junior problems - awesomemath · 2017. 1. 21. · junior problems j385. iftheequalities 2(a+b) 6c...

Junior problems

J385. If the equalities2(a+ b)− 6c− 3(d+ e) = 6

3(a+ b)− 2c+ 6(d+ e) = 2

6(a+ b) + 3c− 2(d+ e) = −3

hold simultaneously, evaluate a2 − b2 + c2 − d2 + e2.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution by AN-anduud Problem Solving Group, Ulaanbaatar, MongoliaGiven equations are equivalent to the following equations:

2(a+ b)− 6c− 3(d+ e) = 63(a+ b)− 2c+ 6(d+ e) = 2

6(a+ b) + 3c− 2(d+ e) = −3

⇔2(a+ b)− 6c− 3(d+ e) = 6

7c+ 10.5(d+ e) = −7−22.5(d+ e) = 0

Hence we get,

d+ e = 0, c = −1, a+ b = 0 ⇔ a2 = b2, d2 = e2, c2 = 1

⇔ a2 − b2 + c2 − d2 + e2 = 1.

Also solved by Daniel Lasaosa, Pamplona, Spain; Alessandro Ventullo, Milan, Italy; Andreas Chara-lampopoulos, 4th Lyceum of Glyfada, Glyfada, Greece; Prithwijit De, HBCSE, Mumbai, India; Adnan Ali,A.E.C.S-4, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; Alok Kumar, New Delhi, India; Andrian-na Boutsikou, High School Of Nea Makri, Athens, Greece; Ángel Plaza, University of Las Palmas de GranCanaria, Spain; David E. Manes, Oneonta, NY, USA; Duy Quan Tran, University of Health Science, HoChi Minh City, Vietnam; Orgilerdene Erdenebaatar, National University of Mongolia, Mongolia; JenniferJohannes, College at Brockport, SUNY, NY, USA; Evgenidis Nikolaos, M.N.Raptou High School, Larissa,Greece; Robert Bosch, USA and Jorge Erick, Brazil; Tamoghno Kandar; Polyahedra, Polk State College, FL,USA; Dolsan Zheksheev, Karakol High School, Kyrgyzstan; Bekjol Joldubai, Kadamcay High School, Kyrgyz-stan; Konstantinos Metaxas, 1st High School Ag. Dimitrios, Athens, Greece; Joehyun Kim, Bergen CountyAcademies, Hackensack, NJ, USA; A.S.Arun Srinivaas, Chennai, India; Sushanth Sathish Kumar, JefferyTrail Middle School, CA, USA; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon,MA, USA.

Mathematical Reflections 5 (2016) 1

J386. Find all real solutions to the system of equations

x+ yzt = y + ztx = z + txy = t+ xyz = 2.

Proposed by Mohamad Kouroshi, Tehran, Iran

Solution by Alessandro Ventullo, Milan, ItalySubtracting the second equation to the first equation, the third equation to the second equation, the fourthequation to the third equation and the first equation to the fourth equation, we obtain

(x− y)(1− zt) = 0(y − z)(1− tx) = 0(z − t)(1− xy) = 0(t− x)(1− yz) = 0.

We have four cases.

(i) x− y = 0 and z − t = 0, i.e. x = y and z = t. Substituting these values into the second and the fourthequation, we get (x − z)(1 − zx) = 0. If x = z, then x + x3 = 2, which gives x = y = z = t = 1. Ifzx = 1, then y + t = 2, i.e. x+ z = 2, which gives x = 1, z = 1, so x = y = z = t = 1.

(ii) x− y = 0 and 1− xy = 0, i.e. x = y and xy = 1. We obtain x = y = ±1. If x = y = 1, then zt = 1 andz+ t = 2, which gives z = t = 1. If x = y = −1, then zt = −3 and z+ t = 2, which gives z = 3, t = −1or z = −1, t = 3.

(iii) 1 − zt = 0 and z − t = 0, i.e. z = t and zt = 1. We obtain z = t = ±1. If z = t = 1, then xy = 1and x + y = 2, which gives x = y = 1. If z = t = −1, then xy = −3 and x + y = 2, which givesx = 3, y = −1 or x = −1, y = 3.

(iv) 1 − zt = 0 and 1 − xy = 0, i.e. zt = 1 and xy = 1. We obtain x + y = 2 and z + t = 2, which givesx = y = z = t = 1.

In conclusion, the real solutions to the given system of equations are

(x, y, z, t) ∈ {(1, 1, 1, 1), (1, 1, 3,−1), (1, 1,−1, 3), (3,−1, 1, 1), (−1, 3, 1, 1)}.

Also solved by Daniel Lasaosa, Pamplona, Spain; Polyahedra, Polk State College, FL, USA; Dolsan Zhek-sheev, Karakol High School, Kyrgyzstan; Bekjol Joldubai, Kadamcay High School, Kyrgyzstan; KonstantinosMetaxas, 1st High School Ag. Dimitrios, Athens, Greece; Joehyun Kim, Bergen County Academies, Hac-kensack, NJ, USA; Andreas Charalampopoulos, 4th Lyceum of Glyfada, Glyfada, Greece; David E. Manes,Oneonta, NY, USA; A.S.Arun Srinivaas, Chennai, India; Adnan Ali, A.E.C.S-4, Mumbai, India; AndriannaBoutsikou, High School Of Nea Makri, Athens, Greece; Erdenebayar Bayarmagnai, National University ofMongolia, Mongolia; Robert Bosch, USA; Albert Stadler, Herrliberg, Switzerland; Hyun Min Victoria Woo,Northfield Mount Hermon School, Mount Hermon, MA, USA.


J387. Find all digits a, b, c, x, y, z for which abc, xyz, and abcxyz are all perfect squares (no leading zerosallowed).

Proposed by Adrian Andreescu, Dallas, Texas

Solution by Polyahedra, Polk State College, USALetm2 = abc, n2 = xyz, and k2 = abcxyz. Then 1000m2+n2 = k2, with 10 ≤ m,n ≤ 31 and 317 ≤ k ≤ 999.There are three cases.

Case I: Suppose that n and k are not divisible by 5. Since 53|(k2 − n2), we must have either 53|(k − n)or 53|(k + n). Thus k = 125q ± n, where 3 ≤ q ≤ 8. Then 8m2 = 125q2 ± 2qn, so q must be even, that is,q ∈ {4, 6, 8}.

