Junior problems
J355. Let a Let a,, b, c be positive real numbers such that a + a + b b + + c c = = 3. Prove that 4(a 4(a2 + b2 + c2 )
− (a3 + b3 + c3) ≥ 9 Proposed by Anant Mudgal, India
Solution by Michael Tang, Edina High School, MN, USA By AM-GM, a AM-GM, a 2 b + ab + ab2 + ac2 + a2 c + b + b2 c + bc + bc2 (a + b + b + + c c))3
abc, so ≥ 6abc,
abc + (a ( a2 b + ab + ab2 + ac2 + a2 c + b + b2 c + bc + bc2 ) ≥ (a + b + b + + c c))3 . − 6abc +
Expanding ( Expanding (a a + b + b + + c c))3 on the left-hand side, we get (a3 + b3 + c3 ) + 4(a 4(a2 b + ab + ab2 + ac2 + a2 c + b + b2 c + bc + bc2 )
≥ (a + b + b + + c c))3
or 4(a 4(a3 + b3 + c3 + a2 b + ab + ab2 + ac2 + a2 c + b + b2 c + bc + bc2 )
− 3(a 3(a3 + b3 + c3 ) ≥ (a + b + b + + c c))3 .
Factoring the first term, we have 4(a 4(a + b + b + + c c)( )(a a2 + b2 + c2 )
3(a3 + b3 + c3 ) ≥ (a + b + b + + c c))3 − 3(a
Since a Since a + + b b + + c c = = 3, this is equivalent to 12(a 12(a2 + b2 + c2 )
− 3(a 3(a3 + b3 + c3 ) ≥ 27 27..
Therefore, 4(a 4(a2 + b2 + c2 )
− (a3 + b3 + c3) ≥ 9
as requested. Equality holds if and only if a if a = b = b = = c c = = 1.
Also solved by Brian Bradie, Christopher Newport University, Newport News, VA, USA; Seung Hwan An, The Taft School; YunJin Jeong, Emma Willard School, Troy, NY, USA; Yeonjune Kang, Peddie School; Kyoung A Lee, The Hotchkiss School, Lakeville, CT, USA; Joseph Lee, Loomis Chattee School, Windsor, CT, USA; Ji Eun Kim, Tabor Academy, MA, USA; Kwonil Ko, Cushing Academy, Ashburham, MA, USA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA; Hyunsun (Heidi) Kim, Trinity School, NY, USA; Jamshid Yakhshiv Yakhshivev, ev, Academi Academicc Lyceum Lyceum No. 3, Bukhar Bukhara, a, Uzbekist Uzbekistan; an; Zafar Ahmed, Ahmed, BARC, BARC, India; India; Nikos Kalapodis, Patras, Greece; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Edgar Wang; George Gavrilopoulos; Henry Ricardo, New York Math Circle, Tappan, NY, USA; Lucie Wang, Lycée Louis le Grand, Paris, France; Vincelot Ravoson, Lycée Henri IV, Paris, France; Arkady Alt, San Jose, California, USA; David E. Manes, Oneonta, NY, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Georgios Tsapakidis, Panagia Prousiotissa Private School, Agrinio, Greece; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Duy Quan Tran, Nguyen Binh Khiem high school for the gifted, Vinhlong, Vietnam; Sewon Park, Peddie School, Hightstown, NJ, USA; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA,USA; Madhurima Mondal, Kalyani University Experimental High School, India; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Albert Stadler, Herrliberg, Switzerland; Neculai Stanciu and Titu Zvonaru, Romania; Paul Revenant; Prishtina Math Gymnasium Problem Solving Group, Republic of Kosova; SooYoung Choi, ChungDam Middle School, Seoul, South Korea; Joel Schlosberg, Bayside, NY; Daniel Lasaosa, Pamplona, Spain; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Polyahedra, Polk State College, FL, USA; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Eeshan Banerjee and Ajay Kumar Banerjee, West Bengal, India; Tristan Shin, San Diego, California, USA; Stefan Petrevski, Pearson College UWC, Metchosin, British Columbia, Canada. Mathematical Mathematical Reflections
6 (2015)
1
J356. Find all positive integer integerss n such that 2(6 + 9i 9i)n
n
− 3(1 + 8i8i)
= 3(7 + 4i 4i)n .
Proposed by Titu Andreescu, University if Texas at Dallas, USA Solution by Gabriel Chicas Reyes, El Salvador Rearranging the given equation and taking absolute values yields n
But 3(1 + 8i 8i)n
|
n
n
4i) |. |2(6 + 9i9i) | = |3(1 + 8i8i) + 3(7 + 4i + 3(7 + 4i 4i) | ≤ |3(1 + 8i 8i) | + |3(7 + 4i 4i) by the triangle inequality. Thus we have |2(6 + 9i9i) | ≤ |3(1 + 8i8i) | + |3(7 + 4i4i) | n
n
n
n
n
n
and computing the absolute values yields 2 which further simplifies to
n
n
≤ 62 + 92
3
12 + 82
9n−1
+3
72 + 42
n
n
≤5 .
But this inequality is satisfied only for n = 1, 2, 3. Indeed Indeed,, as 94−1 = 729 > 625 = 54 and 9m 5m for all n 0, we have 9m+3 > 5 m+4 for all m all m 0. Hence 9n−1 > 5 n for every n 4. It remains to verify whether 2( whether 2(66 + 9i)n = 3( 3(1 + 8i)n +3(7+4i +3(7+4i)n holds for n for n = = 1, 2, 3. For n For n = = 1, 3 both sides of the given equation respectively become
≥
≥
≥
≥
12 + 18i 18i = 24 + 36i 36i 2484 + 486i 486i = 552 + 108i. 108i.
− For n or n = 2 both sides equal
−
216i, so this is the only solution. −90 + 216i
Also solved by Brian Bradie, Christopher Newport University, Newport News, VA; Seung Hwan An, The Taft School; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA; Ji Eun Kim, Tabor Academy, MA; Kyoung A Lee, The Hotchkiss School, Lakeville, CT; Yeonjune Kang, Peddie School; YunJin Jeong, Emma Willard School, Troy, NY, USA; Joseph Lee, Loomis Chattee School, Windsor, CT; Kwonil Ko, Cushing Academy, Ashburham, MA; Hyunsun (Heidi) Kim, Trinity School, NY, USA; Albert Stadler, Herrliberg, Switzerland; Moubinool Omarjee, Lycée Henri IV, Paris, France; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA; Sewon Park, Peddie School, Hightstown, NJ, USA; Michael Tang, Edina High School, MN, USA; Nikos Kalapodis, Patras, Greece; Jamshid Yakhshivev, Academi cademicc Lyceum Lyceum No. 3, Bukhar Bukhara, a, Uzbekist Uzbekistan; an; Polyahe Polyahedr dra, a, Polk State College, FL, USA; Trist Tristan an Shin, San Diego, California, USA; Arkady Alt, San Jose, California, USA; Joel Schlosberg, Bayside, NY; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City; Daniel Lasaosa, Pamplona, Spain; Prishtina Math Gymnasium Problem Solving Group, Republic of Kosova.
Mathematical Mathematical Reflections
6 (2015)
2
J357. Prove Prove that for any any z
∈ C such that |z + 1 | = √ 5, √ 5 − 1 2 √ 5 + 1 ≤ |z | ≤ 2 2 z
2
Proposed by Mihály Bencze, Braşov, România Solution by Tristan Shin, San Diego, CA, USA
√ | |
The condition is equivalent to z 2 + 1 = 5 z . By the Trian Triangle gle Inequali Inequality ty,, z 2 + 1 + 5 z = z2 + 1 z2 1. Solving this inequality gives
√ | |
≥ −
√ | |≤ − ≥ | | √ | | ≥ − √ √ − z
√ 3+ 5
2
Now, by the Triangle Inequality again, z 1 + 1 + inequality gives
|z| ≥
3
z2
5
2
=
=
5+1 2
2
z 2 , so
.
