Junior problems
J373. Let a Let a,, b, c be real numbers greater than
−1. Prove that
(a2 + b2 + 2)(b 2)(b2 + c2 + 2)(c 2)(c2 + a2 + 2)
≥ (a + 1)2(b + 1)2(c + 1)2.
Proposed by Adrian Andreescu, Dallas, TX, USA Solution by Arkady Alt, San Jose, California, USA From Cauchy-Schwarz, we know that a2 + b2 + 2 = a2 + 1 + b + b2 + 1
≥ (a +2 1)
2
+
(b + 1)2 . 2
By AM-GM, this is at least ( least (a a + 1) (b + 1) . Thus
·
a2 + b2 + 2
cyc
≥ cyc
(a + 1) (b + 1) = (a (a + 1)2 (b + 1)2 (c + 1)2 .
·
Equality holds if and only if a a = b = b = = c c = = 1.
Also Also solved solved by Vincelot Vincelot Ravoson, avoson, Franc rance, e, Paris, Paris, Lycé Lycée Henri IV; Rithvik Rithvik Pasumarty Pasumarty,, Wayzata ayzata High School, Plymouth, MN, USA; Ji Eun Kim, Tabor Academy, Marion, MA, USA; Sushanth Sathish Kumar, Jeffery Trail Middle School, Irvine, CA, USA; Byeong Yeon (Jackie) Ryu, The Hotchkiss School, Lakeville, CT, USA; Stanescu Florin, Serban Cioculescu School, Gaesti, Romania; Rajarshi Kanta Ghosh, Kolkata, India; Corneliu Mănescu-Avram, Transportation High School, Ploieşti, Romania; Joachim Studnia, Lycee Condorcet, Paris, France; A.S. Arun Srinivaas, Chennai, India; WSA; Albert Stadler, Herrliberg, Switzerland; Nicuşor Zlota‚ ”Traian Vuia” Technical College, Focşani, Romania; Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Nermin Hodzic, Dobosnica, Bosnia and Herzegovina; Alessandro Ventullo, Milan, Italy; Daniel Lasaosa, Pamplona, Spain; Bazar Tumurkhan, National University of Mongolia; Sutanay Bhattacharya, Bishnupur High School, West Bengal, India; Paul Revenant, Lycee du Parc, Lyon, France; Arpon Basu, AECS-4, Mumbai, India; Stefan Petrevski, Pearson College UWC, Victoria, Canada; Alysson Espíndola de Sá Silveira, Fortaleza, Ceará, Brazil; Catalin Prajitura, Student, College at Brockport, SUNY, USA; Polyahedra, Polk State College, FL, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Ángel Plaza, Department of Mathematics, University of Las Palmas de Gran Canaria, Spain; Joel Schlosberg, Bayside, NY, USA; Robert Bosch, Archimedean Academy, USA; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
1
J374. Let a Let a,, b, c be positive real numbers such that a + a + b b + + c c abc + abc + 2
≥ 3. Prove that
≥ a3 + b93 + c3 . Proposed by Mehmet Berke, İsler, Denizli, Turkey
Solution by Robert Bosch, Archimedean Academy, USA Since a Since a + + b b + + c c
≥ 3, by Cauchy-Schwarz we have that 2
2
a + b + c
2
a3 + b3 + c3
≥ ≥
(a + b + b + + c c))2 3, and 3 (a2 + b2 + c2 )2 (a + b + b + + c c))3 a + b + b + + c c 9
≥
≥
≥ 3.
Now recall the third-degree Schur’s inequality a3 + b3 + c3 + 3abc 3abc
+ c)) + b + b2 (c + a + a)) + c + c2 (a + b + b)), ≥ a2(b + c
or equivalently 2(a 2(a3 + b3 + c3 ) + 3abc 3abc
≥ (a + b + b + + c c)( )(a a2 + b2 + c2 ).
The inequality to be proved is abc( abc(a3 + b3 + c3 ) + 2(a 2(a3 + b3 + c3 )
≥ 9,
or equivalently abc( abc(a3 + b3 + c3
2(a3 + b3 + c3 ) + 3abc 3 abc ≥ 9. − 3) + 2(a
Note that abc( abc(a3 + b3 + c3
2(a3 + b3 + c3 ) + 3abc 3 abc ≥ − 3) + 2(a ≥ ≥
2(a 2(a3 + b3 + c3 ) + 3abc, 3 abc, (a + b + b + + c c)( )(a a2 + b2 + c2 ), 9.
Equality holds if and only if a a = b = b = = c c = = 1.
Also solved by Vincelot Ravoson, France, Paris, Lycée Henri IV; WSA; Nicuşor Zlota, ”Traian Vuia” Technic echnical al College, College, Focşani ocşani,, Romania; omania; A.S. Arun Arun Srinivaas Srinivaas,, Chennai, Chennai, India; India; Nermin Nermin Ho Hodzic dzic,, Dobosni Dobosnica ca,, Bosnia and Herzegovina; Sutanay Bhattacharya, Bishnupur High School, West Bengal, India; Polyahedra, Polk State College, FL, USA; Daniel Lasaosa, Pamplona, Spain; Arkady Alt, San Jose, CA, USA; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
2
J375. Solve Solve in real numbers numbers the equation
√ x + √ y = 1 + 3
3
2
1 x + y + y + + . 4
Proposed by Adrian Andreescu, Dallas, TX, USA Solution by Stefan Petrevski, Pearson College UWC, Victoria, Canada Let a Let a =
√ x and b = √ y. After transferring 1 to the left-hand side and squaring both sides, we obtain that 3
3
2
a3 + b3 = a 2 + b2 + 2ab 2ab
− a − b.
But this is equivalent to (a + b + b)( )(a a2 + b2 + 1 a b ab) ab) = 0, so one of the factors must be 0. Howeve However, r, if a + b = 0, the left-hand side of the initial equation is 0, 0 , while the right-hand side is positive, a contradiction.
− − −
Therefore, a Therefore, a 2 + b2 + 1 = a + b + ab. ab. This rearranges nicely to ( to (a a 1)2 + (b (b we easily see that the only solution is (a, b) = (1, (1, 1). 1). Thus ( Thus (x, x, y) = (1, (1, 1). 1).