If q = 4, then m2 = 250± n. But 250 + n ∈ [261, 281] which contains no perfect square; and 250− n ∈[219, 239] \ {225} which contains no perfect square either.

If q = 6, then 2m2 = 125 · 9 ± 3n. So m = 3b, n = 3a, and 2b2 = 125 ± a. But 125+a2 ∈ [65, 67] which

contains no perfect square; and 125−a2 ∈ [58, 60] which contains no perfect square either.

If q = 8, then m2 = 1000 − 2n. So m = 2d, n = 2c, and d2 = 250 − c ∈ [236, 244] which contains noperfect square.

Case II: Suppose that 5|n but 25 does not divide n. Then 5|k but 25 does not divide k. Write n = 5aand k = 5b, with a ∈ {2, 3, 4, 6}. Then 40m2 + a2 = b2.

If a = 2, then b = 2c and c2−10m2 = 1. This Pell’s equation has fundamental solution (c1,m1) = (19, 6)and all solutions (ci,mi) given by ci +mi

√10 = (19 + 6

√10)i. Hence no such mi ∈ [10, 31].

If a = 3, then b2− 10(2m)2 = 9. This Pell’s equation has all solutions generated by three distinct sets offundamental solutions (b, 2m) = (7, 2), (13, 4), and (57, 18). Since (7 + 2

√10)(19 + 6

√10) = 253 + 80

√10,

we have either 2m ≤ 18 or 2m ≥ 80, thus no solution m ∈ [10, 31].If a = 4, then b = 4d and m = 2e. So d2 − 10e2 = 1, thus (d, e) = (19, 6), and (m,n, k) = (12, 20, 380).If a = 6, then b = 2f and f2 − 10m2 = 9. So (f,m) = (57, 18) and (m,n, k) = (18, 30, 570).Case III: Finally, consider 25|n. Then n = 25, k = 25a, and m = 5b. So a2 − 10(2b)2 = 1. Hence

(a, 2b) = (19, 6) and (m,n, k) = (15, 25, 475).In conclusion, there are three solutions: 144400, 324900, and 225625.

Also solved by Daniel Lasaosa, Pamplona, Spain; Alessandro Ventullo, Milan, Italy; David E. Manes,Oneonta, NY, USA; Albert Stadler, Herrliberg, Switzerland; Joel Schlosberg, Bayside, NY, USA; RobertBosch, USA.


J388. Let ABCD be a cyclic quadrilateral with AB = AD. Points M and N are taken on sides CD andBC, respectively, such that DM +BN =MN . Prove that the circumcenter of triangle AMN lies onsegment AC.

Proposed by Hayk Sedrakyan, Paris, France

Solution by Polyahedra, Polk State College, USAAs in the figure, locate E on MN such that EM = MD. Then BN = NE. So ∠ABE + ∠ADE =∠AEB + ∠AED. If ∠ABE > ∠AEB, then ∠ADE < ∠AED, which imply that AB < AE < AD, acontradiction. Likewise, we cannot have ∠ABE < ∠AEB. Hence AB = AE = AD, thus 4ABN ∼= 4AENand 4AEM ∼= 4ADM . Therefore,

∠ANM = ∠ANB = π − ∠ABN − ∠BAN = π − ∠ABD − ∠CAD − ∠BAN

=π

2+

1

2∠BAD − ∠CAM − ∠MAD − ∠BAN =

π

2− ∠CAM,

from which the conclusion follows.

EM

N

D

A

B

C

Also solved by Daniel Lasaosa, Pamplona, Spain; Dolsan Zheksheev, Karakol High School, Kyrgyzstan;Bekjol Joldubai, Kadamcay High School, Kyrgyzstan; Andrianna Boutsikou, High School Of Nea Makri,Athens, Greece; Erdenebayar Bayarmagnai,National University of Mongolia, Mongolia; Evgenidis Nikolaos,M.N.Raptou High School, Larissa, Greece; Robert Bosch, USA.


J389. Solve in real numbers the system of equations(x2 − y + 1

) (y2 − x+ 1

)= 2

[(x2 − y

)2+(y2 − x

)2]= 4.

Proposed by Alessandro Ventullo, Milan, Italy

Solution by Robert Bosch, USALet x2 − y = a, y2 − x = b. The system becomes

a2 + b2 = 2,

ab+ a+ b = 3.

After multiply by 2 the second equation and adding we obtain (a + b)2 + 2(a + b) − 8 = 0. So a + b = −4or a + b = 2. In the first case a2 +

49

a2= 2. This equation is equivalent to a4 − 2a2 + 49 = 0, that can be

considered a quadratic on a2, its discriminant is negative, thus there are not real solutions. Now, if a+b = 2,

the equation to be solved is a2 +1

a2= 2, or (a2 − 1)2 = 0, so a = ±1. Hence the pairs (a, b) to consider

are (1, 1) and (−1,−1). Anyways, x2 − y = y2 − x, or (x − y)(x + y + 1) = 0, so x = y or x + y = −1. If

x = y, the resulting equation is x2− x− 1 = 0, so x = y =1±√5

2. If x+ y = −1, and a = b = 1, we obtain

x2 + y2 = 1, and after plugging the value y = −x − 1, the quadratic 2x2 + 2x = 0, and x = 0, y = −1 orx = −1, y = 0. Finally if a = b = −1, then x2 + y2 = −3 < 0, impossible for real numbers. The solutions(x, y) to the original system are (

1±√5

2,1±√5

2

),

(0,−1),(−1, 0).

Also solved by Daniel Lasaosa, Pamplona, Spain; Polyahedra, Polk State College, FL, USA; BekjolJoldubai, Kadamcay High School, Kyrgyzstan; Konstantinos Metaxas, 1st High School Ag. Dimitrios, Athens,Greece; Joehyun Kim, Bergen County Academies, Hackensack, NJ, USA; A.S.Arun Srinivaas, Chennai,India; Sushanth Sathish Kumar, Jeffery Trail Middle School, CA, USA; Hyun Min Victoria Woo, NorthfieldMount Hermon School, Mount Hermon, MA, USA; Andreas Charalampopoulos, 4th Lyceum of Glyfada,Glyfada, Greece; Prithwijit De, HBCSE, Mumbai, India; Adnan Ali, A.E.C.S-4, Mumbai, India; AlbertStadler, Herrliberg, Switzerland; David E. Manes, Oneonta, NY, USA; Duy Quan Tran, University of HealthScience, Ho Chi Minh City, Vietnam; Erdenebayar Bayarmagnai, National University of Mongolia, Mongolia;Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Vincelot Ravoson, Lycée Henri IV, France,Paris; Wada Ali, Ben Badis College, Algeria.