5 z = z2 + 1
1 , so
5 1 2
|−1| ≥
1
− z2.
Solvin Solvingg this
2
.
Combining these two inequality gives the desired result.
Also Also solved solved by Jamshid Jamshid Yakhshive Yakhshivev, v, Academic cademic Lyceum Lyceum No. 3, Bukhara Bukhara,, Uzbekis Uzbekistan; tan; Arkady Arkady Alt, San Jose, Californi California, a, USA; Joel Schlosb Schlosber erg, g, Bayside, Bayside, NY; Daniel Daniel Lasaosa Lasaosa,, Pamplona, Pamplona, Spain; Spain; Paolo Paolo Perfetti Perfetti,, Università degli studi di Tor Vergata Roma, Roma, Italy; Polyahedra, Polk State College, FL, USA; Nikos Kalapodis, Patras, Greece; Prishtina Math Gymnasium Problem Solving Group, Republic of Kosova; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City; Michael Tang, Edina High School, MN, USA; Zafar Ahmed, BARC, India; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Seung Hwan An, The Taft School; Sewon Park, Peddie School, Hightstown, NJ, USA; Albert Stadler, Herrliberg, Switzerland; Ji Eun Kim, Tabor Academy, MA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA; Kwonil Ko, Cushing Academy, Ashburham, MA; Yeonjune Kang, Peddie School; Kyoung A Lee, The Hotchkiss School, Lakeville, CT; Hyunsun (Heidi) Kim, Trinity School, NY, USA; YunJin Jeong, Emma Willard School, Troy, NY, USA; Joseph Lee, Loomis Chattee School, Windsor, CT, USA.
Mathematical Mathematical Reflections
6 (2015)
3
J358. Prove Prove that for x for x
∈ R, the equations, 22
1
x−
=
1 22x
+1
and 2 and 2 2
x
−1
=
1 1
22
x−
−1
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Tolibjon Ismoilov, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan First of all we denote n denote n = 22
1
x−
n3
, (n > 1) 1 ) and rewrite the given equations:
1)(n3 − n − 1) = 0. 0. − n4 − 1 = 0 ⇔ (n2 − n + 1)(n n2 − n + 1 = 0 has no solutions in real numbers, and n 3 − n − 1 = 0 is equivalent to the first equation. Then
−n−1 =0
n5
and
22
1
x−
=
1 22
x
−1
and
22
+1
x
=
1 22
1
x−
−1
are equivalent.
Also Also solved solved by Jamshid Jamshid Yakhshivev akhshivev,, Academic ademic Lyceum Lyceum No. 3, Bukhar Bukhara, a, Uzbekist Uzbekistan; an; Duy Quan Tran, Tran, Nguyen Binh Khiem high school for the gifted, Vinhlong, Vietnam; Seung Hwan An, The Taft School; Sewon Park, Peddie School, Hightstown, NJ, USA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA; Moubinool Omarjee, Lycee Henri IV, Paris, France; Albert Stadler, Herrliberg, Switzerland; Edgar Wang; Lucie Wang, Lycée Louis le Grand, Paris, France; YunJin Jeong, Emma Willard School, Troy, NY, USA; Hyunsun (Heidi) Kim, Trinity School, NY, USA; Yeonjune Kang, Peddie School; Joseph Lee, Loomis Chattee School, Windsor, CT; Kyoung A Lee, The Hotchkiss School, Lakeville, CT; Kwonil Ko, Cushing Academy, Ashburham, MA; Ji Eun Kim, Tabor Academy, MA; Vincelot Ravoson, Lycée Henri IV, Paris, France; Polyahedra, Polk State College, FL, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Arkady Alt, San Jose, California, USA; Michael Tang, Edina High School, MN, USA; Joel Schlosberg, Bayside, NY; Tristan Shin, San Diego, California, USA; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City; Latofat Bobojonova, Bobojonova, Academic Academic Lyceum S.H.Sirojiddinov, S.H.Sirojiddinov, Tashkent, Uzbekistan; Prishtina Math Gymnasium Problem Solving Solving Group, Group, Republic epublic of Kosova; Kosova; Daniel Daniel Lasaosa Lasaosa,, Pamplona Pamplona,, Spain; Spain; David David E. Manes, Oneonta, Oneonta, NY, USA; Brian Bradie, Christopher Newport University, Newport News, VA, USA; Khurshid Juraev, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzekistan.
Mathematical Mathematical Reflections
6 (2015)
4
J359. The midline midline of triangle ABC , ABC , parallel to side B side B C , intersects the triangle’s circumcircle at B and C . Evaluate the length of segment B C in terms of triangle AB triangle AB C ’s ’s side-lengths.
Proposed by Dorin Andrica and Dan Ştefan Marinescu, România Solution by Polyahedra, Polk State College, FL, USA Let M and N N be the midpoints of AB and AC . Let x = B M and y = N C . Then Then by Power Power of a Poin Point, t, a c2 a b2 c2 −b2 x 2 + y = 4 and x + 2 y = 4 . Subtractin Subtractingg the two two equations we get x y = 2a ; adding the two a c2 +b2 equations we get 2 (x + y + y)) + 2xy 2xy = = 4 . Since 4 Since 4xy xy = = (x + y + y))2 (x y)2 , we then have
−
− −
c2 + b2 = a( a (x + y + y)) + (x ( x + y + y))2 2
− (x −
a y)2 = + x + x + y y 2
a 2 4
− − 2
c2
− b2
2a
2
.
Hence, a B C = + x + x + y y = = 2
2a2 (c2 + b2 ) + a + a4 + (c (c2 4a2
− b2)2 =
[a2 + (b (b + c + c))2 ] [a2 + (b (b 2a
− c)2] .
Also solved by Sewon Park, Peddie School, Hightstown, NJ, USA; Neculai Stanciu and Titu Zvonaru, Romania; Yeonjune Kang, Peddie School; YunJin Jeong, Emma Willard School, Troy, NY, USA; Hyunsun (Heidi) Kim, Trinity School, NY, USA; Joseph Lee, Loomis Chattee School, Windsor, CT; Kyoung A Lee, The Hotchkiss School, Lakeville, CT; Jamshid Yakhshivev, Academic Lyceum No. 3, Bukhara, Uzbekistan; Kwonil Ko, Cushing Academy, Ashburham, MA; Ji Eun Kim, Tabor Academy, MA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA; Seung Hwan An, The Taft School; Nikos Kalapodis, Patras, Greece; Tristan Shin, San Diego, California, USA; Daniel Lasaosa, Pamplona, Spain; Prishtina Math Gymnasium Problem Solving Group, Republic of Kosova; Tolibjon Ismoilov, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Arkady Alt, San Jose, California, USA; Lucie Wang, Lycée Louis le Grand, Paris, France.
Mathematical Mathematical Reflections
6 (2015)
5
J360. In triangle triangle AB AB C , let AA let AA and B and B B be the angle bisectors of ∠A and A B ab sin C 2
≤
1
B (b + c + c)sin )sin A + 2
∠B .
Prove that
1
+
(c + a + a)sin )sin
A + B 2
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Nikos Kalapodis, Patras, Greece
Let I Let I be be the incenter of triangle ABC . ABC . Applying Applying Ptolemy’s Ptolemy’s Inequalit Inequality y to quadrilater quadrilateral al IB CA we obtain A B
· IC ≤ ≤
I B
·
A C + I + IA A
·
B C ,
i.e.
A B
≤
IB IA A C + + B C (1) IC IC
·
·
From the law of sines in triangles I triangles I B C and I and I A C we C we obtain C 2 = B I C sin A + 2 sin
I B
IA
,
IC
Also, by the angle bisector theorem we have A C = Substituting (2), (3) to (1) we have A B
i.e.