−
(a − b)2 = 0, from where − 1)2 + (a
Also solved by Rithvik Pasumarty, Wayzata High School, Plymouth, MN, USA; Ji Eun Kim, Tabor Academy, Marion, MA, USA; Sushanth Sathish Kumar, Jeffery Trail Middle School, Irvine, CA, USA; Byeong Yeon (Jackie) Ryu, The Hotchkiss School, Lakeville, CT, USA; Duy Quan Tran, University of Health Science, Ho Chi Minh City, Vietnam; Corneliu Mănescu-Avram, Transportation High School, Ploieşti, Romania; Albert Stadler, Herrliberg, Switzerland; WSA; Nicuşor Zlota, ”Traian Vuia” Technical College, Focşani, Romania; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Polyahedra, Polk State College, FL, USA; Marissa Meehan, Student, College at Brockport, SUNY, USA; Arpon Basu, AECS-4 , Mumbai, India; Joel Schlosberg, Bayside, NY, USA; Nermin Hodzic, Dobosnica, Bosnia and Herzegovina; Alessandro Ventullo, Milan, Italy; Daniel Lasaosa, Pamplona, Spain; Bazar Tumurkhan, National University of Mongolia; Sutanay Bhattacharya, Bishnupur High School, West Bengal, India; Nguyen Viet Hung, Hanoi University of Science, Vietnam; Robert Bosch, Archimedean Academy, USA; Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Arkady Alt, San Jose, California, USA; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
3
J376. Let α Let α,, β,γ be be the angles of a triangle. Prove that 5
−
1 + 4cos α 5
−
1 + 4cos β 5
−
1 4cos γ
≥ 1.
Proposed by Dragoljub Milošević, Gornji Milanovac, Serbia Solution by Polyahedra, Polk State College, FL, USA Let x Let x = = s s
− a, y = s = s − b, and z and z = s − c. Then by the law of cosines, 5
−
1 = 4cos α 5bc
−
bc 2(b 2(b2 + c2
− a2)
=
bc bc + bc + 8(s 8(s
)(s − c) − b)(s
=
(z + x + x)( )(x x + y + y)) . (z + x + x)( )(x x + y + y)) + 8yz 8yz
Now by the AM-GM inequality, (x + y + y + + z z)( )(zz + x + x)( )(x x + y + y)) so
5
−
1 4cos α
− x[(z [(z + x + x)( )(x x + y + y)) + 8yz 8 yz]] = (x + y + y)( )(yy + z + z)( )(zz + x + x)) − 8xyz ≥ 0,
x ≥ x + y . Adding this with the other two analogous inequalities gives the desired result. + y + + z z
Equality holds if and only if x = x = y y = = z z,, or α or α = β = β = = γ .
Also solved by Rithvik Pasumarty, Wayzata High School, Plymouth, MN, USA; Sushanth Sathish Kumar, Jeffery Trail Middle School, Irvine, CA, USA; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Albert Stadler, Herrliberg, Switzerland; Nicuşor Zlota‚ ”Traian Vuia” Technical College, Focşani, Romania; Adnan Ali, A.E.C.S-4, Mumbai, India; Arkady Alt, San Jose, CA, USA; Arpon Basu, AECS-4, Mumbai, India; Joel Schlosberg, Bayside, NY, USA; Daniel Lasaosa, Pamplona, Spain; Michel Faleiros Martins, Petrobras University, Brazil; Sutanay Bhattacharya, Bishnupur High School, West Bengal, India; Bazar Tumurkhan, National University of Mongolia; Robert Bosch, Archimedean Academy, USA; Nermin Hodzic, Dobosnica, Bosnia and Herzegovina; Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; WSA; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
4
J377. Let AB Let AB C be be a triangle with
∠A
≤ 90◦. Prove that sin2
A 2
≤ m2R ≤ cos2 A2 . a
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Nermin Hodzic, Dobosnica, Bosnia and Herzegovina Let d be the distance from the circumcenter of triangle ABC AB C to the side a . Then a
d = R = R cos A. a
Using the triangle inequality, we have
R
− d ≤ m ≤ R + d + d ⇔ a
R(1
a
a
− cos A) ≤ m ≤ R(1 + cos A) ⇔ 1 − cos A ≤ m ≤ 1 + cos A ⇔ a
a
2
sin2
2R
A 2
2
≤ m2R ≤ cos A2
Equality holds on the RHS if and only if b = c = c or if ∠A =
a
2
π π . Equality holds on the LHS if only if ∠A = . 2 2
Also solved by WSA; Nicuşor Zlota‚ ”Traian Vuia” Technical College, Focşani, Romania; Robert Bosch, Archimedean Academy, USA; Sutanay Bhattacharya, Bishnupur High School, West Bengal, India; Adnan Ali, A.E.C.S-4, Mumbai, India; Arpon Basu, AECS-4 , Mumbai, India; Daniel Lasaosa, Pamplona, Spain; Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Polyahedra, Polk State College, FL, USA; Arkady Alt, San Jose, CA, USA; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
5
J378. Let P Let P be be a point in the interior of the triangle AB triangle ABC C such such that ∠BAP = BAP = 105◦ , and let D let D,, E,F be be the intersections of BP,CP,DE of BP,CP,DE with with the sides AC,AB,BC , respectively respectively.. Assume Assume that the poin p ointt B lies between C and F and F and and that ∠BAF = ∠CAP. Find CAP. Find ∠BAC . BAC .
Proposed by Marius Stănean, Zalău, România Solution by Polyahedra, Polk State College, FL, USA Suppose that AP AP intersects BC at Q at Q.. By Ceva’s and Menelaus’s theorems, AE B Q C D AE F B CD =1= . EB QC DA EB F C DA
·
·
·
·
Hence, F Hence, F B QC = BQ B Q F C = F Q F C F B (F Q + QC ) = F Q BC Let x Let x = = ∠BAF . BAF . Then by the Law of Sines,
·
·
·
FB AB = , sin x sin F Thus
−
QC AQ = , sin x sin C
√
·
FQ AQ = , sin(x sin(x + 105◦ ) sin F
2sin x = sin(x sin(x + 105◦ ) =
√ √
√ 2 − √ 6
√ 2 = √ 1 and x = 30◦ . Therefore, that is, tan is, tan x = 3√ 6+ 2+ 6 3
4
sin x +
so 2F F B · QC = F Q · BC . BC . − F B · QC , so 2 BC AB = . sin(x sin(x + 105◦ ) sin C
√ 6 + √ 2 4
x + 105◦ = ∠BAC = x +
cos x,
135◦ .
Also solved by Rithvik Pasumarty, Wayzata High School, Plymouth, MN, USA; WSA; Joel Schlosberg, Bayside, NY, USA; Robert Bosch, Archimedean Academy, USA; Daniel Lasaosa, Pamplona, Spain; Nermin Hodzic, Dobosnica, Bosnia and Herzegovina; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
6
Senior problems problems
S373. Let x Let x,, y, z be positive real numbers. Prove that
cyc
1 xy + xy + 2z 2z2
xy + xy + yz yz + + zx zx ≤ xyz( . xyz (x + y + y + + z z))
Proposed by Tolibjon Ismoilov, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan Solution by Daniel Lasaosa, Pamplona, Spain Multiplying throughout by the product of the denominators and rearranging terms, the proposed inequality is equivalent to 4(y 4(y + z + z))2 x4 4x4 yz 3x2 y 2 z2 (y z )2 0.