J390. Let ABC be a triangle. Points D,D′ lie on side BC, points E,E′ lie on side AC and points F, F ′ lie onside AB such that AD = AD′ = BE = BE′ = CF = CF ′. Prove that if AD,BE,CF are concurrent,then so are AD′, BE′, CF ′.

Proposed by Josef Tkadlec, Vienna, Austria

Solution by Robert Bosch, USA and Jorge Erick, BrazilLet P be the feet of altitude from A, then PD = PD′ and

BD ·BD′ = (BP − PD)(BP + PD),

= BP 2 − PD2,

= BP 2 +AP 2 −AD2,

= AB2 −AD2,

= AB2 −BE2.

By the same argument AB2 −BE2 = AE ·AE′. In a similar way we obtain

CE · CE′ = BF ·BF ′,AF ·AF ′ = CD · CD′.

ThereforeBD

DC· CEEA· AFFB· BD

′

D′C· CE

′

E′A· AF

′

F ′B=BD ·BD′

AE ·AE′· CE · CE

′

BF ·BF ′· AF ·AF

′

CD · CD′= 1.

The conclusion isBD

DC· CEEA· AFFB

= 1⇔ BD′

D′C· CE

′

E′A· AF

′

F ′B= 1.

Thus by Ceva’s theorem the result follows.

Also solved by Daniel Lasaosa, Pamplona, Spain.


Senior problems

S385. Let a, b, c be positive real numbers. Prove that

1

a3 + 8abc+

1

b3 + 8abc+

1

c3 + 8abc≤ 1

3abc.

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution by Evgenidis Nikolaos, M.N.Raptou High School, Larissa, GreeceSince abc > 0 the given inequality is equivalent to

abc

a3 + 8abc+

abc

b3 + 8abc+

abc

c3 + 8abc≤ 1

3.

Then, we can rewrite the inequality as follows:

1a3+8abcabc

+1

b3+8abcabc

+1

c3+8abcabc

≤ 1

3⇔ 1

a2

bc + 8+

1b2

ca + 8+

1c2

ab + 8≤ 1

3.

Now, we set a2

bc = x, b2

ca = y, c2

ab = z with xyz = 1.Therefore, it suffices to prove that

1

x+ 8+

1

y + 8+

1

z + 8≤ 1

3.

Doing the maths we transform the above mentioned inequality to

3[(xy + yz + zx) + 16(x+ y + z) + 192] ≤ xyz + 8(xy + yz + zx) + 64(x+ y + z) + 512

or equivalently because xyz = 1,

5(xy + yz + zx) + 16(x+ y + z) ≥ 63(1).

By AM-GM inequality, we obtain that xy + yz + zx ≥ 3 3√

(xyz2) = 3 and x+ y + z ≥ 3 3√xyz = 3. Hence,

(1) holds.Equality holds if and only if x = y = z = 1⇔ a = b = c.

Also solved by Daniel Lasaosa, Pamplona, Spain; Joehyun Kim, Bergen County Academies, Hackensack,NJ, USA; A.S.Arun Srinivaas, Chennai, India; Sushanth Sathish Kumar, Jeffery Trail Middle School, CA,USA; Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA; AlessandroVentullo, Milan, Italy; Andreas Charalampopoulos, 4th Lyceum of Glyfada, Glyfada, Greece; Arkady Alt,San Jose, CA, USA; Henry Ricardo, New York Math Circle, NY, USA; Prithwijit De, HBCSE, Mumbai,India; Adnan Ali, A.E.C.S-4, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; AN-anduud ProblemSolving Group, Ulaanbaatar, Mongolia; Andrianna Boutsikou, High School Of Nea Makri, Athens, Greece;Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; David E. Manes, Oneonta, NY, USA;Duy Quan Tran, University of Health Science, Ho Chi Minh City, Vietnam; Erdenebayar Bayarmagnai,National University of Mongolia, Mongolia; Jamal Gadirov, Istanbul University, Istanbul, Turkey; RobertBosch, USA.


S386. Evaluatecos π42

+cos 2π

4

22+ · · ·+

cos nπ42n

Proposed by Mohamad Kouroshi, Tehran, Iran

Solution by Daniel Lasaosa, Pamplona, SpainLet ρ = e

iπ4 and let <{z} denote the real part of complex number z. Clearly, cos kπ4 = <

{ρk}, or the sum

that we need to evaluate rewrites as

n∑k=1

<{ρk

2k

}= <

{ρ2 −

ρn

2n

1− ρ2

}= <

{ρ (2− ρ∗)− ρn(2−ρ∗)

2n−1

(2− ρ) (2− ρ∗)

}.

Now,(2− ρ) (2− ρ∗) = 4− 2 (ρ+ ρ∗) + |ρ|2 = 4− 4 cos

π

4+ 1 = 5− 2

√2,

<{ρ (2− ρ∗)} = 2<{ρ} − |ρ|2 =√2− 1,

<{ρn (2− ρ∗)} = 2<{ρn} − |ρ|2<{ρn−1

}= 2 cos

nπ

4− cos

(n− 1)π

4,

or the proposed sum equals √2− 1− 1

2n−2 cosnπ4 + 1

2n−1 cos(n−1)π

4

5− 2√2

.

Also solved by Joehyun Kim, Bergen County Academies, Hackensack, NJ, USA.


S387. Find all nonnegative real numbers k such that∑a(a− b)(a− kb) ≥ 0

for all nonnegative numbers a, b, c.

Proposed by Mehtaab Sawhney, Commack High School, New York, USA

Solution by Adnan Ali, Student in A.E.C.S-4, Mumbai, India

We show that k = k0 is the maximum possible value of k, where k0 =−1+(3+

√2)(√

2√2−1)

2 . Assume that theinequality holds for all nonnegative reals a, b, c for k ≤ k0. Then let c = 0 and W.L.O.G. let a > b. Then the

inequality is a3+b3−a2b ≥ kab(a−b). Let x = a/b > 1, then the inequality is same as k ≤ x+1

x(x− 1). Let

f(x) = x +1

x(x− 1), then by routine calculus, we get fmin = −1+(3+

√2)(√

2√2−1)

2 for x = (√2+1)+

√2√2−1

2 .