C 2 B sin A + 2 sin
≤
A B C ab sin 2
≤
·
sin =
(2)
A sin + B 2
ab ab , B C = (3) b + c + c c + a + a
ab + b + c + c
1
B (b + c + c)sin )sin A + 2
Equality holds iff quadrilateral I quadrilateral I B C A is cyclic, i.e.
C 2
sin sin
+
∠C =
C 2
· A + B 2
ab c + a + a
1
A (c + a + a)sin )sin + B 2
60o .
.
Also solved by Kyoung A Lee, The Hotchkiss School, Lakeville, CT; Yeonjune Kang, Peddie School; Joseph Lee, Loomis Chattee School, Windsor, CT; YunJin Jeong, Emma Willard School, Troy, NY, USA; Hyunsun (Heidi) Kim, Trinity School, NY, USA; Neculai Stanciu and Titu Zvonaru, Romania; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA; Sewon Park, Peddie School, Hightstown, NJ, USA; Seung Hwan An, The Taft School; Kwonil Ko, Cushing Academy, Ashburham, MA; Ji Eun Kim, Tabor Academy, MA; Polyahedra, Polk State College, FL, USA; Tristan Shin, San Diego, California, USA; Latofat Bobojonova, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Daniel Lasaosa, Pamplona, Spain; Tolibjon Ismoilov, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Adnan Ali, Student in A.E.C.S4, Mumbai, India. Mathematical Mathematical Reflections
6 (2015)
6
Senior problems problems
S355. Let a Let a,, b, c be nonnegative real numbers such that ab + bc + ca = ca = 1 and min( min(a,b,c) a,b,c) Prove that (a + b + b + + c 16(a + b + b + + c 1)(abc + + a + b + c 2). c 2)4 16(a c 1)(abc a + b + c 2).
−
≥
−
≤ √ 2|a + b + c − 2|.
−
Proposed by Marcel Chiriţă, Bucharest, România Solution by Li Zhou, Polk State College, USA The ”min( ” min(a,b,c a,b,c))” should be ” max(a,b,c max(a,b,c))”, otherwise the conclusion is false for (a,b,c) a,b,c) = (3, (3, 15 , 18 ). Without loss of generalit generality y, assume assume that 0 c b a 2 a + b + c 2 . The The condi conditi tion on ab + ab + bc + ca = 1 implies a + b + b + + c c 3(ab 3(ab + + bc bc + + ca ca)) > 1 and abc + abc + a a + + b b + + c c 2 = (a 1)(b 1)(b 1)(c 1)(c 1). 1). If a 1, then (a + b + c 1)(a 1)(a 1)(b 1)(b 1)(c 1)(c 1) 0. Hence it suffices to consider a consider a > 1. 1. Then b Then b < 1, 1, and by the AM-GM inequality,
≥
−
−
−
≤ ≤ ≤ ≤ √ | − ≤
−
− |
−
−
−
≤
16(a 16(a + b + b + + c c
1)(a − 1)(1 − b)(1 − c) − 1)(a b + b + c c 2 b + c + c 2 1− = (2a (2a − 2 + b + b + + c c))2 (2 − b − c)2 ≤ 16 a 1 + 2 2 2 2 = a2 − (a + b + b + + c c − 2)2 ≤ 2(a 2(a + b + b + + c c − 2)2 − (a + b + b + + c c − 2)2 = (a + b + b + + c c − 2)4 .
−
Mathematical Mathematical Reflections
6 (2015)
7
S356. Let a,b Let a,b,, c, d, e be real numbers such that sin a + sin b + sin c + sin d + sin e cos b + cos c + cos d + cos 4.
≤
≥ 3.
Prove Prove that that cos a +
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia Using the Cauchy-Schwartz inequality, we have:
cos a
cyc
≤ · · − − · − · √ 12
cyc
=
cos2 a =
5
(1
cyc
25
sin2 a)
cyc
sin2 a =
5
12
25
cyc
cyc
sin2 a
cyc
2
≤
25
−
sin a
cyc
≤
25
Equality holds only when sin a = sin b = sin c = sin d = sin e =
− 9 = 4.
3 . 5
Also solved by Ioan Viorel Codreanu, Satulung, Maramures, Romania; Moubinool Omarjee, Lycee Henri IV, Paris, France; Nicusor Zlota, Traian Vuia Technical College, Focsani ,Romania; Albert Stadler, Herrliberg, Switzerland; Seung Hwan An, The Taft School; Sewon Park, Peddie School, Hightstown, NJ, USA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA; Kwonil Ko, Cushing Academy, Ashburham, MA; Yeonjune Kang, Peddie School; Joseph Lee, Loomis Chattee School, Windsor, CT; YunJin Jeong, Emma Willard School, Troy, NY, USA; Kyoung A Lee, The Hotchkiss School, Lakeville, CT; Ji Eun Kim, Tabor Academy, MA; SooYoung Choi, ChungDam Middle School, Seoul, South Korea; Nikos Kalapodis, Patras, Greece; Li Zhou, Polk State College, USA; Brian Bradie, Christopher Newport University, Newport News, VA; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Daniel Lasaosa, Pamplona, Spain; Spain; Latofat Latofat Bobojonova, Bobojonova, Academic Academic Lyceum S.H.Sirojiddinov, Tashkent, Tashkent, Uzbekistan; Khurshid Juraev, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Tolibjon Ismoilov, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan.
Mathematical Mathematical Reflections
6 (2015)
8
S357. Prove Prove that in any triangle, triangle,
a(ha 2r) (3a (3a + b + b + + c c)( )(h ha + 2r 2 r)
−
≤ 43
Proposed by Mihály Bencze, Braşov, România Solution by Arkady Alt, San Jose, California, USA Since h Since h a =
2sr , then a
2sr 2r ha 2r s a a = = , 2sr ha + 2r 2r s + a + a + 2r 2r a
−
−
−
and by AM-GM,
a (ha 2r) a (s a) a (s a) 2a (s a) = = = 2 (3a (3a + b + b + + c c)) (ha + 2r 2 r) (2a (2a + 2s 2s) (s + a + a)) 2 (s + a + a)) 4 (s + a + a))2
−
Hence,
cyc
−
a (ha 2r ) (3a (3a + b + b + + c 2 r) c)) (ha + 2r
−
−
− ≤
2a + (s ( s a) 2 4 (s + a + a))2
−
2
=
1 . 16
≤ cyc
1 3 = . 16 4
Also solved by Edgar Wang; Albert Stadler, Herrliberg, Switzerland; Seung Hwan An, The Taft School; Neculai Stanciu and Titu Zvonaru, Romania; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA; Ji Eun Kim, Tabor Academy, M; Joseph Lee, Loomis Chattee School, Windsor, CT; Yeonjune Kang, Peddie School; Kwonil Ko, Cushing Academy, Ashburham, MA; Kyoung A Lee, The Hotchkiss School, Lakeville, CT; SooYoung Choi, ChungDam Middle School, Seoul, South Korea; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Li Zhou, Polk State College, USA; Khurshid Juraev, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Brian Bradie, Christopher Newport University, Newport News, VA; Latofat Bobojonova, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Daniel Lasaosa, Pamplona, Spain; Tolibjon Ismoilov, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Nikos Kalapodis, Patras, Greece; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania.
Mathematical Mathematical Reflections
6 (2015)
9
S358. Prove that for each each integer n, there are eighteen integers integers such that both their sum and the sum of their fifth powers are equal to n.
Proposed by Nairi Sedrakyan, Armenia Solution by Li Zhou, Polk State College, USA First, consider n = 2m 1. Notice Notice that that (2m (2m 1)5 (2m (2m 1) = 8m 8m(m 1)(2m 1)(2m 1)(2m 1)(2m2 2m + 1). 1). If m 1 (mod 3) then 3) then m m 1 0 (mod 3); 3); if m 1 (mod 3) then 3) then 2m 1 0 (mod 3). 3). If m 1 (mod 5) 2 then m then m 1 0 (mod 5); 5); if m 2, 1 (mod 5) then 5) then 2m 2m + 1 0 (mod 5); 5); if m 2 (mod 5) then 5) then 5 2m 1 0 (mod 5). 5). Since 120 Since 120 = 8 3 5, we conclude that (2 that (2m m 1) 2m 1 (mod 120). 120). So we can write 5 (2m (2m 1) = 120q 120q + + 2m 2 m 1.