−
cyc
Now, clearly
−
4(y 4(y + z + z))2 x4
−
≥
− 4x4yz − 3x2y2z2 = 4x4 y2 + yz + yz + z z 2 − 3x2 y 2 z 2 = = (y − z )2 + 3x 3x2 (xy + xy + yz yz + + zx zx)( )(xy xy − yz + yz + zx zx)),
or defining a defining a = y = yzz , b = z = zx x and c = c = xy xy,, it suffices to show that
(a + b + b
cyc
− c)(a )(a − b)2 ≥ 0,
which is in turn equivalent to a3 + b3 + c3 + 3abc 3abc
≥ a2b + ab + ab2 + b2 c + bc + bc2 + c2 a + ca + ca2 .
This is a well-known form of Schur’s inequality, and since x, x, y,z are y,z are positive reals, so are a are a,, b, c, and equality holds iff a = a = b b = = c c,, or iff x = x = y y = = z. z .
Also solved by Vincelot Ravoson, France, Paris, Lycée Henri IV; WSA; Nicuşor Zlota, ”Traian Vuia” Technical College, Focşani, Romania; Albert Stadler, Herrliberg, Switzerland; Nermin Hodzic, Dobosnica, Bosnia and Herzegovina; Sutanay Bhattacharya, Bishnupur High School, West Bengal, India; Michel Faleiros Martins, Petrobras University, Brazil; Arkady Alt, San Jose, California, USA; Angel Plaza, Department of Mathematics, University of Las Palmas de Gran Canaria, Spain; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
7
S374. Let a Let a,, b, c be positive real numbers. Prove that at least one of the numbers a + b + b b + c + c c + a + a , , a + b + b c b + c + c a c + a + a b
−
−
−
is not in the interval ( interval (11, 2). 2).
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Sutanay Bhattacharya, Bishnupur High School, West Bengal, India If at least one of the numbers a + b c, b + c a, c + a b, is negative, then the corresponding fraction(s) will be negative, and we are done immediately. So let us assume that a + b c, b + c a, c + a b > 0. 0 . Without loss of generality, assume c = max a,b,c . Then
− {
−
−
}
a + b + b 2c a b 2= a + b + b c a + b + b c (c a) + (c ( c b) = 0 = a + b + b c
− − − − − − − ≥ ⇒ −
−
−
−
a + b + b a + b + b c
− ≥ 2,
so we are done.
Also solved by Vincelot Ravoson, France, Paris, Lycée Henri IV; WSA; Duy Quan Tran, University of Health Science, Ho Chi Minh City, Vietnam; Rithvik Pasumarty, Wayzata High School, Plymouth, MN, USA; Ji Eun Kim, Tabor Academy, Marion, MA, USA; Sushanth Sathish Kumar, Jeffery Trail Middle School, Irvine, CA, USA; Byeong Yeon (Jackie) Ryu, The Hotchkiss School, Lakeville, CT, USA; Albert Stadler, Herrliberg, Switzerland; Corneliu Mănescu-Avram, Transportation High School, Ploieşti, Romania; Arkady Alt, San Jose, California, USA; Robert Bosch, Archimedean Academy, USA; Nermin Hodzic, Dobosnica, Bosnia and Herzegovina; Tan Qi Huan, Universiti Sains Malaysia, Malaysia; Arpon Basu, AECS-4, Mumbai, India; Christine Izyk, Student, College at Brockport, SUNY; Catalin Prajitura, Student, College at Brockport, SUNY, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Li Zhou, Polk State College, Winter Haven, FL, USA; Ángel Plaza, Department of Mathematics, University of Las Palmas de Gran Canaria, Spain; Andreas Charalampopoulos, 4th Lyceum of Glyfada, Glyfada, Greece; Evgenidis Nikolaos, M.N.Raptou High School, School, Larissa, Larissa, Greec Greece; e; Daniel Lasaosa, Pamplona, Spain; Spain; Alessandro Alessandro Ventullo, Milan, Italy; Toshihiro oshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
8
S375. Let a Let a,, b, c be nonnegative real numbers such that ab + ab + bc bc + + ca ca = = a a + + b b + + c c > 0. 0 . Prove that a2 + b2 + c2 + 5abc 5abc
≥ 8.
Proposed by An Zhen-Ping, Xianyang Normal University, China Solution by Li Zhou, Polk State College, USA Let k = ab + ab + bc bc + + ca ca = = a + a + b b + + c c.. Then k 2 = (a + b + b + + c c))2 3(ab 3(ab + + bc bc + + ca ca)) = 3k . So k 3. If k > 4, 4 , then 2 2 2 2 a + b + c = (a + b + c) 2(ab 2(ab + bc + ca) ca) = k( k (k 2) > 2) > 8 8.. Assume thus that k that k 4. By the Cauchy-Schwarz (a2 +b2 +c2 )2 3 3 3 2 inequality, a + b + c = k( k (k 2) . Hence, a+b+c
≥
−
−
−
≥
≤
≥
6abc = (a + b + b + + c c))3 + 2(a 2(a3 + b3 + c3 )
− 3(a 3(a + b + b + + c c)( )(a a2 + b2 + c2 ) ≥ k3 + 2k 2k (k − 2)2 − 3k2 (k − 2) = −2k2 + 8k. 8k.
Therefore, 3(a 3(a2 + b2 + c2 + 5abc 5abc
− 8) ≥ 3k(k − 2) − 5k2 + 20k 20k − 24 = 2(k 2(k − 3)(4 − k ) ≥ 0,
completing completing the proof.
Also Also solved solved by WSA; Ioan Ioan Viorel Viorel Codr Codrea eanu, nu, Satulung, Satulung, Maramur Maramures, es, Romania; omania; Nicuşor Nicuşor Zlota‚ Zlota‚ ”Tr ”Traian Vuia” Technical College, Focşani, Romania; Michel Faleiros Martins, Petrobras University, Brazil; Nermin Hodzic, Dobosnica, Bosnia and Herzegovina; Sutanay Bhattacharya, Bishnupur High School, West Bengal, India; Albert Stadler, Herrliberg, Switzerland; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Andrea Fanchini, Cantù, Italy; Nguyen Viet Hung, HSGS, Hanoi University of Science, Vietnam; Arkady Alt, San Jose, California, USA; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
9
S376. Solve Solve in integers integers the equation x equation x 5
− 2xy + xy + y y 5 = 2016. 2016. Proposed by Adrian Andreescu, Dallas, TX, USA
Solution by Robert Bosch, Archimedean Academy, USA and Richard Stong, Rice University, USA If x = is not a fifth-power. If x If x and y and y are negative then x = 0 or y = 0, the equation is unsolvable because 2016 2016 is the equation is unsolvable because the left side is negative. Suppose that 0 < 0 < x y. We have
≤
2016 = x = x 5 Hence 1
xy + y y 5 ≥ x5 − x2 + y 5 − y 2 ≥ y5 − y 2 . − 2xy +
≤ y ≤ 4. So the possible pairs (x, y) to test are (1, (1, 1);(1, 1);(1, 2);(2, 2);(2, 2);(1, 2);(1, 3);(2, 3);(2, 3);(3, 3);(3, 3);(1, 3);(1, 4);(2, 4);(2, 4);(3, 4);(3, 4);(4, 4);(4, 4). 4).
