 2 + 12β
≤ Γ n+12 + 1
1 n+1
2
n+1 2 + 1β2
(12)
By Lemma 2, inequality (12) holds for n ≥ 2 and β2 ≤ ππ22+1+3 or β ≥ 2ππ22+3+1 ∼= 2.368 0.
Now we check inequality (12) for n = 1. If n = 1 then (Γ(12+1))2
(12+1)β2 ≤ (Γ(1+1))
(1+1)
2 β
implies that β2 ≤ ln(4π)
ln(43) or β ≥ 2ln(43)
ln(4π) ∼= 2.381 8.
We conclude that inequality (12) holds for n ≥ 1 when β ≥ 2ln(43)
ln(π4) ∼= 2.381 8 and 2ln(43)
ln(π4)is the minimal constant value.
Let us continue with the left hand side of the inequality (10):
α
sn
2 + 1
n
2 +32 ≤ Γ n2+ 1n1 Γ n2 +32n+11 Γ n2 +12+ 1n+11
n
2 +12+ 1α1
≤ Γ n2 + 1n1
n 2 + 1α1 Squaring both sides,
Γ n+12 + 1
1 n+1
2
n+1 2 + 1α2
≤ Γ n2 + 1n1 2
n 2 + 1α2
(13) By Lemma 2, inequality (13) holds for α2 ≥ 1 or α ≤ 2.
The asymptotic expansion of the gamma function leads to Γ (x) →√
2πxx−12e−x as x → ∞.
x→∞lim
(Γ (x + 1))1x (x + 1)v = lim
x→∞
√
2π1x (x + 1)1+2x1 e−x+1x (x + 1)v
= 1 e lim
x→∞x2x1x1−v =1 e lim
x→∞x1−v
=
0 if v > 1,
1
e if v = 1,
∞ if v < 1.
If v < 1 then (Γ(x+1))
1 x
(x+1)v tends to infinity and this fact implies that α cannot be greater than 2.
MATHCONTEST SECTION
This section of the Journal offers readers an opportunity to solve interesting and ele-gant mathematical problems mainly appeared in Math Contest around the world and most appropriate for training Math Olympiads. Proposals are always wel-comed. The source of the proposals will appear when the solutions be published.
Proposals
46. Let S = {1, 2, 3, . . . , 2743}. Find the smallest positive integer n such that the product of any n distinct elements in S is divisible by 2743.
47. Let p, n be positive integers such that p is prime and p < n. If p divides n + 1 and n
p
, (p − 1)!
= 1, then prove that p · n p
2
dividesn p
− n p
. (Here [x]
represents the integer part of the real number x.)
48. Let Γ be the circumcircle of a triangle ABC and let E and F be the intersections of the bisectors of ∠ABC and ∠ACB with Γ. If EF is tangent to the incircle γ of 4ABC, then find the value of ∠BAC.
49. For all positive integer n we consider the numbers an = 46n+ 1943. Prove that an is divisible by 2013 for all n ≥ 1, and find all values of n for which an− 207 is the cube of a positive integer.
50. Let n be a positive integer. Prove that for all x with 0 ≤ x ≤ 1/2, it holds sin2nx + cos2nx ≥ (1 − 2x2)n+ (2x2)n
Solutions
41. Find a cubic polynomial which zeros are the square of the zeros of p(x) = x3+ 2x2+ 3x + 4.
(First Stage OME 2013) Solution 1 by Adrian Naco, Polytechnic University, Tirana, Albania. Let a, b, c the roots of the given polynomial and x, y, z the roots of the required polynomial. We have that
x + y + z = a2+ b2+ c2= (a + b + c)2− 2(ab + bc + ca)
= (−2)2− 2 · 3 = −2
xy + yz + zx = (ab)2+ (bc)2+ (ca)2= (ab + bc + ca)2− 2(abbc + bcca + caab) =
= 32− 2abc(b + c + a) = 9 − 2(−4)(−2) = −7
xyz = (abc)2= (−4)2= 16 Finally the required polynomial is
q(x) = x3+ 2x2− 7x − 16
Solution 2 by Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain. Let r, s and t be the zeros of p(x). Then, p(x) = (x − r)(x − s)(x − t). We are searching a polynomial of the form q(x) = (x − r2)(x − s2)(x − t2). We have,
q(x2) = (x2− r2)(x2− s2)(x2− t2) = (x − r)(x + r)(x − s)(x + s)(x − t)(x + t) Since
p(−x) = (−x − r)(−x − s)(−x − t) = −(x + r)(x + s)(x + t), then
q(x2) = (x − r)(x + r)(x − s)(x + s)(x − t)(x + t) = p(x)[−p(−x)]
= (x3+ 2x2+ 3x + 4)(x3− 2x2+ 3x − 4) = x6+ 2x4− 7x2− 16 Hence, q(x) = x3+ 2x2− 7x − 16 is a solution and also all polynomials obtained
by it after multiplying it by a constant. 2
Also solved by Bruno Salgueiro Fanego, Viveiro, Spain; Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain; Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria.