Hence k ≤ fmin = −1+(3+√2)(√

2√2−1)

2 = k0.Now to complete the proof we note that the inequality rearranges to

(a3 + b3 + c3) + k(ab2 + bc2 + ca2) ≥ (k + 1)(a2b+ b2c+ c2a)

⇔ (b+ c)(b− c)2 + (c+ a)(c− a)2 + (a+ b)(a− b)2 ≥ (2k + 1)(a− b)(b− c)(a− c) (1)

Denote the left and right hand sides of (1) by L(a, b, c) and R(a, b, c) respectively. W.L.O.G assume thata ≥ b ≥ c and let a1 = a − c ≥ 0, b1 = b − c ≥ 0, c1 = 0, then L(a, b, c) ≥ L(a1, b1, c1) and R(a, b, c) =R(a1, b1, c1). Thus, to prove (1), it suffices to prove that

L(a1, b1, c1) ≥ R(a1, b1, c1),

which is the same as a31 + b31 − a21b1 ≥ ka1b1(a1 − b1). But we have already proved this at the start. Thus,

the inequality holds true for all a, b, c ≥ 0 for all k ≤ k0, where k0 = −1+(3+√2)(√

2√2−1)

2 .

Also solved by Daniel Lasaosa, Pamplona, Spain; Albert Stadler, Herrliberg, Switzerland; Robert Bosch,USA and Jorge Erick, Brazil.


S388. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that

11a− 6

c+

11b− 6

a+

1ac− 6

b≤ 15

abc.

Proposed by Marius Stănean, Zalău, România

Solution by Ángel Plaza, University of Las Palmas de Gran Canaria, SpainThe given inequality may be written equivalently as∑

cyclic

ab(11a− 6) ≤ 15.

by the rearrangement inequality∑cyclic

ab(11a− 6) ≤∑cyclic

a2(11a− 6) =∑cyclic

(11a3 − 6a2) =∑cyclic

(11(a2)3/2 − 6(a2)

).

Since function 11x3/2 − 6x is convex, then

∑cyclic

(11(a2)3/2 − 6(a2)

)≤ 3

(11

(a2 + b2 + c2

3

)3/2

− 6

(a2 + b2 + c2

3

))= 15.

Also solved by Arkady Alt, San Jose, CA, USA.


S389. Let n be a positive integer. Prove that for any integers a1, a2, . . . , a2n+1 there is a rearrangementb1, b2, . . . , b2n+1 such that 2nn! divides

(b1 − b2) (b3 − b4) · · · (b2n−1 − b2n) .

Proposed by Cristinel Mortici, Valahia University, Târgovişte, România

Solution by Evgenidis Nikolaos, M.N.Raptou High School, Larissa, GreeceLet S = {a1, a2, ..., a2n+1} be the set of the statement. Then, it is obvious that S = {b1, b2, ..., b2n+1} becauseb1, b2, ..., b2n+1 is a rearrangement of the elements a1, a2, ..., a2n+1.

Define a sequence of sets S2k+1 = {b1, b2, ..., b2k+1} for 1 ≤ k ≤ n and n ∈ N.Observe that there are 2k different residues mod (2k). Since S2k+1 has 2k + 1 elements, applying the

Pigeonhole Principle, we deduce that there are at least 2 elements of this set , say b2k−1, b2k, that have thesame residue mod(2k).Then, 2k | (b2k−1 − b2k) (∗).

Hence, by (∗) we obtain the following relations

2n | (b2n−1 − b2n) for k = n

2(n− 1) | (b2n−3 − b2n−2) for k = n− 1

2(n− 2) | (b2n−5 − b2n−4) for k = n− 2

· · · · · · · · ·

2 · 2 | (b3−1 − b4) for k = 2

2 | (b1 − b2) for k = 1.

Finally, we can conclude that

2 · (2 · 2) · · · [2(n− 2)] · [2(n− 1)] · (2n) | (b1 − b2)(b3 − b4) · · · (b2n−1 − b2n)

which is obviously equivalent to

2nn! | (b1 − b2)(b3 − b4) · · · (b2n−1 − b2n)

as we desired.

Also solved by Daniel Lasaosa, Pamplona, Spain; Bekjol Joldubai, Kadamcay High School, Kyrgyzstan;Hyun Min Victoria Woo, Northfield Mount Hermon School, Mount Hermon, MA, USA; Andreas Charalam-popoulos, 4th Lyceum of Glyfada, Glyfada, Greece; AN-anduud Problem Solving Group, Ulaanbaatar, Mon-golia; Bekjol Joldubai, Kadamcay High School, Kyrgyzstan; Joel Schlosberg, Bayside, NY, USA; AlessandroVentullo, Milan, Italy; Robert Bosch, USA and Jorge Erick, Brazil.


S390. Let ABC be a triangle and G be its centroid. Lines AG,BG,CG meet the circumcircle of triangleABC at A1, B1, C1, respectively. Prove that√

a2 + b2 + c2 ≤ GA1 +GB1 +GC1 ≤ 2R+1

6

(a2

ma+

b2

mb+

c2

mc

).


Solution by AN-anduud Problem Solving Group, Ulaanbaatar, MongoliaLet AA1 ∩BC = A2, BB1 ∩ CA = B2, CC1 ∩AC = C2. By the Power of the Pointing theorem, we get:

AA2 ·A2A1 = BA2 ·A2C ⇔ ma(GA−1

3ma) =

a2

4

⇔ GA1 =1

3ma +

a2

4ma=

4m2a + a2

12ma+

1

6· a

2

ma.

Similarly, we get

GB1 =1

3mb +

b2

4mb, GC1 =

1

3mc +

c2

4mc.

Hence we get

GA1 +GB1 +GC1 =1

12

(4m2

a + a2

ma+

4m2b + b2

mb+

4m2c + c2

mc

)+

1

6

(a2

ma+

b2

mb+

c2

mc

).

From the other hand,

2R ≥ AA1 = AA2 +A2A1 = ma +a2

4ma=

4m2a + a2

4ma

or4m2

a + a2

ma≤ 8R.