Obtaining the solution ( solution (x, x, y) = (4, (4, 4). 4). Note that if x = x = y y the equation is x is x 5 4x3 + 16x 16x2 + 63x 63x + 252) = 0, 0, thus x = y = y = 4.
− x2 − 1008 = 0 or 0 or (x ( x − 4)(x 4)(x4 +
It only remains to consider when x > 0 and y < 0. 0 . In this case the equation becomes x 5 + 2xz 2xz z 5 = 2016 where z where z = = y > 0. 0. Denoting by s by s = = x x z and p and p = = xz xz the the equation becomes 5 becomes 5sp sp2 +(5s +(5 s3 +2) p +2) p+( +(ss5 2016) = 0. 0. If s s < 0, 0 , then the left side is negative. Suppose now s 0.
−
−
≥
− −
If s 5 then the left side is positive, so s = 0, 1, 2, 3, 4. For s = 0 we obtain x2 = 1008, 1008, which is not a perfect perfect square. square. For the other other values values consider consider the equati equation on as a quadra quadratic tic on p, thus the discriminant 6 3 ∆(s ∆(s) = 5s + 20 20ss +40320 +40320ss + 4 have to be a perfect square, but ∆(1) but ∆(1) = 40349, 40349, ∆(2) = 81124, 81124, ∆(3) = 125149 and ∆(4) and ∆(4) = 183044 are 183044 are not.
≥
Finally, the only solution to the original equation is ( is (x, x, y ) = (4, (4, 4). 4).
Also Also solved solved by Nermin Nermin Ho Hodzic, dzic, Dobosnic Dobosnica, a, Bosnia Bosnia and Herzeg Herzegovina ovina;; Joseph Joseph Currier; Currier; Albe Albert rt Stadler, Herrliberg, Switzerland; Daniel Lasaosa, Pamplona, Spain; Michel Faleiros Martins, Petrobras University, Brazil; Li Zhou, Polk State College, Winter Haven, FL, USA; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
10
S377. If z z is a complex number with z
| | ≥ 1, prove that |2z −√ 1|5 ≥ |z − 1|4 . 4
25 5
Proposed by Florin Stănescu, Găesti, România Solution by Nermin Hodzic, Dobosnica, Bosnia and Herzegovina 2
Let z = a = a + + bi bi. We have z
| | ≥ 1 ⇒ a
+ b2
≥ 1. Now we have |2z −√ 1| ≥ |z − 1| ⇔ 4 25 5 5
4
10
8
|2z − 1| ≥ |z − 1| ⇔ 5 16 5
From a 2 + b2
(2a (2a
2
− 1)
+ 4b 4 b2
5
≥ 5
(a
2
− 1)
+ b2
2
4
.
≥ 1, we have 2
a + b
2
−a≥
a2 + b2 1 + 2 2
− a =
(a
− 1)
2
+ b2
2
≥0
So it suffices to prove
Since a 2 + b2
4(a 4(a2 + b2 5
− a ≥ 0, from AM-GM we have 4(a 4(a + b − a) + 1 ≥ 5
2
2
5
− a) + 1 ≥
5
Equality holds if and only if a 2 + b2 = 1 and a 2 + b2
5
5
(a2 + b2 5
a2 + b2
− a)
4
−a
4
5
= (a2 + b2
4
− a)
− a = 1, which imply z = ±i.
Also solved by Nicuşor Zlota‚ ”Traian Vuia” Technical College, Focşani, Romania; Ji Eun Kim, Tabor Academy, Marion, MA, USA; Albert Stadler, Herrliberg, Switzerland; Byeong Yeon (Jackie) Ryu, The Hotchkiss School, Lakeville, CT, USA; Michel Faleiros Martins, Petrobras University, Brazil; Li Zhou, Polk State College, Winter Haven, FL, USA; Robert Bosch, Archimedean Academy, USA; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
11
S378. In a triangle, let ma , mb , mc be the lengths of the medians, wa, wb , wc be the lengths of the angle bisectors, and r and r and R and R be the inradius and circumradius, respectively. Prove that ma m b m c + + wa wb wc
≤ R + r
r R
2
.
Proposed by Dragoljub Milošević, Gornji Milanovac, Serbia Solution by Michel Faleiros Martins, Petrobras University, Brazil We will prove the following stronger statement ma m a m c + + wa wa wc
≤ 1 + Rr .
By the Figure we see quickly that wa c sin
A A + wa b sin = 2K 2 2
2K ⇒ w = (b + c ⇒ + c)sin )sin a
A
.
2
Using the similar expressions for wb and wc the inequality becomes A B C ma(b + c + c)sin )sin + mb (c + a + a)sin )sin + mc (a + b + b)sin )sin 2 2 2
≤ 2K
1+
R = 2K + + 2sR. r
By the triangle inequality we obtain ma
≤ R + u + u
a
and (b + c + c)sin )sin
A 2
≤ a,
and the other similar expressions mb
+ u , ≤ R + u
Mathematical Mathematical Reflections
b
3 (2016)
(a + c + c)sin )sin
B 2
≤b
and mc
+ u , ≤ R + u c
(a + b + b)sin )sin
C 2
≤ c. 12
We conclude that LHS LH S
+ bm + cm + cm ≤ au + bu + bu + cu + cu + (a ( a + b + b + + c c))R = 2K + + 2sR. 2 sR. ≤ ≤ am + bm a
b
c
a
b
c
Also solved by Nicuşor Zlota ‚”Traian Vuia” Technical College, Focşani, Romania; Dorina Mormocea, National College of Informatics, Piatra Neamt, Romania; Nermin Hodzic, Dobosnica, Bosnia and Herzegovina; Robert Bosch, Archimedean Academy, USA; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
13
Undergraduate problems
U373. Prove Prove the following following inequality inequality holds for all positiv p ositivee integers integers n
1 1+ 1+2
1 1+ 1+2+3
···
≥ 2,
1 1+ 1+2+
· · · + n
< 3 < 3..
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by Albert Stadler, Herrliberg, Switzerland n
1+
j =2
≤ − − ≤
1 = 1 + 2 + ... + ... + + j j
n
1+
j =2
n
exp
2
j =2
2 = exp j( j ( j + j + 1) 1 j
1 j + j + 1
n
log 1 +
j =2
= exp 1
2 j( j ( j + j + 1)
2 n + 1
n
exp
2
j =2
1 j( j ( j + j + 1)
=
e < 3. 3 .