− − ≡
− − ≡−
− −
− − ≡ ≡ − ≡ −
−
− ≡
≡ − ≡ ≡ − ≡− − ≡ · · − − Let a Let a 1 = 2m 2 m − 1, a 2 = 0, a 3 = −(q − 2), 2), a 4 = · · · = a 7 = q − 1, a 8 = · · · = a13 = −q , a 14 = · · · = a 17 = q + q + 1, and a18 = −(q + + 2). 2) . Then a Then a 3 + · · · + a18 = 0 and 5 a53 + · · · + a18 = −(q − 2)5 + 4(q 4(q − 1)5 − 6q 5 + 4(q 4(q + + 1) 5 − (q + + 2) 5 = −120 120q. q. 5 . Hence a1 + · · · + a18 = 2m − 1 = a 51 + · · · + a18 5 . Now if n = change a2 from 0 from 0 to to 1 that a1 +· · ·+a18 = 2m = a n = 2m, then we simply change a 1,, and it is obvious that a = a 51 +· · ·+a18
Mathematical Mathematical Reflections
6 (2015)
10
S359. Prove Prove that in any triangle, triangle, ma
1 2ra
−
R bc
+ mb
1 2rb
−
R ca
+ mc
1 2rc
−
R ab
≥
0.
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Brian Bradie, Christopher Newport University, Newport News, VA With
abc , s a s b s c 4∆ where ∆ where ∆ is the area of the triangle and s is the semi-perimeter, it follows that ra =
∆
−
,
rb =
1 2ra 1 2rb 1 2rc
∆
,
−
− bcR
=
− acR
=
− abR
=
rc =
2(s 2(s
∆
−
,
and R =
− a) − a ,
4∆ 2(s 2(s b) 4∆ 2(s 2(s c) 4∆
− − b , and − − c.
Thus, ma is equivalent to
1 2ra
−
R bc
+ mb
1 2rb
−
R ac
+ mc
(a + b + b + + c c)( )(m ma + m + mb + m + mc )
1 2rc
−
R ab
≥
0
3(am + bm + bm + cm + cm ). (1) ≥ 3(am Without loss of generality, suppose that a ≥ b ≥ c. Then m ≤ m ≤ m , and (1) follows from Chebyshev’s a
a
c
b
b
c
sum inequality. inequality.
Also solved by Sewon Park, Peddie School, Hightstown, NJ, USA; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA; YunJin Jeong, Emma Willard School, Troy, NY, USA; Kyoung A Lee, The Hotchkiss School, Lakeville, CT; Kwonil Ko, Cushing Academy, Ashburham, MA; Ji Eun Kim, Tabor Academy, MA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA; Seung Hwan An, The Taft School; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Nikos Kalapodis, Patras, Greece; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Li Zhou, Polk State College, Winter Haven, FL; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Arkady Alt, San Jose, California, USA; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City; Daniel Lasaosa, Pamplona, Spain; Khurshid Juraev, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Uzbekistan; Latofat Latofat Bob Bobojonova, ojonova, Academic Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan.
Mathematical Mathematical Reflections
6 (2015)
11
S360. Let ABC be ABC be a triangle with orthocenter H H and circumcenter O. The parallels parallels through through B and C to at D and E and E ,, respectively. Prove that BE,CD, that BE,CD, AH AO intersect AO intersect the external angle bisector of ∠BAC at D are concurrent.
Proposed by Iman Munire Bilal, University of Cambridge and Marius Stanean, România Solution by SooYoung Choi, ChungDam Middle School, Seoul, South Korea Let H Let H be the second intersection intersection of AH with AH with the circumcircle of ABC . ABC . Since AE Since AE is is the external bisector of ∠BAC , BAC , ∠DAB = DAB = ∠EAC . EAC . Since D Since DB B AO and AH and AO and AO are isogonal,
∠DBA DB A = ∠OAB
= ∠H AC = ∠H BC,
using the fact that H ABC is ABC is cyclic. In a similar way, ∠DAB
= ∠EAC, ∠DBA DB A =
∠ACD = ∠H CB .
Therefore,
∠H BC, ∠ACE = ∠BC H ,
so by Jacobi’s theorem, AH,CD, theorem, AH,CD, and and B E are are concurrent.
Also solved by Li Zhou, Polk State College, Winter Haven, FL; Daniel Lasaosa, Pamplona, Spain; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Gabriel Chicas Reyes, El Salvador; Khurshid Juraev, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Philippe Fondanaiche, Paris, France; George Gavrilopoulos.
Mathematical Mathematical Reflections
6 (2015)
12
Undergraduate problems
U355. Let a be a real number such that a = 0 and a = 1, and let n be an integer greater than 1. Find all polynomials P ( P (X ) with real coefficients such that
±
a2 X 2 + 1 P ( P (aX ) = a2n X 2 + 1 P ( P (X ).
Proposed by Marcel Chiriţă, Bucharest, România
Solution by Li Zhou, Polk State College, USA Notice that for 1 for 1 j < k n, a 2 j X 2 + 1 has complex zeros i/a j , which are not zeros of a a 2k X 2 + 1. Hence, the given functional equation forces P ( P (X ) = (a2 X 2 + 1)Q 1)Q1 (X ). Then P ( P (aX ) = (a4 X 2 + 1)Q 1)Q1 (aX ), which 4 2 in turn forces Q forces Q 1 (X ) = (a X + 1)Q 1)Q2 (X ). Inductively, we see that
≤
≤
P ( P (X ) = (a2 X 2 + 1)(a 1)(a4 X 2 + 1)
±
1)Q −1 (X ). · · · (a2 −2X 2 + 1)Q n
n
Now the functional equation is satisfied if and only if Qn−1 (aX ) = Q n−1 (X ), thus Q thus Q n−1 (X ) must be a real constant c. Therefore, P ( P (X ) = c( c (a2 X 2 + 1)(a 1)(a4 X 2 + 1)
1), · · · (a2 −2X 2 + 1), n
c
∈ R.
Also solved by Daniel Lasaosa, Pamplona, Spain; Albert Stadler, Herrliberg, Switzerland; Shohruh Ibragimov, National University of Uzbekistan, Tashkent, Uzbekistan; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Kwonil Ko, Cushing Academy, Ashburham, MA; Ji Eun Kim, Tabor Academy, MA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA; Seung Hwan An, The Taft School.
Mathematical Mathematical Reflections
6 (2015)
13
U356. Let ( Let (x xn )n≥1 be a monotonic sequence, and let a lim
n→∞
∈ (−1, 0). 0). Find x1 a −1 + x2 a −2 + · · · + x n
n
n
.
Proposed by Mihai Piticari and Sorin Rădulescu, România
Solution by Daniel Lasaosa, Pamplona, Spain Denote by y by y n the expression whose limit we are asked to find. Note first that yn
− ay −1 = x = x 1 a −1 + x2 a −2 + · · · + x − a n
n
x1 an−2 + x2 an−3 +
· · · + x −1 = x . Now, if y converges to limit L, for any > 0 there is an N such N such that for all n ≥ N we N we have |y − L| < . Then, x Then, x +1 = y +1 − ay −1 ∈ ((1 − a)L − 2, (1 − a)L + 2 2), or sequence (x ) needs to converge (in fact, it must converge to ( to (11 − a)L where L where L is the limit of (y )) for ( for (x x ) ≥1 to converge. Since (x ( x ) is monotonic, it n
n
n
n
n
n
n
n
n
n
n
n n
n
converges iff it is bounded.