Similarly, we have4m2

b + b2

mb≤ 8R,

4m2c + c2

mc≤ 8R.

Thus we have,

GA1 +GB1 +GC1 ≤1

12(8R+ 8R+ 8R) +

1

6

(a2

ma+

b2

mb+

c2

mc

)= 2R+

1

6

(a2

ma+

b2

mb+

c2

mc

).

Hence right side inequality is proved. Use the median length formula, we get

4m2a = 2b2 + 2c2 − a2, 4m2

b = 2c2 + 2a2 − b2, 4m2c = 2a2 + 2b2 − c2 :

GA1 +GB1 +GC1 =1

12

(2b2 + 2c2

ma+

2c2 + 2a2

mb+

2a2 + 2b2

mc

)+

1

6

(a2

ma+

b2

mb+

c2

mc

)=

1

6(a2 + b2 + c2)

(1

ma+

1

mb+

1

mc

).


Applying AM-GM inequality three times and using following formula, we get

m2a +m2

b +m2c =

3

4(a2 + b2 + c2) :

GA1 +GB1 +GC1 ≥1

6(a2 + b2 + c2) · 3 · 3

√1

ma· 1

mb· 1

mc

≥ 1

2(a2 + b2 + c2) · 1

3√mambmc

≥ 1

2(a2 + b2 + c2) · 3

ma +mb +mc

≥ 1

2(a2 + b2 + c2) · 3√

3(m2a +m2

b +m2c)

=3

2· 23·√a2 + b2 + c2 =

√a2 + b2 + c2.

LHS is proved. Equality holds only when a = b = c.

Also solved by Daniel Lasaosa, Pamplona, Spain; Robert Bosch, USA; Nikos Kalapodis, Patras, Greece;Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Adnan Ali, A.E.C.S-4, Mumbai, India; ArkadyAlt, San Jose, CA, USA; Sushanth Sathish Kumar, Jeffery Trail Middle School, CA, USA.


Undergraduate problems

U385. Evaluate

limn→∞

√n

(√(n+ 1)n

nn−1−

√nn−1

(n− 1)n−2

).

Proposed by Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain

Solution by Daniel Lasaosa, Pamplona, SpainNote first that, since ln(1 + x) = x− x2

2 + x3

3 − . . . , we then have

ln

((n+ 1)n

nn−1

)= ln(n) + n ln

(1 +

1

n

)= ln(n) + 1− 1

2n+

1

3n2− . . . ,

or(n+ 1)n

nn−1= en · exp

(− 1

2n

)· exp

(1

3n2

)· · · = e

(n− 1

2+

11

24n

)+O

(1

n2

),

where Landau notation has been used. Similarly,

nn−1

(n− 1)n−2= e

(n− 3

2+

11

24(n− 1)

)+O

(1

(n− 1)2

)= e

(n− 3

2+

11

24n

)+O

(1

n2

).

Now,

n− 1

2+

11

24n+O

(1

n2

)=

(√n− 1

4√n+O

(1

n√n

))2

,

or √(n+ 1)n

nn−1=√en−

√e

4√n+O

(1

n√n

).

Similarly, √nn−1

(n− 1)n−2=√en− 3

√e

4√n+O

(1

n√n

),

or√n

(√(n+ 1)n

nn−1−

√nn−1

(n− 1)n−2

)=

√e

2+O

(1

n

),

whose limit when n→∞ is clearly√e2 , and we are done.

Also solved by Arkady Alt, San Jose, CA, USA; Adnan Ali, A.E.C.S-4, Mumbai, India; Albert Stadler,Herrliberg, Switzerland; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Erdenebayar Bayar-magnai, National University of Mongolia, Mongolia; Moubinool Omarjee, Lycée Henri IV, Paris, France;Robert Bosch, USA.


U386. Given a convex quadrilateralABCD, denote by SA, SB, SC , SD the area of trianglesBCD,CDA,DAB,ABC,respectively. Determine the point P in the plane of the quadrilateral such that

SA ·−→PA+ SB ·

−−→PB + SC ·

−−→PC + SD ·

−−→PD = 0.

Proposed by Dorin Andrica, Babeş-Bolyai University, Cluj-Napoca, România

Solution by Daniel Lasaosa, Pamplona, SpainNote first that the area of ABCD is S = SA + SC = SB + SD. Note also that, denoting by O the pointwhere the diagonals AC and BD meet, and by hA, hC the respective distances from A,C to BD, we havethat the sine of the angle formed by the diagonals is hA

|OA| =hC|OC| , or hA ·

−−→OC + hC ·

−→OA = 0 becase

−→OA,−−→OC

are vectors in opposite directions on line AC. Note finally that 2SA = BD · hC and 2SC = 2BD · hA, orSC ·−−→OC + SA ·

−→OA = 0, hence for any point P on the plane,

SA ·−→PA+ SC ·

−−→PC = −S ·

−−→OP.

Similarly,SB ·

−−→PB + SD ·

−−→PD = −S ·

−−→OP,

or for any point P in the plane, we have

SA ·−→PA+ SB ·

−−→PB + SC ·

−−→PC + SD ·

−−→PD = −2S

−−→OP,

where S > 0 because ABCD is convex, hence P is the intersection of diagonals AC and BD, and no otherpoint in the plane may satisfy the condition given in the problem statement. The conclusion follows.

Also solved by Adnan Ali, A.E.C.S-4, Mumbai, India.


U387. A polynomial with complex coefficients is called special if all its roots lie on the unit circle. Is anycomplex polynomial the sum of two special polynomials?

Proposed by Gabriel Dospinescu, Ecole Normale Supérieure, Lyon, France

Solution by Robert Bosch, USAThe answer is yes. We understand (on the unit circle) as inside or on the unit disk. The idea is to useRouché’s theorem. Let’s see, let

p(z) = anzn + an−1z

n−1 + · · ·+ a1z + a0,

this polynomial can be written as

p(z) = (azn+1) + (−azn+1 + anzn + an−1z

n−1 + · · ·+ a1z + a0),

= (f(z)) + (g(z)),

where|a| > |an|+ |an−1|+ · · ·+ |a0|.