Also solved by Daniel Lasaosa, Pamplona, Spain; Sutanay Bhattacharya, Bishnupur High School, India; Vincelot Ravoson, Lycée Henri IV, Paris, France; Rithvik Pasumarty, Wayzata High School, Plymouth, MN, USA; Ji Eun Kim, Tabor Academy, MA, USA; Byeong Yeon Ryu, Hotchkiss School, Lakeville, CT, USA; Li Zhou, Polk State College, USA; Arpon Basu, AECS-4, Mumbai, India; Adnan Ali, A.E.C.S-4, Mumbai, India; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Joel Schlosberg, Bayside, NY, USA; Moubinool Omarjee, Lycée Henri IV, Paris, France; Michel Faleiros Martins, Petrobras University, Brazil; Robert Bosch, Archimedean Academy, USA; Shohruh Ibragimov, National University of Uzbekistan, Tashkent, Uzbekistan; Alessandro Ventullo, Milan, Italy; Zafar Ahmed, BARC, Mumbai, India and Timilan Mandal, SVNIT, Surat, India; Nermin Hodžić, Dobošnica, Bosnia and Herzegovina; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
14
U374. Let p Let p and q and q be be complex numbers such that two of the zeros a, a, b, c of the polynomial x polynomial x 3 + 3 px2 + 3qx + 3 pq = = 0 are equal. Evaluate a Evaluate a 2 b + b + b2 c + c + c2 a.
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Alessandro Ventullo, Milan, Italy Assume WLOG that a that a = b = b.. Then, a Then, a 2 b + b + b2 c + c + c2 a = b = b3 + b2 c + bc + bc2 . By Viète’s Formulas, we have abc = 3 pq ab + ab + bc bc + + ca ca = 3q a + b + b + + c c = 3 p.
− −
Since a Since a = b = b,, we have
b2 c = 3 pq 2 b + 2bc 2bc = 3q 2b + c + c = 3 p.
Multiplying side by side the last two equations, we get
− −
(b2 + 2bc 2bc)(2 )(2bb + c + c)) =
−9 pq.
−9 pq = = 3(−3 pq ) = 3b2 c, we get
Since
(b2 + 2bc 2bc)(2 )(2bb + c + c)) = 3b 3 b2 c, i.e. b3 + b2 c + bc + bc2 = 0. It follows that a2 b + b + b2 c + c + c2 a = 0.
Also solved by Daniel Lasaosa, Pamplona, Spain; Rithvik Pasumarty, Wayzata High School, Plymouth, MN, USA; Ji Eun Kim, Tabor Academy, MA, USA; Li Zhou, Polk State College, USA; Arpon Basu, AECS4, Mumbai, India; Adnan Ali, A.E.C.S-4, Mumbai, India; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Corneliu Mănescu-Avram, Transportation High School, Ploieşti, Romania; Catalin Prajitura, College at Brockport, SUNY, NY, USA; Joel Schlosberg, Bayside, NY, USA; Michel Faleiros Martins, Petrobras University, Brazil; Nermin Hodžić, Dobošnica, Bosnia and Herzegovina; Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Robert Bosch, Archimedean Academy, USA; Shohruh Ibragimov, National University of Uzbekistan, Tashkent, Uzbekistan; Albert Stadler, Herrliberg, Switzerland; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
15
U375. Let
n
an =
(k2 + 1)2 , k 4 + k 2 + 1
k
k=1
Determine an and evaluate limn→∞
n = 1, 2, 3, . . . .
an . n
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by Albert Stadler, Herrliberg, Switzerland We note that 1
≤
k
(k2
k4
+ 1)2
+ k 2
+1
=
k
k2 1+ 4 k + k 2 + 1 =1 +
Therefore,
2((k 2((k
and thus,
≤
1 1)2 + (k (k
n
n
−
1 1 = ≤ 1 + k4 + kk 2 + 1 = 1 + 2(k − 2 2 2(k − k + 1) 2(k 2(k + k + k + 1)
k
k=1
1 − . 2(k2 + k + k + 1) − 1) + 1) 2(k
(k 2 + 1)2 k4 + k 2 + 1
1 ≤ n + 12 − 2(n , 2 2(n + n + n + 1)
a = n, n
an . n→∞ n lim
Also solved by Daniel Lasaosa, Pamplona, Spain; Rithvik Pasumarty, Wayzata High School, Plymouth, MN, USA; Li Zhou, Polk State College, USA; Adnan Ali, A.E.C.S-4, Mumbai, India; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Juan Manuel Sánchez Gallego, University of Antioquia, Medellín, Colombia; Arkady Alt, San Jose, CA, USA; Arpon Basu, AECS-4, Mumbai, India; Henry Ricardo, New York Math Circle; Joel Schlosberg, Bayside, NY, USA; Moubinool Omarjee, Lycée Henri IV, Paris, France; Michel Faleiros Martins, Petrobras University, Brazil; Nermin Hodžić, Dobošnica, Bosnia and Herzegovina; Nicusor Zlota ‚Traian Vuia Technical College, Focsani, Romania; Robert Bosch, Archimedean Academy, USA; Shohruh Ibragimov, National University of Uzbekistan, Tashkent, Uzbekistan; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
16
U376. Evaluate Evaluate
1 lim 1 + sin n→∞ n + 1
1 1 + sin n + 2
···
1 1 + sin n + n + n
.
Proposed by Marius Cavachi, Constanţa, România Solution by Henry Ricardo, New York Math Circle Letting P Letting P ((n) denote the given product, we have, since ln(1 ln(1 + x) = x + O(x2 ) for x close to 0 and sin(1 and sin(1//(n + k )) = O(1 O (1/n /n)), n
ln P ( P (n) =
k=1
1 ln 1 + sin n + k + k
n
=
k=1 n
=
sin
k=1 n
=
∗
1 1 sin + O sin2 n + k + k n + k + k
sin
k=1
1 + O n + k + k
1 + O n + k + k
1 n2
1 n
.
( )
Since sin x = x = x + + O O((x3 ) for small values of x, x , we see that n
k=1
1 sin = n + k + k
n
k=1
Using the well-known result l result lim imn→∞
1 + n + k + k
n
O
k=1
n 1/(n + k + k)) k=1 1/ n
lim
n
→∞
n
sin
k=1
1 n3
=
k=1
1 + O n + k + k
1 n2
.
= ln ln 2, we have 1 = l n 2. 2. n + k + k
Finally, equation ( equation ( ) yields
∗
n
ln( lim P ( P (n)) = lim lim (ln P ( P (n)) = lim lim n
n
→∞
→∞
n
→∞
k=1
1 sin + l im O n→∞ n + k + k
1 = ln ln 2, n
so limn→∞ P ( P (n) = 2.