Case 1: If (x ( xn )n≥1 is bounded, then it is convergent with some limit L. L. Consider any any > 0, 0 , and note that n
| |
{
}
(1−a) since a < 1, 1 , then a 0 when n when n , ie., there exists some N some N 1 such that, for all n all n N 1 , a < (1− . 3L At the same time, since the sequence ( sequence (x xn ) is convergent with limit L limit L,, there exists some M some M 1 such that, for all (1−| (1−|a|) n M 1 , we have xn L < . On the the other other hand, hand, since (xn ) is monotonic and bounded, there exists 3 a difference D = x1 L xn L for all n. Using Using again again that a n converges to 0, there exists some M 2 (1−|a|) such that, for all n all n M 2, we have a n+1 < (1−| Taking M = = max M 1 , M 2 and N = N = max N 1 , M + + 2 , 3D . Taking M we have for all n all n N N that n
||
|| →
≥
→∞
≥
| − | | − | ≥ | − | ≥ | | ≥
− − L
1
while
| |
2
L 1 + a + a + + a a +
a
xn + ax + axn−1 + a + a2 xn−2 +
· · · + a
· · · + a
M
xn−M
− L
M
n−1
{
an L = < , 1 a 3
−
1 + a + a + + a a2 +
≤ |x − L| + |a| |x −1 − L| + · · · + |a| |x − − L| < n
n
n M
and
· · · + a +1 1 − |a| +1 M
3
M
≤
< , 3
+1 · · · + a −1x1 − L a +1 + · · · + a −1 ≤ +1 ≤ |a| +1 |x − −1 − L| + · · · + |a| − −2 |x1 − L| ≤ 1 − |a| − −1 +1 +1 − −2 < < , ≤ |a| D 1 + |a| + · · · + |a|
+1 aM +1 xn−M −1 +
}
M
M
or using these three equations,
yn
−1
≤ | − L
a
xn
n
n M
n M
M
n
n M
n M
3
− L| + |a| |x −1 − L| + · · · + |a| −1 |x1 − L| + n
n
3
an L < . 1 a
−
It then follows that, if ( if (x xn )n≥1 is bounded hence convergent to some limit L, then lim x1 an−1 + x2 an−2 +
n→∞
· · · + x
n
=
L
1
− a.
If (xn )n≥1 is increasing but not bounded, note that xi > xi−1 for all i 2, whereas for some N 0 we have xn 0, and for some M 0 we have xn x1 + x + x2 + + xN −1 for all n M , M , or for all even 2 n M + + 2, and since 0 since 0 < < a < 1, 1, we have
≥
≥
≥ ≥
x1 an−1 + x2 an−2 + Mathematical Mathematical Reflections
6 (2015)
≥ |
+ a)) · · · + x ≥ (1 + a n
≥ |
···
xn + a + a2 xn−2 +
≥
≥
+ a))x , · · · + a −2x2 ≥ (1 + a n
n
14
whereas for all odd n odd n
≥ M + 2,2, we have x1 a −1 + x2 a −2 + · · · + x ≥ (1 + a + a)) x + a + a2 x −2 + · · · + a −2 x2 + a −1 x1 ≥ (1 + a + a))x + a + a −1 x1 , both of which diverge to +∞ because a because a −1 converges to 0 to 0,, or lim x1 a −1 + x2 a −2 + · · · + x = +∞. →∞ n
n
n
n
n
n
n
n
n
n
n
n
n
n
An analogous argument allows us to conclude that, if ( if (x xn )n≥1 is decreasing but not bounded, then lim x1 an−1 + x2 an−2 +
n→∞
· · · + x
n
=
−∞.
= 0. If D = 0, then (xn ) is Note: The previous manipulations in Case 1 assume that D > 0 and L
constant with value L value L,,
n
· · · + a −1 = 1 −L a − 1La− a , and the second term in the RHS converges to 0 because |a| < 1, 1 , with identical result as in the general case. 0 but L If D D = but L = = 0, we may skip the definition of N N 1 , define M define M as as in the general case, and take N take N = M + + 2, 2 or for all n all n ≥ N , we have |y | < 3 < , and (y ) converges to 0 = −1 , again as in the general case.
yn = L = L 1 + a + a + +
n
n
n
L
a
Also solved by Moubinool Omarjee, Lycée Henri IV, Paris, France; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy.
Mathematical Mathematical Reflections
6 (2015)
15
U357. Evaluate Evaluate lim
n→∞
· · · 1+
1
1+
n2
2
n n2
1+
n2
√ e
n
Proposed by Dorin Andrica, Babeş-Bolyai University, Cluj Napoca, România Solution by Arkady Alt, San Jose, California, USA Let a Let a n := ln
1 1+ 2 n
2 1+ 2 n e
Since ln(1 + x + x)) = x Hence,
n
√ − ∈ − − − − − − − − − − − n ... 1 + 2 n
x2 x3 k + + o x3 then ln then ln 1 + 2 2 3 n an = n = n
k=1
k=1
k n
k2 k3 + + no 2n3 3n5 n (n + 1) 2n
1 n3
k n2
,n
N.
k = 2 n
k2 k3 + + o 2n4 3n6
1 n3
k2 k3 + + o 2n4 3n6
n = 2
n
k=1
k n
1 n3
1 2
n
n 2
1 2
k=1
(n ( n + 1) 1) (2n (2n + 1) + o 12 12n n2
for k for k = = 1, 2,...,n.
=
n
k2 + 2n3
k=1
k3 + o 3n5
1 n2
=
=
√ e.
1 n2
=
1 n
− 61 = 31 .
n→∞
n
√ 1 1+ 2 n
then
lim
k n2
n (n + 1)(2n 1)(2n + 1) n2 (n + 1)2 + + o 12 12n n3 12 12n n5
n 2
1 2
1 and, therefore, lim an = n→∞ 2 Since
ln 1 +
k=1
n
n
n
= n
2 1+ 2 n e
n ... 1 + 2 n
√ 1 1+ 2 n
2 1+ 2 n e
n ... 1 + 2 n
= e a
n
n
= e
lim a
n
n→∞
3
Also solved by Daniel Lasaosa, Pamplona, Spain; Albert Stadler, Herrliberg, Switzerland; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Moubinool Omarjee, Lycée Henri IV, Paris, France; Joshua Siktar,Carnegie Mellon University, USA; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Li Zhou, Polk State College, USA; Zafar Ahmed, BARC, India; ANanduud Problem Solving Group, Ulaanbaatar, Mongolia; Kwonil Ko, Cushing Academy, Ashburham, MA, USA; Ji Eun Kim, Tabor Academy, MA, USA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA; Brian Bradie, Christopher Newport University, Newport News, VA, USA.
Mathematical Mathematical Reflections
6 (2015)
16
U358. Let ( Let (x xn )n≥1 be an increasing sequence of real numbers for which there is a real number a > 2 such 2 such that xn+1
1)x −1 , ≥ ax − (a − 1)x
for all n all n
n
n
≥ 1. Prove that ( that (x x ) ≥1 is divergent. n n
Proposed by Mihai Piticari and Sorin Rădulescu, România Solution by Daniel Lasaosa, Pamplona, Spain Note that xn+1
> 2 (x − x −1 ) , x +1 − x > x − x −1 . − x −1 ≥ a (x − x −1) > 2 Denote y Denote y = x +1 − x , or ( or (yy ) ≥1 is an increasing sequence of positive reals, hence it either converges to a positive positive real, or it diverges. diverges. In either case, there there is a positive real r and a positive integer N such N such that, for all n all n ≥ N , then y then y ≥ r, where r where r can be taken as half the limit of y in the case it converges, and r and r can take any positive real value in the case that y diverges. In either case, for every n ≥ N , we have x have x ≥ (n − N )r, n
n
n
n
n
n
n
n
n
n
n
n
n n
n
n
n
n
which clearly diverges. The conclusion follows.