Clearly the polynomial f(z) has all its roots inside the unit disk, we shall prove g(z) is special too. ByRouché’s theorem it’s enough to verify the following inequality

|p(z)| < | − f(z)|,

on the curve γ : |z| = 1, sinceg(z) = p(z) + (−f(z)).

Finally|anzn + an−1z

n−1 + · · ·+ a1z + a0| < |azn+1|,

is true by Triangle Inequality and by the condition imposed on the constant a.

Also solved by Adnan Ali, Student in A.E.C.S-4, Mumbai, India.


U388. Evaluate

∑∞n=1

sin2n θ + cos2n θ

n2.

Proposed by Li Zhou, Polk State College, USA

Solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia1) Let θ = π

2 + πk, k ∈ Z or θ = πm,m ∈ Z. Then we have

∞∑n=1

sin2n θ + cos2n θ

n2=

∞∑n=1

1

n2=π2

6.

2) Let θ 6= π2 + πk, k ∈ Z and θ 6= πm,m ∈ Z.

Substitute that sin2 θ = t, then we have:

∞∑n=1

sin2n θ + cos2n θ

n2=

∞∑n=1

tn + (1− t)n

n2

= −(∫ t

0

ln(1− z)z

dz +

∫ 1−t

0

ln(1− z)z

dz

)= Li2(t) + Li2(1− t) =

π2

6− ln t ln(1− t).

Where Li2(t) is Dilogarithm function and

Li2(t) + Li2(1− t) =π2

6− ln t ln(1− t)

equality is Landen’s formula. Hence we have:

∞∑n=1

sin2n θ + cos2n θ

n2=π2

6− ln

(sin2 θ

)ln(1− sin2 θ

).

Also solved by Arkady Alt, San Jose, CA, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; AlbertStadler, Herrliberg, Switzerland; Robert Bosch, USA and Jorge Erick, Brazil.


U389. Let P be a nonconstant polynomial whose zeros x1, x2, ..., xn are all real. Prove that

exp

(∫ b

a

P ′′′(x)P (x)

P ′(x)dx

)<

∣∣∣∣P (a)2P ′(b)3P ′(a)3P (b)2

∣∣∣∣ ,whenever a < b < min(x1, x2, ..., xn).


Solution by the authorUse the identity

1

x− x1+

1

x− x2+ ...+

1

x− xn=P ′(x)

P (x).

Take the derivative:

P ′′(x)P (x)− P ′(x)2

P (x)2=∑− 1

(x− xk)2

Take the derivative one more time:

(P ′′′(x)P (x) + P ′′(x)P ′(x)− 2P ′(x)P ′′(x))P (x)2 − (P ′′(x)P (x)− P ′(x)2)2P (x)P ′(x)P (x)4

=∑ 2

(x− xk)3< 0

for a ≤ x ≤ b < min(x1, ..., xn).Then

P ′′′(x)P (x)3 + P ′′(x)P ′(x)P (x)2 − 2P”(x)P ′(x)P (x)2 − 2P ′′(x)P ′(x)P (x)2 + 2P ′(x)3P (x) < 0,

which is equivalent to

P ′′′(x)P (x)3 − 3P ′′(x)P ′(x)P (x)2 + 2P ′(x)3P (x) ≤ 0.

Dividing by P (x)2P ′(x)2 yields

P ′′′(x)P (x)

P ′(x)2≤ 3P ′′(x)

P ′(x)− 2P ′(x)

P (x)

.By integration,∫ b

a

P ′′′(x)P (x)

P ′(x)2≤ 3

∣∣lnP ′(b)− lnP ′(a)∣∣− 2 |lnP (b)− lnP (a)| = lnP (a)2P ′(b)3

P ′(a)3P (b)2,

hence

exp

(∫ b

a

P ′′′(x)P (x)

P ′(x)2

)≤∣∣∣∣P (a)2P ′(b)3P ′(a)3P (b)2

∣∣∣∣ ,as desired.


U390. Prove that there is a unique representation of sin(πz) as a series of the form

sin(πz) =

∞∑k=1

akzk(1− z)k

that converges for all complex numbers z, wherein the coefficients ak are real number satisfying |ak| ≤c · π2k

(2k)! for some absolute constant c.

Proposed by Albert Stadler, Herrliberg, Switzerland

Solution by Daniel Lasaosa, Pamplona, SpainNote first that if such an expression exists, it must be unique. Indeed, let j be the lowest value of k such

that ak is different in both expression, or the difference between both expressions would be a polynomialwith nonzero coefficient for zj , thus nonzero.

Define now v = z(1− z), or

sin(πz) = sin

(π

(z − 1

2

)+π

2

)= cos

(π

(z − 1

2

))=

∞∑n=0

(−1)nπ2n(z − 1

2

)2n(2n)!

=

=∞∑n=0

π2n(v − 1

4

)n(2n)!

,

where we have used that(z − 1

2

)2= z2−z+ 1

4 = 14−v. Note therefore that the coefficient ak of vk = zk(1−z)k

is

ak =

∞∑n=k

π2n

(2n)!

(n

k

)(−1

4

)n−k=

π2k

(2k)!

∞∑d=0

(−1)dπ2d(2k)!4d(2k + 2d)!

(k + d

k

)=

=π2k

(2k)!

∞∑d=0

(−2)d(2k − 1)!!

(2k + 2d− 1)!!

1

d!

(π4

)2d,

where we have defined d = n− k, and (2m− 1)!! = (2m− 1)(2m− 3) · · · 3 · 1. Note now first that, since theseries is unique and sin(πz) = 0, we must have a0 = 0, or we will henceforth obviate k = 0. Note next thatfor all k ≥ 1, we have 2 < 2k + 1, 2k + 3, . . . , 2k + 2d− 1, or 2d(2k − 1)!! < (2k + 2d− 1)!!, hence

|ak| ≤π2k

(2k)!

∞∑d=0

2d(2k − 1)!!

(2k + 2d− 1)!!

1

d!

(π4

)2d<

π2k

(2k)!

∞∑d=0

1

d!

(π4

)2d= exp

(π2

16

).

The conclusion follows, and it suffices to take c = exp(π2

16

).


Olympiad problems

O385. Let f(x, y) = x3−y36 +3xy+48. Let m and n be odd integers such that |f(m,n)| ≤ mn+37. Evaluate

f(m,n).