42. Let n be an odd positive integer. A knight is placed at random in one of the n2 squares of a chessboard. It is possible that knight back to its initial position after passing once through each square?
(Training Sessions of Spanish Team for Second Stage OME 2013)
Solution by Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain. Taking as origin in the plane the initial position of the knight the 8 possible new positions after jumping once are ±1 ± 2i and ±2 ± i, respectively. When knight runs any polygonal on the chessboard to a point P we can get the coordinates of P drawing from the origin (initial position of the knight) vectors equivalents to the vectors that represent its movements and adding up their components. When the polygonal describe by the knight be closed the sum of the vectors drawn from the origin representing its movements will be zero.
h g
f
e d
c b a Y
X
-1-2i 1-2i -2+i
-2-i 2-i
2+i
-1+2i 1+2i
Let us denote by a, b, c, d, e, f, g, h the number of times that knight moves in the directions of the positions ±1 ± 2i and ±2 ± i, respectively. Suppose that chess knight passes once through each square of the chess board and go back to its initial position. Then, we have a + b + c + d + f + g + h = n2 (odd). On the other hand, since the polygonal is closed it sum is zero. That is,
(1+2i)a+(2+i)b+(2−i)c+(1−2i)d+(−1−2i)e+(−2−i)f +(−2+i)g+(−1+2i)h = 0 from which follows
(a + 2b + 2c + d − e − 2f − 2g − h) + (2a + b − c − 2d − 2c − f + g + 2h)i = 0 Identifying real and imaginary parts, yields a + 2b + 2c + d − e − 2f − 2g − h = 0 and 2a + b − c − 2d − 2c − f + g + 2h = 0 and
(a + d) − (e + h) = 2(f + g − b − c) and (b + g) − (c + f ) = 2(d + c − a − h) Then, we have that a + d + e + h and b + g + c + f are even on account that the sum of two positive integers whose difference is even is also even. Finally, we get that a + b + c + d + e + f + g + h is even. Contradiction. So, It is not possible that knight back to its initial position after passing once through each square of a
chessboard with an odd number os squares. 2
Also solved by Bruno Salgueiro Fanego, Viveiro, Spain; Francesc Gis-pert S´anchez, BARCELONA TECH, Barcelona, Spain and Jos´e Luis D´ıaz-Barrero, BARCELONA TECH, Barcelona, Spain; Omran Kouba, Higher Institute for Applied Sciences and Technology, Damascus, Syria 43. Compute
sin 1◦+ sin 2◦+ . . . + sin 133◦+ sin 134◦ cos 1◦+ cos 2◦+ . . . + cos 133◦+ cos 134◦
(Training Sessions of Spanish Team for First Stage OME 2013) Solution by Francesc Gispert S´anchez, BARCELONA TECH, Barcelona, Spain. From the identity
cos(135◦− n◦) = cos 135◦cos n◦+ sin 135◦sin n◦ we get
sin n◦− cos n◦=√
2 cos(135◦− n◦) Then,
sin 1◦− cos 1◦ = √
2 cos 134◦ sin 2◦− cos 2◦ = √
2 cos 133◦ . . . .
sin 133◦− cos 133◦ = √ 2 cos 2◦ sin 134◦− cos 134◦ = √
2 cos 1◦
Adding up the preceding equalities, and after rearranging terms, yields
(sin 1◦+ sin 2◦+ . . . + sin 133◦+ sin 134◦) − (cos 1◦+ cos 2◦+ . . . + cos 133◦+ cos 134◦)
=√
2 (cos 1◦+ cos 2◦+ . . . + cos 133◦+ cos 134◦) From the preceding immediately follows
sin 1◦+sin 2◦+. . .+sin 133◦+sin 134◦= (1+√
2)(cos 1◦+cos 2◦+. . .+cos 133◦+cos 134◦) and
sin 1◦+ sin 2◦+ . . . + sin 133◦+ sin 134◦
cos 1◦+ cos 2◦+ . . . + cos 133◦+ cos 134◦ = 1 +√ 2
2 Also solved by Daniel Vacaru, Pitesti, Romania and Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain and Jos´e Luis D´ ıaz-Barrero, BARCELONA TECH, Barcelona, Spain; Bruno Salgueiro Fanego, Viveiro, Spain; Omran Kouba, Higher Institute for Applied Sciences and
2 Also solved by Daniel Vacaru, Pitesti, Romania and Jos´e Gibergans-B´aguena, BARCELONA TECH, Barcelona, Spain and Jos´e Luis D´ ıaz-Barrero, BARCELONA TECH, Barcelona, Spain; Bruno Salgueiro Fanego, Viveiro, Spain; Omran Kouba, Higher Institute for Applied Sciences and