Also solved by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Moubinool Omarjee, Lycée Henri IV, Paris, France; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City; Francisco Javier Martínez Aguinaga, Universidad Complutense de Madrid, Spain; Brian Bradie, Christopher Newport University, Newport News, VA, USA; Arkady Alt, San Jose, CA, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Kwonil Ko, Cushing Academy, Ashburham, MA, USA; Ji Eun Kim, Tabor Academy, MA, USA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA; Seung Hwan An, The Taft School.
Mathematical Mathematical Reflections
6 (2015)
17
U359. Let a1 , . . . , an and b1 , . . . , bm be sequences sequences of nonnegativ nonnegativee real numbers. numbers. Furthermore urthermore,, let c1 , . . . , cn and d1 , . . . , dm be sequences of real numbers. Prove that n
n
m
ci c j min(a min(ai , a j ) +
i=1 j =1
m
n
di d j min(b min(bi , b j )
≥2
i=1 j =1
m
ci d j min(a min(ai , b j ).
i=1 j =1
Proposed by Mehtaab Sawhney, Commack High School, New York, USA Solution by Daniel Lasaosa, Pamplona, Spain Note first that if ak = 0 for some k 1, 2, . . . , n , then then all all term termss wher wheree ck appear appear do vanish, anish, since since min(a min(ai , ak ) = min(a min(ak , b j ) = ak = 0 because ai , b j 0 for all i, j , hence hence the problem problem is equiv equivale alent nt to eliminating ak , ck , reducing n by 1 and renumbering the i’s in the ai ’s and ci ’s from 1 to n 1. In oth other er words, we may assume wlog that all ai ’s and b j ’s are positive, and all ci , d j such that ai = 0 or b j = 0 are irrelevant since they do not participate because the terms in which they appear, vanish.
∈{
} ≥
−
Let u be the numbe numberr of distin distinct ct positiv positivee values alues in a1 , a2 , . . . , an , b1 , b2 , . . . , bm . We will will prov prove by induction on u the following Claim: The proposed inequality always holds, with equality iff the ai ’s and b j ’s can be partitioned in groups such that their values are equal, and such that the sum of the c the ci ’s and the sum of the d the d j ’s corresponding to each group of equal-valued a i ’s and b and b j ’s are equal. Proof: When u = 1, a1 = a2 = = an = b1 = b2 = = bm = K K for some positive real K , and denoting S c = c 1 + c + c2 + + cn and S and S d = d = d 1 + d + d2 + + dm , the inequality rewrites as
{
· · ·
···
n
0
≤ K
n
m
m
ci c j +
i=1 j =1
· · ·
···
− n
di d j
i=1 j =1
}
m
2
= K S c 2 + S d 2
ci d j
i=1 j =1
= K ( K (S c
− 2S S
c d
=
− S )2, d
and the Claim is clearly clearly true in this this case. case. Assume Assume that the result result is true true for 1, 2, . . . , u 1, and let = min a1 , a2 , . . . , an , b1 , b2 , . . . , bm . Define a Define a i = a i and b and b j = b j . Note that the inequality inequality then rewrites as
{
}
n
−
n
m
m
ci c j +
i=1 j =1
i=1 j =1
m
m
2
n
ci d j
di d j min(b min(bi , b j )
i=1 j =1
ci c j min(a min(ai , a j )+
i=1 j =1
n
n
+
i=1 j =1
m
+
−
− ≥ n
di d j
−
m
ci d j min(a min(ai , b j ),
2
i=1 j =1
where there are u are u distinct distinct values of the a i ’s and b and b j ’s, but one of them is zero, and all zero a i ’s and b and b j ’s and their corresponding c corresponding c i ’s and d and d j ’s can be removed removed,, yielding yielding u u 1 distinct nonzero values of the a i ’s and b and b j ’s. Therefore, by hypothesis of induction for u 1, the last two terms in the LHS are no less than the RHS. At the same time, by hypothesis of induction for u = u = 1, the first term in the LHS is non-negative, being zero iff the sum of all ci ’s and the sum of all d j ’s are equal. The Claim follows follows for all u. Note Note that for equalit equality y to hold in each step in the induction process, the sum of each set of ci ’s and d j ’s that remain after removing each must be equal, or the sum of the ci ’s and d j ’s that disappear after each is removed must also be equal.
−
−
The conclusion follows directly from the Claim, equality holds if, for every nonzero value taken by some of the a the a i ’s and b and b j ’s, the sum of the c i ’s and the sum of the d j ’s corresponding to the a i ’s and b and b j ’s with equal value, are equal, and the c the c i ’s and the d j ’s corresponding to zero a zero a i ’s and b and b j ’s are irrelevant.
Also solved by Moubinool Omarjee, Lycée Henri IV, Paris, France.
Mathematical Mathematical Reflections
6 (2015)
18
U360. Let f Let f : [ 1, 1]
−
→ [0, [0, ∞) be b e C 1 monotone increasing function. Prove that 1
(f (x))
1
1 2015
≤ 2015
−1
f ( f (x) 1 x
1 2015
dx.
−
−1
Proposed by Oleksiy Klurman, University College London Solution by the author Let q Let q = =
1 . Changing f Changing f ((x) 2015 1
S = =
−1
f (x) − f (0) f (0) we we can assume f ( f (−1) = 0. 0. Integration by parts yields → f (
f q (x) 1 dx = dx = f q (x)(1 q (1 x) q 1
−
−
−
x)1−q 1−1 +
|
1
q 1
− q
f (x)f q−1 (x)(1
−1
− x)1− dx. q
Since f ( f ( 1) = 0 we 0 we have
−
S 1 =
1
− q S = q
1
f (x)f q−1 (x)(1
−1
− x)1− dx. q
We now estimate S estimate S 1 + S + S to to get the result: 1 S 1 + S + S = S = q since
1
−1
f q + f (x)f q−1 (x)(1 q (1 x)
−
f q (x) + f ( f (x)f q−1 (x)(1 q (1 x)
−
pointwise. Indeed, if
f q (x) (1 x)q
1−q
− x)
1
≥ dx
[f (x)]q dx
−1
− x)1− ≥ [f (x)] . q
f (x)] − ≥ [f (
q
q
the inequality clearly holds. In the other case, if
f q (x) < [f [f (x)]q , q (1 x)
−
then
[f (x)]q−1 < (1
− x)1− f −1(x) q q
and the second term dominates the right-hand side.
Mathematical Mathematical Reflections
6 (2015)
19
Olympiad problems
O355. Let AB Let AB C be be a triangle with incenter I incenter I . Prove that (IB + I + IC C )2 (IC ( IC + + I IA A)2 (I A + I + IB B )2 + + a(b + c + c)) b(c + a + a)) c(a + b + b))
≤ 2.
Proposed by Nguyen Viet Hung, High School for Gifted Students, Hanoi University of Science, Vietnam Solution by Arkady Alt, San Jose, CA, USA Since
2 bcs (s la = b + c + c then
IA + IA + I IB B =
bcs (s s
− a) +
cas (s s
− a) , 2
and c (IA + IA + I IB B) = s By Cauchy Inequality
b (s
Thus,
− a) +
(IA + IA + I IB B )2
b)
b (s
2
− ≤ a (s
cas (s c + a + a
− b)
la (b (b + c + c)) lb (c (c + a + a)) + = a + b + b + + c c a + b + b + + c c
− b) =
2
c s
b (s
− a) +
(b + a + a)) (s
2
+ b)) ⇐⇒ ≤ cs (a + b
− a) +
− a (s
− a (s
b)
2
b)
.
− a + s + s − b) = c (a + b + b)) .
(I A + I + IB B )2 c(a + b + b))
≤ sc
and, therefore,
cyc
(I A + I + IB B )2 c(a + b + b))
≤ cyc
c = 2. s
Also solved by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Adnan Ali, Student in A.E.C.S4, Mumbai, India; Khurshid Juraev, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Latofat Bobojonova, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Nicusor Zlota ‚Traian Vuia Technical College, Focsani, Romania; Neculai Stanciu and Titu Zvonaru, Romania; Tolibjon Ismoilov, Academic Lyceum S.H.Sirojiddinov, S.H.Sirojiddinov, Tashkent, Uzbekistan; Uzbekistan; Ioan Viorel Codreanu, Codreanu, Satulung, Maramures, Maramures, Romania; Li Zhou, Polk State College, USA.