Also solved by Albert Stadler, Herrliberg, Switzerland; Daniel Lasaosa, Pamplona, Spain; Rithvik Pasumarty, Wayzata High School, Plymouth, MN, USA; Stanescu Florin, Serban Cioculescu School, Gaesti, Romania; Li Zhou, Polk State College, USA; Zafar Ahmed, BARC, Mumbai, India and Timilan Mandal, SVNIT, Surat, India; Adnan Ali, A.E.C.S-4, Mumbai, India; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Arkady Alt, San Jose, CA, USA; Juan Felipe Buitrago Velez, University of Antioquia, Colombia; Moubinool Omarjee, Lycée Henri IV, Paris; Michel Faleiros Martins, Petrobras University, Brazil; Nermin Hodžić, Dobošnica, Bosnia and Herzegovina; Shohruh Ibragimov, National University of Uzbekistan, Tashkent, Uzbekistan; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
17
U377. Let m Let m and n and n be positive integers and let f k (x) = sin(sin(. sin(sin(. . . (sin x) . . . )) .
Evaluate
k times
f m (x) . x→0 f n (x)
lim
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam Solution by Daniel Lasaosa, Pamplona, Spain It is well known (or it can be easily proved by considering the Taylor expansion of sin x at x = 0) that limx→0 sinx x = 1, and consequently limx→0 sinx x = 1. We may generalize this result into the following Claim: For every positive integer k, we have f k (x) x = lim = 1. 1. x→0 x→0 f k (x) x lim
Proof: The initial result is clearly the Claim for k for k = = 1. If the Claim is true for k − 1, denote y denote y = f k−1 (x), or clearly l clearly lim imx→0 y = 0, and f k (x) = sin sin (f k−1 (x)), )), or f k (x) sin y f k−1 (x) = , x y x
·
where the limit of both factors is 1 when x 0 by hypothesis of induction, and hence the limit of their product, and the limit of the inverse of their product, is also 1. The Claim follows.
→
If n = n = m m,, the expression whose limit is asked is clearly 1, 1 , and so is trivially trivially its limit. Otherwise, Otherwise, if m m > n, define y define y = f = f n (x) and k and k = m = m n, or f or f m (x) = f k (y), and
−
f m (x) f k (y) = lim = 1, x→0 f n (x) y →0 y lim
and similarly when m when m < n defining y defining y = f = f m (x) and k and k = n
− m. It follows that
f m (x) = 1. x→0 f n (x) lim
Also solved by Bhattacharya, Bishnupur High School, India; Rithvik Pasumarty, Wayzata High School, Plymouth, MN, USA; Byeong Yeon Ryu, Hotchkiss School, Lakeville, CT, USA; Li Zhou, Polk State College, USA; Zafar Ahmed, BARC, Mumbai, India and Timilan Mandal, SVNIT, Surat, India; Juan Manuel Sánchez Gallego, University of Antioquia, Medellín, Colombia; Adam Krause, College at Brockport, SUNY, NY, USA; Adnan Ali, A.E.C.S-4, Mumbai, India; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Arkady Alt, San Jose, CA, USA; Henry Ricardo, New York Math Circle; Joel Schlosberg, Bayside, NY, USA; Moubinool Omarjee, Lycée Henri IV, Paris, France; Nermin Hodžić, Dobošnica, Bosnia and Herzegovina; Robert Bosch, Archimedean Academy, USA; Shohruh Ibragimov, National University of Uzbekistan, Tashkent, Uzbekistan; Albert Stadler, Herrliberg, Switzerland; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
18
U378. Let f Let f : [0, [0, 1]
→ R be a continuous function. Prove that 1 1 (−1) −1 1 f ( f (x) ln −1 xdx = xdx = ... (n − 1)! 0 0 0
n
n
1
f (x ( x1 x2
0
···x
n ) dx1 dx2
· · · dx
n.
Proposed by Albert Stadler, Herrliberg, Switzerland Solution by Li Zhou, Polk State College, USA
1
For n or n = = 1, both sides become 0 f ( f (x)dx. dx. As an induction hypothesis, assume that the claim is true for some n 1. Then integrating by parts we get
≥
( 1)n n!
−
where
1
0
( 1)n n I (x) = ln x n!
−
f ( f (x) lnn xdx = xdx = I I (1) (1)
x
f ( f (t)dt,
0
−
→0+
x
= =
lim I (x) + J, + J,
→0+
x
( 1)n−1 J = = (n 1)!
Now I Now I (1) (1) = 0, 0, and by L’Hôpital’s rule, lim I (x) =
−
−
lnn−1 x x
1
0
x
f ( f (t)dtdx.
0
( 1)n−1 xf ( xf (x) lnn+1 x ( 1)n−2 (n + 1)f 1)f (0) (0) x lnn x lim = lim n! n n n! x→0+ x→0+ ( 1)n−3 (n + 1)f 1)f (0) (0) x lnn−1 x lim = n 1)! x→0+ (n (n + 1)f 1)f (0) (0) (n + 1)f 1)f (0) (0) lim x ln x = lim x = 0. n n x→0+ x→0+
−
−
−
···
−
−
Finally, using the substitution t = xy = xy and applying the induction hypothesis to g to g (x) = f ( f (yx) yx ), we get
− 1
J =
0
( 1)n−1 (n 1)!
−
1
0
n 1
f ( f (yx) yx ) ln − xdxdy = xdxdy =
· · · 1
0
1
0
1
0
f ( f (yx 1
···x
n )dx1
· · · dx dy, n
completing the induction.
Also solved by Daniel Lasaosa, Pamplona, Spain; Adnan Ali, A.E.C.S-4, Mumbai, India; Juan Manuel Sánchez Gallego, University of Antioquia, Medellín, Colombia; Moubinool Omarjee, Lycée Henri IV, Paris, France; Michel Faleiros Martins, Petrobras University, Brazil; Nermin Hodžić, Dobošnica, Bosnia and Herzegovina; Shohruh Ibragimov, National University of Uzbekistan, Tashkent, Uzbekistan; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
19
Olympiad problems
O373. Let n 3 be a natu natura rall number number.. On a n n table table we perform perform the followin followingg operati operation: on: choose choose a (n 1) (n 1) square 1) square and add or subtract 1 subtract 1 to all its entries. At the beginning all the entries in the table are 0 are 0.. Is it possible after a finite number of operations to obtain all the numbers from 1 to n 2 in the table?
