Junior problems - AwesomeMath€¦ · J483. Leta;b;cberealnumberssuchthat13a+41b+13c=2019 and max‰V 41 13 a−bV;V 13 41 b−cV;Sc−aS’≤1: Provethat2019 ≤a2 +b2 +c2 ≤2020

Junior problems

J481. Find all triples (p, q, r) of primes such that

p2 + 2q2 + r2 = 3pqr.

Proposed by Adrian Andreescu, University of Texas at Austin, USA

Solution by Polyahedra, Polk State College, USAIf neither p nor r is divisible by 3, then p2+2q2+r2 ≡ 1+2q2+1 ≡ 2,1 (mod 3), an impossibility. Suppose thatr = 3. Then p2 + 2q2 = 9(pq − 1). If q is odd, then p2 + 2q2 and 9(pq − 1) have opposite parity. So q = 2. Then0 = p2 − 18p+ 17 = (p− 1)(p− 17), thus p = 17. By symmetry in p and r, the solutions are (p, q, r) = (17,2,3)and (3,2,17).

Also solved by Daniel Lasaosa, Pamplona, Spain; Albert Stadler, Herrliberg, Switzerland; Corneliu Mănescu-Avram, Ploies,ti, Romania; Daniel Văcaru, Pites,ti, Romania; Dumitru Barac, Sibiu, Romania; Joe Simons,Utah Valley University, UT, USA; Joel Schlosberg, Bayside, NY, USA; Ioannis D. Sfikas, Athens, Greece;Joonsoo Lee, Dwight Englewood School, NJ, USA; Nguyen Viet Hung, Hanoi University of Science, Viet-nam; Nick Iliopoulos, 8th General Highschool of Trikala, Greece; Sushanth Sathish Kumar, PHS, SoCal,USA; Titu Zvonaru, Comănes,ti, Romania; Jeffrey Roh, St. Andrew’s School, DE, USA; Andrew Yang,Hotchkiss School, Lakeville, CT, USA; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; SebastianFoulger, Charters School, Sunningdale, England, UK; Jeewoo Lee, Townsend Harris HS, NY, USA; JennaPark, Blair Academy, Blairstown, NJ, USA; Min Jung Kim, Tabor Academy, MA, USA; Kelvin Kim, BergenCatholic High School, NJ, USA; Chirita Andrei-Giovani, The Greek Catholic Timotei Cipariu High School,Bucharest, Romania; Bryant Hwang, Korea International School; Jiho Lee, Canterbury School, New Milford,CT, USA.

Mathematical Reflections 3 (2019) 1

J482. Find all positive integers less than 10,000 which are palindromic both in base 10 and base 11.

Proposed by Mircea Becheanu, Montreal, Canada

Solution by the authorWhen we mention the number of digits of a number, we consider this in base 10. The digits in base 11 are1,2, . . . ,9, ζ, where ζ = 10.. The one digit numbers 1,2, . . . ,9 are obviously palindromic in both bases. Apalindromic four digits numbers is divisible by 11, so it can not be palindromic in base 11. Hence we haveto consider only three digits numbers.

Let N , 100 < N < 999, be a number which is palindromic in base 11 and let

N = 112a + 11b + a

be its representation in base 11. It is clear that a ≤ 8, as N is a three digits number. We try to obtain itsrepresentation in base 10.

N = (10 + 1)2a + (10 + 1)b + a = 102a + 10(2a + b) + (2a + b).

If 2a + b < 10, the number N can not be palindromic in base 10, because we can not have a = 2a + b. Also,2a + b can not be divisible by 10.

We consider first the case 10 < 2a + b < 20. Put 2a + b = 10 + c where 1 ≤ c ≤ 9. Then

N = 102(a + 1) + 10(c + 1) + c (1)

If 1 ≤ c ≤ 8, the formula (1) gives a representation of N in base 10. In order to have N palindromic weput the condition c = a + 1, which gives a + b = 11 and a ≤ 7 So, we obtain the possibilities (a = 1, b = ζ);(a = 2, b = 9), ...,(a = 7, b = 4). They give numbers 1ζ1, 292, 383, 474, 565, 656, 747 in base 11, which havein base 10 representation 232,343,454,565,676,787,898, respectively.

If c = 9, by plugging this in (1) we obtain

N = 102(a + 2) + 9. (2)

Hence, a = 7, b = 5 and we have in base 11 the number 757, which is 909 in base 10.Let consider now the case 2a + b > 20. Let denote 2a + b = 20 + c. Then

N = 102(a + 2) + 10(c + 2) + c.