Mathematical Mathematical Reflections
6 (2015)
20
O356. We take out an even even number number from the set 1, 2, 3, . . . , 25 . Find this number knowing that the remaining set has precisely 1 precisely 124 24 subsets subsets with three elements that form an arithmetic progression.
{
}
Proposed by Marian Teler, Costeşti and Marin Ionescu, Piteşti, România Solution by Adnan Ali, Student in A.E.C.S-4, Mumbai, India First we prove by iteration that the set 1, 2, , 25 has 144 subsets with three elements that form an AP. Let f Let f ((m) denote the number of such subsets for the set 1, 2, , m (m 3). Consider the sets
{
···
}
{
··· } ≥ A = {1, · · · , 2n + 1 } B = {1, · · · , 2n + 1, 1, 2n + 2, 2 , 2n + 3},
where n Consid ider er 2n + 2 in B . Any Any threethree-te term rm AP with with 2n + 2 as a member is of the form Z+ . Cons 2i, i + n + 1, 1, 2n + 2 for any integer 1 integer 1 i n, with another three-term AP 2n + 1, 1, 2n + 2, 2, 2n + 3 . Since there are n choices for i in total, and one extra AP of which 2n + 2 is a member, there are exactly n + 1 three-term AP’s of which 2 which 2n n + 2 can be a member. Next we consider 2n 2 n + 3 in B . Any AP with 2 with 2n n + 3 as a member is of the form 2 j 1, n + j + j + 1, 2n + 3 for any integer 1 integer 1 j n + 1. Thus j Thus j having n having n + 1 choices, there are only n + 1 three-term 1 three-term AP’s of which 2 which 2n n + 3 can 3 can be a member. But here we have over-counted the three-term AP 2n + 1, 1 , 2n + 2, 2, 2n + 3 , and so we must subtract it once from the total. Thus we have
∈
{
}
≤ ≤
{ −
{
}
{
}
≤ ≤
}
f (2 f (2n n + 3) = f = f (2 (2n n + 1) + n + n + + 1 + n + n + + 1
− 1 = f (2 f (2n n + 1) + 2n 2n + 1. 1.
Knowing that f (3) f (3) = 1, 1, we conclude that f (25) f (25) = 122 = 144 144.. Now consider consider any any any even number number 2k 1, 2, , 25 . Since Since its remov removal has left left the remaini remaining ng set with only 124 124 such such subsets, we conclude that 2k is a member of exactly 144 124 = 20 three20 three-eleme element nt AP’s. But any three-elem three-element ent AP involving involving 2k is of the form 2u, u + k, + k, 2k for any integer 1 u 12 12,, u = k , (which implies that there are 11 11 choices choices for u) and also of the form 2k v, 2k, 2k + v for 1 2k v 2k 1 and 2k + 1 2k + v + v 25 25 which gives 1 v min.(2 min.(2k k 1, 25 2k). In case case min. min.(2 (2k k 1, 25 2k) = 2k 1, v has 2k 1 choices and so in total, 11 total, 11 + 2k 2k 1 = 20 20 which yields 2k = 10 10,, (and clearly 10 1 < 25 10 10)) and the other case when min.(2 min.(2k k 1, 25 2k) = 25 2k , there are 25 are 25 2k choices for v for v.. Thus 11+ Thus 11+ 25 2k = 20 20 which which yields 2 yields 2k k = 16 (and clearly 25 clearly 25 16 16 < < 16 16 1). Thus we conclude that the even number removed is either 10 or 16. 16 .
{
} { ≤ ≤ − − − −
∈
···
−
} { − − − − −
≤ ≤ } ≤ − ≤ − − − − −
−
≤
− −
−
≤
Also solved by Marian Teler, Costesti and Marin Ionescu, Pitesti, Romania; Li Zhou, Polk State College, USA; Prishtina Math Gymnasium Problem Solving Group, Republic of Kosova; Daniel Lasaosa, Pamplona, Spain; Khurshid Juraev, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzekistan; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City; Adithya Bhaskar, Atomic Energy Central School-2, Mumbai, India; Kwonil Ko, Cushing Academy, Ashburham, MA, USA; Ji Eun Kim, Tabor Academy, MA, USA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA.
Mathematical Mathematical Reflections
6 (2015)
21
O357. Prove Prove that in any triangle ab + ab + 4m 4ma mb bc + bc + 4m 4mb mc ca + ca + 4m 4mc ma + + c a b
16K K ≥ 16 . R
Proposed by Titu Andreescu, USA and Oleg Mushkarov, Bulgaria Solution by Nikos Kalapodis, Patras, Greece
−
By the AM-GM inequality we have that a = (s b) + (s ( s c) Similarly b Similarly b 2 (s c)(s )(s a) and c 2 (s a)(s )(s b).
− − ≥2 ≥ − − ≥ − − 2(b 2(b2 + c2 ) − a2 (b + c + c))2 − a2 2 ≥ We also have m have m = = s( s (s − a). 4 4 Thus, m Thus, m ≥ s(s − a), and similarly m ≥ s(s − b) and m and m ≥
(s
b)(s )(s
− c).
a
a
Therefore
b
cyc
ab + ab + 4m 4 ma mb c
a + b + b c
−
4
cyc
−
24 3 (s
≥
4(s 4(s
c
−
− c)
cyc
− b) ≥ 4 · 3
(s
a)(s )(s
s(s
3
a)(s )(s
a)(s )(s
− b)(s )(s − c) ≥ 24 ·
− c).
−
4s − b) + 4s c
(s
(s
(a + b + b)( )(bb + c + c)( )(cc + a + a)) (s abc
1
s 72 72K K
3 1
−a + s−
a)(s )(s
− b) =
)(s − b)(s )(s − c) ≥ − a)(s
72 = r + r + rc = 1 a + rb + r + K b s c
−
1 2 16 16K K · r + r ≥ 72 72K K · = . + r + r + r 9R R a
b
c
Where we used successively the AM-GM Inequality, the well-known inequality (a ( a + b + b)( )(bb + c + c)( )(cc + a + a)) 9R the AM-HM Inequality, and finally the inequality ra + r + rb + r + rc = r + r + 4R 4R 2 .
≤
abc, ≥ 8abc,
Also solved by Daniel Lasaosa, Pamplona, Spain; Li Zhou, Polk State College, USA; Albert Stadler, Herrliberg, Switzerland; Nicusor Zlota, Traian Vuia Technical College, Focsani, Romania; Latofat Bobojonova, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA; Arkady Alt, San Jose, California, USA; Ioan Viorel Codreanu, Satulung, Maramures, Romania; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Kwonil Ko, Cushing Academy, Ashburham, MA, USA; Ji Eun Kim, Tabor Academy, MA, USA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA.