≥ − × −
×
Proposed by Alessandro Ventullo, Milan, Italy Solution by Li Zhou, Polk State College, USA It is possible if and only if n = n = 6. Let a Let a,, b, b , c, c , and d and d be be the numbers of times when the ( the (n n 1) (n 1) squares 1) squares on the upper-left (UL), upper-right (UR), lower-left (LL), and lower-right (LR) are chosen, respectively. Then in modulo 2, the result will be: a, a, b, c, d appear in 1 in 1 cell each; a each; a + b, b + d, d + c, c + a appear in n in n 2 cells 2 2 2 each; and a and a + b + c + d appears in ( in (n n 2) cells. Note that the set 1, 2, . . . , n has n /2 odd entries and 2 n /2 even entries. If a b c d 0, 1 (mod 2), 2), then we have at least 4(n 4(n 2) + (n (n 2)2 even cells and at most 4 odd o dd 2 2 cells, cells, and 4 4(n 4(n 2) (n 2) = 8 n < 0. 0 . If a b c 0 and d 1 (mod 2), 2), then we have 2n 1 even cells and (n 1)2 odd cells, and (n 1)2 (2n (2n 1) / 0, 1 . If a b c 1 and d 0 (mod 2), 2), then we have 2n 3 even cells and (n 1)2 + 2 odd cells, and (n 1)2 + 2 (2n (2n 3) / 0, 1 . If a 2), then there are (n 2)2 more even cells than odd cells. a b 1 and c d 0 (mod 2), If a d 1 and b c 0 (mod 2), 2), then there are (n 2)2 + 2 even cells and 4(n 4(n 2) + 2 odd 2 odd cells, 2 and 4(n 4(n 2) + 2 (n 2) 2 = (n 2)(6 n) 0, 1 if and only if n = n = 6. Finally, we can see that the result is achievable for n = 6 by letting a = d = d = = 35 35,, b = 2, and c = 4. Also, the b = 2 times we choose UR, we use 1 + 1 = 2 for its upper-right corner cell and 1 1 = 0 for its other (n 1)2 1 cells; the c = 4 times we choose LL, we use 4 for its lower-left corner cell and 2 2 = 0 for its other (n 1)2 1 cells; and the d = 35 35 times times we choose LR, we use 18 17 = 1 for 1 for its (n 2)2 upper-left cells. Note that for any odd integer k, 1 k 35 35,, the system x system x + + y y = = 35 and x y = k always has integer solutions with 0 with 0 y < x 35 35,, from which it is obvious that we can obtain all numbers from 1 to 36 36 in in the table.
− × −
−
≡ ≡ − − ≡ − ≡ ≡ − −
− −
≡ ≡ ≡ − − − − ≡ ≡ ≡ − ∈{ } ≡ ≡ ≡ − − ∈{ } ≡ ≡ ≡ ≡ ≡ ≡ − − −
{
− −
−
}
−
−
−
−
− ∈{ }
−
− −
−
−
−
≤
≤
≤ ≤
−
−
−
− −
Also solved by Daniel Lasaosa, Pamplona, Spain; Sutanay Bhattacharya, Bishnupur High School, India; Arpon Arp on Basu, AECS-4, AECS-4, Mumbai, Mumbai, India; India; Joel Joel Schlosb Schlosber erg, g, Bayside, Bayside, NY, USA; Nermin Ho Hodžić, džić, Dobošni Dobošnica ca,, Bosnia and Herzegovina; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
20
O374. Prove Prove that in any triangle, triangle, max( A
| − B|, |B − C |, |C − A|) ≤ arccos
− 4r R
1 .
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Robert Bosch, Archimedean Academy, USA We can suppose without loss of generality that A that A
≤ B ≤ C . Hence max {|A − B | , |B − C | , |C − A|} = C − A.
So we need to prove the following inequality C or equivalently cos(C cos(C
− A ≤ arccos
− 4r R
1 ,
− A) ≥ 4Rr − 1 = 4(cos A + cos B + cos C ) − 5.
Note that + C )) )) = − cos(A cos(A + C + C ) = 2 cos cos2 − (A + C A + C + C C − A cos ,
cos B = cos( cos(18 1800◦ cos A + cos C = 2 co c os cos(C cos(C
− A)
− −
= 2 co cos2
2 C A 2
− A + C + C 2
1,
2
1.
Finally the inequality to be proved becomes
− − ≥ cos
C
A
2
2cos
A + C + C 2
2
0.
Also solved by Daniel Lasaosa, Pamplona, Spain; Adnan Ali, A.E.C.S-4, Mumbai, India; Arkady Alt, San Jose, CA, USA; Michel Faleiros Martins, Petrobras University, Brazil; Nermin Hodžić, Dobošnica, Bosnia and Herzegovina; Li Zhou, Polk State College, USA; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
21
− bc = ≥ 12 . Prove that bc = 1 and e, f ≥ √ f 2 (a2 + b2 + c2 + d2 ) − f ( f (ac + ac + bd bd)) ≥ (e + f + f )) 2
O375. Let a,b,c,d Let a,b,c,d,, e, f be be real numbers such that ad
e2 (a2 + b2 + c2 + d2 ) + e + e((ac + ac + bd bd)) +
.
Proposed by Marius Stănean, Zalău, România Solution by Michel Faleiros Martins, Petrobras University, Brazil Using substitution = a + + bi w = a bi and z = d + d + ci ci then wz = wz = (ad But ad
bc) + (ac ( ac + + bd bd))i = ρ = ρ(cos (cos θ + i + i sin θ). − bc)
⇒ − ≥ − − √ − − √ − √
π π bc = 1 ⇒ θ ∈ − , − bc = 2 2
,
1 = ρ cos θ
ρ =
1 , cos θ
ac + ac + bd bd = =
sin θ . cos θ
By the AM-GM inequality
a2 + b2 + c2 + d2
2 (a2 + b2 )(c )(c2 + d2 ) = 2 (ad
bc) bc)2 + (ac (ac + + bd bd))2 = 2ρ =
2 . cos θ
It is sufficient to prove that
2e2 + e sin θ + cos θ
or
2e2 + e sin θ +
Let
2f 2
2f 2
f sin sin θ cos θ
2(e 2(e + f + f ))
f sin sin θ
2cos θ(e + f + f ))
2x2 + x sin θ
Ωθ (x) =
≥0 ≥0
()
2cos θ x.
√ 4x + sin θ √ 2cos θ − 2 2x2 + x sin θ For x or x ≥ 12 (x ( x = e = e or x = x = f f )), x ≥ sin4 and x and x ≥ − sin4 . Ω (x) ≥ 0 ⇔ (4x (4x + sin θ)2 ≥ 8cos θ(2x (2x2 + x sin θ) Ωθ (x) =
θ
θ
θ
⇔ (1 − cos θ)(16x )(16x2 + 8x 8x sin θ + 1 + cos θ) ≥ 0 (4x + sin θ)2 + cos θ(1 + cos θ) ≥ 0. ⇔ (4x The last inequality is true for any θ ∈ − 2 , 2 . So Ω (x) is a increasing function and the same occurs for
√ √ ≥ √ √ ≥ − − π π
Ω−θ (x). Thus
LHS LH S = Ωθ (e) + Ω −θ (f ) f )
Ωθ
1 2
θ
+ Ω−θ
2 2
1
1 = 2
sin2 θ
2
1 + sin θ +
√ − 1
sin θ
√
−2
cos θ
2 cos θ = 0
∴ Ω θ (e) + Ω −θ (f ) f )
≥ 0.