If c < 8 we obtain a + 2 = cÔ⇒ a + b = 22 and this is impossible, as a, b are digits. If c = 8 or c = 9 we obtainc = a + 3 and this is again impossible.So, the required integers are (in base 10):

1; 2; 3; 4; 5; 6; 7; 8; 9; 232; 343; 454; 565; 676; 787; 898 and 909.

Also solved by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Albert Stadler, Herrliberg, Swi-tzerland; Daniel Lasaosa, Pamplona, Spain; Jeffrey Roh, St. Andrew’s School, DE, USA; Andrew Yang,Hotchkiss School, Lakeville, CT, USA; Polyahedra, Polk State College, USA; Sebastian Foulger, ChartersSchool, Sunningdale, England, UK; Ioannis D. Sfikas, Athens, Greece; Titu Zvonaru, Comănes,ti, Romania.


J483. Let a, b, c be real numbers such that 13a + 41b + 13c = 2019 and

max(∣41

13a − b∣ , ∣13

41b − c∣ , ∣c − a∣) ≤ 1.

Prove that 2019 ≤ a2 + b2 + c2 ≤ 2020.

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

First solution by the authorFrom Cauchy-Schwarz’s Inequality,

20192 = (13a + 41b + 13c)2 ≤ (132 + 412 + 132)(a2 + b2 + c2),

implying 2019 ≤ a2 + b2 + c2.We have

∣41a − 13b∣ ≤ 13, ∣13b − 41c∣ ≤ 41, ∣13c − 13a∣ ≤ 13,

so

(41a − 13b)2 + (13b − 41c)2 + (13c − 13a)2 ≤ 132 + 412 + 132.

Also,

(13a + 41b + 13c)2 = 20192.

Adding the last two relations yields

(132 + 412 + 132)(a2 + b2 + c2) ≤ (132 + 412 + 132) + 20192.

Because 132 + 412 + 132 = 2019, it follows that

a2 + b2 + c2 ≤ 1 + 2019 = 2020,

as desired.


Second solution by Daniel Lasaosa, Pamplona, SpainDenote a = 13 + 41u and c = 13 + 41v. Note that

b = 2019 − 13(a + c)41

= 41 − 13(u + v).

Therefore,a2 + b2 + c2 = 2019 + 1681 (u2 + v2) + 169(u + v)2 ≥ 2019,

with equality if u = v = 0, ie iff a = c = 13 and b = 41. Moreover,

∣1850u + 169v∣ ≤ 13, ∣169u + 1850v∣ ≤ 41, ∣41u − 41v∣ ≤ 1,

or since 18502 + 1692 + 169 ⋅ 412 = 2019 ⋅ 1850, then

1 = 132 + 412 + 132

2019≥ ∣1850u + 169v∣2 + ∣169u + 1850v∣2 + 169∣41u − 41v∣2

2019=

= 1850 (u2 + v2) + 338uv = 1681 (u2 + v2) + 169(u + v)2.

Now, equality simultaneously requires

1850u + 169v = ±13, 169u + 1850v = ±41, 41u − 41v = ±1.

But from the first two relations we obtain 41u − 41v = ±1 ± 1341 , clearly not equal to ±1, or equality cannot

hold. We conclude that2019 ≤ a2 + b2 + c2 < 2020,

with equality in the lower bound iff a = c = 13 and b = 41.

Also solved by Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Jeffrey Roh, St. Andrew’sSchool, DE, USA; Andrew Yang, Hotchkiss School, Lakeville, CT, USA; Polyahedra, Polk State College, USA;Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Jeewoo Lee, Townsend Harris HS, NY, USA; JennaPark, Blair Academy, Blairstown, NJ, USA; Min Jung Kim, Tabor Academy, MA, USA; Kelvin Kim, BergenCatholic High School, NJ, USA; Chirita Andrei-Giovani, The Greek Catholic Timotei Cipariu High School,Bucharest, Romania; Dumitru Barac, Sibiu, Romania; Joonsoo Lee, Dwight Englewood School, NJ, USA;Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti, Romania; Jiho Lee, Canterbury School,New Milford, CT, USA.


J484. Let a and b positive real numbers such that a2 + b2 = 1. Find the minimum value of

a + b1 + ab.

Proposed by Marius Stănean, Zalău, Romania

Solution by Nguyen Viet Hung, Hanoi University of Science, VietnamFirst, we will show that

(a + b)2(a2 + b2)(a2 + ab + b2)2 ≥ 8

9.