Solution by Robert Bosch, USA and Jorge Erick, BrazilIf α =

m− n2

, then α is integer and m = n+ 2α. Consider the following identity:

m3 − n3

6+ kmn = (α+ k)

[(n+ α)2 +

1

3(α− 2k)2

]− 4

3k3,

due toa3 + b3 + c3 − 3abc = (a+ b+ c)(a2 + b2 + c2 − ab− bc− ca),

for a = m, b = −n, c = 2k. Thus

f(m,n)− (mn+ 37) = (α+ 2)

[(n+ α)2 +

1

3(α− 4)2

]+

1

3≤ 0⇒ α+ 2 < 0,

f(m,n) +mn+ 37 = (α+ 4)

[(n+ α)2 +

1

3(α− 8)2

]− 1

3≥ 0⇒ α+ 4 > 0.

We conclude that α = −3 and

f(m,n) = (α+ 3)

[(n+ α)2 +

1

3(α− 6)2

]+ 12 = f(−5, 1) = 12.

Also solved by Alessandro Ventullo, Milan, Italy; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; AlbertStadler, Herrliberg, Switzerland; Prithwijit De, HBCSE, Mumbai, India.


O386. Find all pairs (m,n) of positive integers such that 3m − 2n is a perfect square.

Proposed by Alessandro Ventullo, Milan, Italy

Solution by David E. Manes, Oneonta, NY, USABy direct computation if m ≤ 4, then (1, 1), (2, 3), (3, 1) and (4, 5) are solutions. We will show that thereare no others.

Assume m ≥ 5. If n = 1, then a simple induction argument shows that 3m − 2n is always strictly betweentwo consecutive squares. Therefore, 3m − 2 is not a perfect square. Thus, n ≥ 2. Assume 3m − 2n = k2

for some integer k and the integer m is odd. Then 3m ≡ 3 mod 4 and 2n ≡ 0 mod 4 imply 3m − 2n ≡ 3mod 4, a contradiction since k2 ≡ 0 or 1 mod 4. Therefore, m is even so that m = 2r for some integerr. Then 32r − k2 = 2n. Factoring, we get (3r + k)(3r − k) = 2n so that 3r + k = 2s and 3r − k = 2t forsome integers s, t, s > t and s+ t+ n. Adding the two equations, one obtains 2 · 3r = 2t(2s−t + 1. Therefore3r = 2t−1(2s−t + 1) implies t = 1. Therefore, 3r − 2s−1 = 1, an equation that has solutions only if r = 1 or2 in which case m ≤ 4, a contradiction. Hence, the only solutions are the ones claimed above.

Also solved by Adnan Ali, A.E.C.S-4, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; PrithwijitDe, HBCSE, Mumbai, India; Robert Bosch, USA and Jorge Erick, Brazil; Minh Pham Hoang, High Schoolfor the Gifted, Vietnam National University, Ho Chi Minh City, Vietnam.


O387. Are there integers n for which 36n−3 + 33n−1 + 1 is a perfect cube?


Solution by Daniel Lasaosa, Pamplona, SpainNote that (

32n−1 + 1)3

= 36n−3 + 34n−1 + 32n + 1 > 36n−3 + 33n−1 + 1 > 36n−3 =(32n−1

)3,

or since 36n−3 +33n−1 +1 is always strictly between two consecutive perfect cubes, it can never be a perfectcube itself, and we are done.

Also solved by Sushanth Sathish Kumar, Jeffery Trail Middle School, CA, USA; Minh Pham Hoang,High School for the Gifted, Vietnam National University, Ho Chi Minh City, Vietnam; Alessandro Ventullo,Milan, Italy; Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Adnan Ali, A.E.C.S-4, Mumbai,India; Albert Stadler, Herrliberg, Switzerland; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia;Arpon Basu, AECS-4, Mumbai, India; Bekjol Joldubai, Kadamcay High School, Kyrgyzstan; David E. Manes,Oneonta, NY, USA; Jamal Gadirov, Istanbul University, Istanbul, Turkey; Ricardo Largaespada, UniversidadNacional de Ingeniería, Managua, Nicaragua; Robert Bosch, USA.


O388. Prove that in any triangle ABC with area S,

mambmc (ma +mb +mc)√ma

2mb2 +mb

2mc2 +mc

2ma2≥ 2S.


Solution by Daniel Lasaosa, Pamplona, SpainWe prove a stronger result, namely that the inequality also holds with 3S in the RHS instead of 2S. Notefirst that, since ma

2 = 2b2+2c2−a24 and similarly for mb,mc, we have

ma2mb

2 +mb2mc

2 +mc2ma

2 =9(a2b2 + b2c2 + c2a2

)16

,

ma4 +mb

4 +mc4 =

9(a4 + b4 + c4

)16

,

or using Heron’s formula,

9S2 =18(a2b2 + b2c2 + c2a2

)− 9

(a4 + b4 + c4

)16

=

= 2(ma

2mb2 +mb

2mc2 +mc

2ma2)−(ma

4 +mb4 +mc

4).

Squaring both sides and multiplying by ma2mb

2 +mb2mc

2 +mc2ma

2, the proposed inequality is equivalentto

ma2mb

2mc2 (ma +mb +mc)

2 +(ma

4 +mb4 +mc

4) (ma

2mb2 +mb

2mc2 +mc

2ma2)≥

≥ 2(ma

2mb2 +mb

2mc2 +mc

2ma2)2,

or after some algebra, to

ma2mb

2((ma +mb)

2 −mc2)(ma −mb)

2 +mb2mc

2((mb +mc)

2 −ma2)(mb −mc)

2+

+mc2ma

2((mc +ma)

2 −mb2)(mc −ma)

2 ≥ 0.

Now, it is well known that

(ma +mb)2 −mc

2 = (ma +mb +mc) (ma +mb −mc) ≥ 0,

since the medians of triangle ABC are the sides of a triangle whose area is 34 the area of ABC. Or, all terms

in the LHS are non-negative, being simultaneoulsy zero iff ma = mb = mc. The conclusion follows, equalityholds iff ABC is equilateral.

Also solved by Evgenidis Nikolaos, M.N.Raptou High School, Larissa, Greece; Arkady Alt, San Jose, CA,USA; Adnan Ali, A.E.C.S-4, Mumbai, India; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia;Pham Ngoc Khanh, Hanoi National University of Education, Vietnam; Robert Bosch, USA and Jorge Erick,Brazil.