Also solved by Nermin Hodžić, Dobošnica, Bosnia and Herzegovina; Toshihiro Shimizu, Kawasaki, Japan. Mathematical Mathematical Reflections
3 (2016)
22
O376. Let a Let a1 , a2 , . . . , a100 be a permutation of the numbers 1 numbers 1,, 2, . . . , 100 100.. Let S Let S 1 = a = a 1 , S 2 = a1 + a2 , . . . , S100 = + a 100 . Find Find the maximum maximum possible possible number number of perfect perfect square squaress amo among ng the numbe numbers rs a1 + a 2 + S 1 , S 2 , . . . , S100 .
·· · · ·
Proposed by Nairi Sedrakyan, Yerevan, Armenia Solution by Li Zhou, Polk State College, USA We show that this maximum number is 60 is 60.. First, 1+2+ First, 1+2+ +100 = 5050 < 5050 < 72 72 2. Next, there are 71 are 71 changes changes of 2 2 2 2 parity in the sequence (0 sequence (0 , 1 , 2 , . . . , 71 ), and each change of parity requires adding at least one odd integer from the set 1, 3, . . . , 99 . Note that removing removing one term from the sequenc sequencee (12 , 22 , . . . , 712 ) eliminates at most two changes of parity. Thus at least 11 terms need to be removed to eliminate 71 50 = 21 changes 21 changes of parity parity. Hence, Hence, at most 60 60 squares squares are possible. Now for 1 i 50 50,, let ai = 2i 1, then S i = i 2 , achieving 50 50 squares. squares. Also, let
···
{
}
≤ ≤
−
−
2 S 53 53 = S 50 50 + 100 + 98 + 6 = 52 , 2 S 56 56 = S 53 53 + 96 + 94 + 22 = 54 , 2 S 59 59 = S 56 56 + 92 + 90 + 38 = 56 , 2 S 62 62 = S 59 59 + 88 + 86 + 54 = 58 , 2 S 65 65 = S 62 62 + 84 + 82 + 70 = 60 , 2 S 69 69 = S 65 65 + 80 + 78 + 76 + 10 = 62 , 2 S 74 74 = S 69 69 + 74 + 72 + 68 + 36 + 2 = 64 , 2 S 79 79 = S 74 74 + 66 + 64 + 62 + 60 + 8 = 66 , 2 S 85 85 = S 79 79 + 58 + 56 + 52 + 50 + 48 + 4 = 68 , 2 S 93 93 = S 85 85 + 46 + 44 + 42 + 40 + 34 + 32 + 26 + 12 = 70 ,
achieving 10 achieving 10 more squares.
Also solved by Daniel Lasaosa, Pamplona, Spain; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
23
O377. Let a1 , a2 , . . . , an , b1 , b2 , . . . , bn be positive real numbers such that ai bi > 1 for all i Denote a1 + a + a2 + + an b1 + b + b2 + + bn a = and b = . n n Prove that 1 1 1 n + + + . a1 b1 1 a2 b2 1 an bn 1 ab 1
···
√ −
∈ {1, 2, . . . , n}.
···
√ −
· · · √
− ≥ √ −
Proposed by Marius Stănean, Zalău, România Solution by Daniel Lasaosa, Pamplona, Spain It is well known that Jensen’s inequality applies in multi-variable functions, as long as the Hessian matrix of the function is either positive definite (in which case the inequality holds as for strictly convex single-variable functions functions)) or negative negative definite (in which which case the inequalit inequality y holds as for strictly concav concavee single-v single-variabl ariablee 1 √ functions). Define f ( f (x, y) = xy−1 , or the Hessian matrix is
∂ 2 f ∂x 2 ∂ 2 f ∂y∂x
∂ 2 f ∂x∂y ∂ 2 f ∂y 2
=
1
4
√ − xy
1
5
3y
−
3y2 3x 2xy + xy + 2
xy + 2 − 2xy + 3y2
.
Now, since the prefactor is positive since xy > 1, 1 , the trace has the same sign as 3x2 + 3y 3 y2 and is therefore positive, and the determinant has the same sign as 9x2 y 2
(3x − 2xy + xy + 2)(3y 2)(3y − 2xy + xy + 2) = 5x 5x2 y2 + 8xy 8xy − 4 + 6(x 6(x + y + y)( )(xy xy − 1), 1), − (3x
also clearly positive since x + y + y > 0 and xy > 1. Or both eigenv eigenvalu alues es of the Hessian Hessian have have positiv positivee sum and positive product, hence both are positive, and the Hessian Hessian is positive positive definite. definite. Multi-v Multi-variabl ariablee Jensen’s Jensen’s inequality therefore holds, and is equivalent to the proposed inequality, where equality holds iff all ai ’s are equal and simultaneously all b all b i ’s are equal.
Also Also solved solved by Rithvik Rithvik Pasumarty, Pasumarty, Wayzata Wayzata High School, School, Plymouth, Plymouth, MN, USA; Adnan Ali, A.E.C.SA.E.C.S4, Mumbai, India; Arkady Alt, San Jose, CA, USA; Joel Schlosberg, Bayside, NY, USA; Michel Faleiros Martins, Petrobras University, Brazil; Nermin Hodžić, Dobošnica, Bosnia and Herzegovina; Robert Bosch, Archimedean Academy, USA; Li Zhou, Polk State College, USA; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
24
O378. Consider Consider a convex convex hexagon hexagon ABCDEF ABCDEF such that AB DE D E , BC E F , F , and CD F A. Let M Let M , N , K be the intersections of lines BD and AE , AC and DF , respectively.. Prove Prove that the DF , CE and BF , BF , respectively perpendiculars from M , N , K to to the lines A lines AB, B, CD,EF respectively, respectively, are concurrent.
Proposed by Nairi Sedrakyan, Yerevan, Armenia Solution by Li Zhou, Polk State College, USA M
B
A Q
F
C
P
D
E K
N
Since the hexagon ABCDEF hexagon ABCDEF has has parallel opposite sides, it is well known that its six vertices lie on a conic. By Pascal’s theorem, N,P,Q are N,P,Q are collinear, where P = CB DE and Q = EA BF . BF . No Now w conside considerr the hexagon BANPEK hexagon BANPEK . The side sidess AN and EK concur EK concur with the diagonal BP (at C ); ); the sides N P and KB concur with the diagonal AE (at Q). Thus, Thus, by Pappus’ Pappus’ theorem, the sides BA and P E must E must also concur with the diagonal N K (at K (at a point at infinity), that is, N K AB. AB . Likewi Likewise, se, KM CD and M N EF . EF . Therefore, the perpendiculars from M , N , K to the lines AB,CD,EF respectively, AB,CD,EF respectively, are concurrent at the orthocenter of M N K .
∩
∩
Also solved by Michel Faleiros Martins, Petrobras University, Brazil; Nermin Hodžić, Dobošnica, Bosnia and Herzegovina; Robert Bosch, Archimedean Academy, USA; Saturnino Campo Ruiz, Salamanca, Spain; Toshihiro Shimizu, Kawasaki, Japan.
Mathematical Mathematical Reflections
3 (2016)
25