Mathematical Mathematical Reflections
6 (2015)
22
O358. Let a,b,c,d be non-negat non-negative ive real numbers such such that a 1 b c d and a + b + b + + c c + + d d = = 4. Prove Prove that 15 9 abcd + abcd + 2(ab 2(ab + + bc bc + + cd cd + + da da + + db db + + ac ac)) a2 + b2 + c2 + d2
≥ ≥ ≥ ≥ ≥
Proposed by Marius Stanean, Zalau, Romania Solution by M.A.Prasad, Mumbai, Maharashtra, India Let S Let S 2 = a2 + b2 + c2 + d2 . Then 2( Then 2(ab ab + bc + cd + da + db + ac) ac) = (a + b + c + d)2 we need to prove that
− S 2 = 16 − S 2. Therefore,
15 2(ab 2(ab + + bc bc + + cd cd + + da da + + db db + + ac ac)) abcd(16 abcd(16 S 2 )S 2 + 24S 24 S 2 144
≥ a2 + b2 +9 c2 + d2 ⇔ − ≥ (∗) Let a Let a = 1 + δ, + δ, b = 1 − δ 1 , c = 1 − δ 2 , d = 1 − δ 3 with 0 ≤ δ 1 ≤ δ 2 ≤ δ 3 , ≤ 1 and δ = δ = δ 1 + δ + δ 2 + δ + δ 3 .... .... S 2 = (1 + δ )2 + (1 − δ 1 )2 + (1 − δ 2 )2 + (1 − δ 3 )2 abcd + abcd +
abcd
x = 4(1 + ) where x where x = = δ δ 2 + δ 12 + δ + δ 22 + δ + δ 32 4 = (1 + δ )(1 )(1 δ 1 )(1 δ 2 )(1 δ 3 )
−
− S 2
−
+ δ 2 δ 3 + δ + δ 3 δ 1 + δ + δ (δ 1 δ 2 + δ + δ 2 δ 3 + δ + δ 3 δ 1 ) − δ 1 δ 2 δ 3 − δδ 1 δ 2 δ 3 − δ 2 + δ 1δ 2 + δ + δ 22 + δ + δ 32 δ 2 δ 12 + δ − = 1 − δ 2 + + δ (δ 1 δ 2 + δ + δ 2 δ 3 + δ + δ 3 δ 1 ) − δ 1 δ 2 δ 3 − δδ 1 δ 2 δ 3 2 2 ≥ 1 − x2 x = 16 − 4 − x = 12(1 − ) 12 =1
16
−
15 ≥ 6, then 16− ≥ 16−
We note that if S S 2 get
S 2
9
S 2 .
Therefore, we assume 0
≤ x ≤ 2 Using these in the LHS of (*), we
x x x x 24S 2 ≥ 48(1 − )(1 + )(1 − ) + 96(1 + ) − S 2) + 24S 2 4 12 4 2 3 2 x 5x x x( x (x − 10 10x x + 16) = 48(3 + − + ) = 144 + 6 48 96 2 x(x − 2)(x 2)(x − 8) = 144 + 2 144 for for 0 ≤ x ≤ 2 ≥ 144
abcdS 2 (16
Also solved by Kwonil Ko, Cushing Academy, Ashburham, MA, USA; Ji Eun Kim, Tabor Academy, MA, USA; Hyun Jin Kim, Stuyvesant High School, New York, NY, USA.
Mathematical Mathematical Reflections
6 (2015)
23
O359. Solve, Solve, in positive positive integers, integers, the equation x6 + x3 y3
3xy − y6 + 3xy
x2
− y2 2 = 1.
Proposed by Titu Andreescu, University of Texas at Dallas, USA Solution by Li Zhou, Polk State College, USA Notice that x that x 6 + x3 y 3 y6 + 3xy( So x 2 y 2 + xy = by 4 we get xy(x2 y 2 )2 = (x2 y 2 + xy) xy)3 . So x xy = 1. Multiplying by 4 2 2 the well-known Pell equation ( equation (22x + y + y)) 5y = 4. 2 2 If y = 2z , then (x + z ) 5z = 1, with with minima minimall soluti solution on (x1 + z 1 , z1 ) = ( 9, 9, 4) and all solutions n given given by xn + z n + z n 5 = (9 + 4 5) for n then easy easy to see see that that (x1 , y1 ) = ( 5, 5, 8) and N. It is then (xn+1 , yn+1 ) = (5x (5xn + 8y 8 yn , 8xn + 13y 13 yn ). If y is odd, then we have the minimal solution (2x (2x1 + y + y1 , y1 ) = (3, (3, 1) and 1) and all solutions given by 2xn + n−1 yn + y + yn 5 = (3 + 5)(9 + 4 5) for n N. Again, Again, it is easy to see that this second family family of solutions solutions are ( are (x (1, 1) and 1) and (xn+1 , yn+1 ) = (5x (5xn + 8y 8 yn , 8xn + 13y 13 yn ). x1 , y1 ) = (1,
−
−
−
√
√
√
√
−
− √
−
∈
∈
Also solved by Daniel Lasaosa, Pamplona, Spain; Tolibjon Ismoilov, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzbekistan; Navid Safei, University of Technoogy in Policy Making of Science and Technology, Iran; M.A.Prasad, Mumbai, Maharashtra, India; Khurshid Juraev, Academic Lyceum S.H.Sirojiddinov, Tashkent, Uzekistan; Jorge Ledesma, Faculty of Sciences UNAM, Mexico City, Mexico; Arkady Alt, San Jose, California, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Lucie Wang, Lycée Louis le Grand, Paris, France.
Mathematical Mathematical Reflections
6 (2015)
24
O360. Find the least positive positive integer n with the following following property: property: for any polynomial polynomial P ( P (x) C [x], there exist polynomials f polynomials f 1 (x), f 2 (x),...,f n (x) C [x] and g and g 1 (x), g2 (x),...,gn (x) C [x] such that
∈
∈
∈
n
P ( P (x) =
(f i (x)2 + gi (x)3 )
i=1
Proposed by Oleksiy Klurman, University College London Solution by M.A.Prasad, Mumbai, Maharashtra, India We note that every polynomial P polynomial P ((x) C [x] can be expressed as the sum of the squares of two polynomials P (x)+1 and f 2 (x) = i P (x2)−1 . Choosi Choosing ng g1 (x) = ωg2 (x) where ω is a cube root of unity and g2 (x) f 1 (x) = 2 is any arbitrary polynomial, the polynomials f 1 (x), f 2 (x), g1 (x), g2 (x) will satisfy the desired relation. Now, we show by a counterexample that there exist polynomials which cannot be expressed as (f 1 (x)2 + g1 (x)3 ). Let P Let P ((x) = x 2 and assume to the contrary that there exist polynomials f 1 (x) and g1 (x) such that
∈
−
f 1 (x)2 + g1 (x)3 = x 2 Therefore, g1 (x)3 = (x f 1 (x))(x ))(x + f + f 1 (x)). )). The only only polynomia polynomiall which which can be a divisor divisor of (x f 1 (x)) as well as ( as (x x + f + f 1 (x)) is )) is x. There are two cases (i) x divides (x f 1 (x)) and )) and divides (x + f + f 1 (x)). )). In this case case x will also divide divide g1 (x). Let f 1 (x) = xs1 (x) and g1 (x) = xt 1 (x). We, then get
−
−
−
xt1 (x)3 = (1 (1 Now, ( Now, (11
+ s1 (x)) − s1(x))(1 + s
)) and (1 + s + s1 (x)) are )) are coprime. Therefore, − s1(x)) and 1 − s1 (x) = xt 3 (x)3 and 1 + s + s2 (x) = t 4 (x)3 ⇒ t4 (x)3 − xt3 (x)3 = 2
Clearly, there is no solution to this equation since the polynomials t 4 (x)3 and xt and xt 3 (x)3 are of different degree. (ii) x does not divide ( divide (x x f 1 (x)) or )) or does not divide (x ( x + f + f 1 (x)). )). In this case, we have
− x − f 1 (x) = t 3 (x)3 and x + f + f 2 (x) = t 4 (x)3 ⇒ t4 (x)3 − t3 (x)3 = 2x
This yields t4 (x)
x and t 4 (x)2 + t4 (x)t3 (x) + t + t3 (x)2 = − t3(x) = 2k and t k
Therefore, 4 (t4 (x) + t + t3 (x)) = (t4 (x)2 + t4 (x)t3 (x) + t + t3 (x)2 ) 3 2
−
1 (t4 (x) 3
−
4x t3 (x)) = 3k 2
−
4k 4 k 2 3
Clearly, this has no solution.
Also solved by Navid Safei, University of Technoogy in Policy Making of Science and Technology, Iran.
Mathematical Mathematical Reflections
6 (2015)
25