O389. Let a, b, c be positive real numbers such that abc = 1. Prove that

a2(b+ c)

b2 + c2+b2(c+ a)

c2 + a2+c2(a+ b)

a+b2≥√3(a+ b+ c).

Proposed by Bazarbaev Sardar, National University of Uzbekistan, Uzbekistan

Solution by Erdenebayar Bayarmagnai, National University of Mongolia, Mongoliaa2(b+ c)

b2 + c2−a+b

2(c+ a)

c2 + a2−b+c

2(a+ b)

a2 + b2−c = ab(a− b) + ac(a− c)

b2 + c2+bc(b− c) + ba(b− a)

c2 + a2+ca(c− a) + cb(c− b)

a2 + b2=

=

(ab(a− b)b2 + c2

− ab(a− b)c2 + a2

)+

(ac(a− c)b2 + c2

− ac(a− c)a2 + b2

)+

(bc(b− c)c2 + a2

− bc(b− c)a2 + b2

)=

=ab(a− b)(a2 − b2)(b2 + c2)(c2 + a2)

+ac(a− c)(a2 − c2)(b2 + c2)(a2 + b2)

+bc(b− c)(b2 − c2)(c2 + a2)(a2 + b2)

=

=ab(a− b)2(a+ b)

(b2 + c2)(c2 + a2)+ac(a− c)2(a+ c)

(b2 + c2)(a2 + b2)+

bc(b− c)2(b+ c)

(c2 + a2)(a2 + b2)≥ 0 ⇒

⇒ a2(b+ c)

b2 + c2+b2(c+ a)

c2 + a2+c2(a+ b)

a2 + b2≥ (a+b+c) =

√(a+ b+ c)(a+ b+ c) ≥

√3

3√abc(a+ b+ c) =

√3(a+ b+ c)

This equality holds only when a = b = c = 1

Also solved by Albert Stadler, Herrliberg, Switzerland; Emil Gasimov, Baku Istek Lyceum, Azerbaijan;Nguyen Viet Hung, Hanoi University of Science, Vietnam; Pham Ngoc Khanh, Hanoi National Universityof Education, Vietnam; Robert Bosch, USA.


O390. Let p > 2 be a prime. Find the number of 4p element subsets of the set {1, 2, · · · , 6p} for which thesum of the elements is divisble by 2p.

Proposed by Vlad Matei, University of Wisconsin, Madison, USA

Solution by Adnan Ali, Student in A.E.C.S-4, Mumbai, IndiaWe shall be using the Root of Unity Filter to solve the problem, so we introduce it first:

Theorem 1 (Root of Unity Filter): Define ε = e2πi/n for a positive integer n. For any polynomialF (x) = f0 + f1x+ f2x

2 + · · · (where fk = 0 if k > deg F ), the sum f0 + fn + f2n + · · · is given by

f0 + fn + f2n + · · · =1

n

(F (1) + (ε) + F (ε2) + · · ·+ F (εn−1)

).

Proof: We use a property of the sum sk = 1+ εk+ ε2k+ · · ·+ ε(n−1)k. If n|k, then εk = 1 and so sk = n,else εk 6= 1 and so sk = 1−εnk

1−εk = 0. Thus

F (1) + (ε) + F (ε2) + · · ·+ F (εn−1) = f0s0 + f1s1 + f2s2 + · · · = n(f0 + fn + f2n + · · · ).

�

Now, to start with, we define a generating function G(x, y) as

G(x, y) =∑

n,k≥0 gn,kxnyk

where gn,k is the number of k-element subsets of {1, 2, · · · , 6p} having a sum n. So, the answer required bythe problem is nothing but A := g2p,4p + g4p,4p + g6p,4p + · · · . Next we observe that if a number m is not ina subset, it doesn’t affect the size or sum of the subset, while if it is present in the subset, it increases thesum by m and size by 1. Thus G must contain the term (1 + xmy) and so

G(x, y) = (1 + xy)(1 + x2y) · · · (1 + x6py).

To get the value of A we must extract two types of terms from G: y4p and powers of x2p. We can do thelatter using Theorem 1 and so we perform it first. Define ε = e2πi/2p. Then the filter tells us that

∑n,k≥02p|n

gn,kyk =

1

2p

(G(1, y) +G(εk, y) +G(ε2k, y) + · · ·+G(ε(2p−1)k, y)

), (2)

so we need to calculate G(εk, y) for 0 ≤ k ≤ 2p − 1. For k = 0, we have G(1, y) = (1 + y)6p. For k = 2`,where 1 ≤ ` ≤ p − 1, ε2` = γ`, where γ = e2πi/p. Since gcd(`, p) = 1 the set {`, 2`, · · · , p`} is a completeresidue set modulo p. Thus we have

G(ε2`, y) = G(γ`, y) = (1 + γ`y)(1 + γ2`y) · · · (1 + γ6p`y)

=((1 + γ`y)(1 + γ2`y) · · · (1 + γp`y)

)6=((1 + γy)(1 + γ2y) · · · (1 + γpy)

)6= (1 + yp)6.


For k = p, ε = −1, and so we have G(−1, y) = (1 − y2)3p. Now for 1 ≤ k ≤ 2p − 1 (2 - k and k 6= p)gcd(2p, k) = 1 and so the set {k, 2k, · · · , 2pk} is a complete residue set modulo 2p. Thus

G(εk, y) = (1 + εky)(1 + ε2ky) · · · (1 + ε6pky)

=((1 + εky)(1 + ε2ky) · · · (1 + ε2pky)

)3=((1 + εy)(1 + ε2y) · · · (1 + ε2py)

)3= (1− y2p)3.

Putting all these values back in (1) gives

∑n,k≥02p|n

gn,kyk =

1

2p

((1 + y)6p + (1− y2)3p + (p− 1)(1 + yp)6 + (p− 1)(1− y2p)3

),

from which we can easily extract the coefficient of y4p. Thus our answer (i.e. A) is

1

2p

((6p4p

)+(3p2p

)+ (p− 1)

(64

)+ (p− 1)

(32

))=

1

2p

((6p4p

)+(3p2p

)+ 18p− 18

).


junior problems - awesomemath · 2017. 1. 21. · junior problems j385. iftheequalities 2(a+b) 6c...

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