Indeed, this is equivalent to9(a2 + b2 + 2ab)(a2 + b2) ≥ 8(a2 + b2 + ab)2,

9[(a2 + b2)2 + 2ab(a2 + b2)] ≥ 8[(a2 + b2)2 + 2ab(a2 + b2) + a2b2],

(a2 + b2)2 + 2ab(a2 + b2) ≥ 8a2b2,

(a2 − b2)2 + 2ab(a − b)2 ≥ 0

which is obvious. Hence our statement has been poven. From this statement and a2 + b2 = 1 we get

a + b1 + ab ≥

2√

2

3.

The equality happens if and only if a = b = 1√2. So the minimum value of the given expression is 2

√2

3 .

Also solved by Daniel Lasaosa, Pamplona, Spain; Jeffrey Roh, St. Andrew’s School, DE, USA; AndrewYang, Hotchkiss School, Lakeville, CT, USA; Polyahedra, Polk State College, USA; Takuji Grigorovich Imaii-da, Fujisawa, Kanagawa, Japan; Sebastian Foulger, Charters School, Sunningdale, England, UK; Jeewoo Lee,Townsend Harris HS, NY, USA; Jenna Park, Blair Academy, Blairstown, NJ, USA; Min Jung Kim, TaborAcademy, MA, USA; Chirita Andrei-Giovani, The Greek Catholic Timotei Cipariu High School, Bucharest,Romania; Bryant Hwang, Korea International School; Duy Quan Tran, University of Medicine and Pharma-cy, Ho Chi Minh, Vietnam; Albert Stadler, Herrliberg, Switzerland; Arkady Alt, San Jose, CA, USA; ArleoRobles, Charters Sixth Form, Sunningdale, England, UK; Corneliu Mănescu-Avram, Ploies,ti, Romania; Du-mitru Barac, Sibiu, Romania; Henry Ricardo, Westchester Area Math Circle, NY, USA; Joel Schlosberg,Bayside, NY, USA; Ioannis D. Sfikas, Athens, Greece; Joonsoo Lee, Dwight Englewood School, NJ, USA;Marin Chirciu, Colegiul Nat,ional Zinca Golescu, Pites,ti, Romania; Moubinool Omarjee, Lycée Henri IV,Paris, France; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Sumanth Ravipati, GeorgeMason University, VA, USA; Sushanth Sathish Kumar, PHS, SoCal, USA; Titu Zvonaru, Comănes,ti, Ro-mania; Brian Bradie, Christopher Newport University, Newport News, VA, USA; Nikos Kalapodis, Patras,Greece; Jiho Lee, Canterbury School, New Milford, CT, USA.


J485. Find the maximum and minimum of

1

sin4 x + cos2 x+ 1

sin2 x + cos4 x

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

Solution by Brian Bradie, Christopher Newport University, Newport News, VA, USANote

sin4 x + cos2 x = sin4 x − sin2 x + 1 = sin2 x(sin2 x − 1) + 1

= 1 − sin2 x cos2 x = 1 − 1

4sin2 2x

and

sin2 x + cos4 x = 1 + cos4 x − cos2 x = 1 + cos2 x(cos2 x − 1)

= 1 − sin2 x cos2 x = 1 − 1

4sin2 2x,

so1

sin4 x + cos2 x+ 1

sin2 x + cos4 x= 2

1 − 14 sin2 2x

.

Now,3

4≤ 1 − 1

4sin2 2x ≤ 1

implies

2 ≤ 2

1 − 14 sin2 2x

≤ 8

3.

Thus, the maximum value of1

sin4 x + cos2 x+ 1

sin2 x + cos4 x

is 8/3, which occurs when x = (2n + 1)π/4 for any integer n; the minimum value of

1

sin4 x + cos2 x+ 1

sin2 x + cos4 x

is 2, which occurs when x = nπ/2 for any integer n.

Also solved by Daniel Lasaosa, Pamplona, Spain; Jeffrey Roh, St. Andrew’s School, DE, USA; AndrewYang, Hotchkiss School, Lakeville, CT, USA; Polyahedra, Polk State College, USA; Takuji Grigorovich Imaii-da, Fujisawa, Kanagawa, Japan; Sebastian Foulger, Charters School, Sunningdale, England, UK; Jeewoo Lee,Townsend Harris HS, NY, USA; Jenna Park, Blair Academy, Blairstown, NJ, USA; Min Jung Kim, TaborAcademy, MA, USA; Chirita Andrei-Giovani, The Greek Catholic Timotei Cipariu High School, Bucharest,Romania; Nikos Kalapodis, Patras, Greece; Duy Quan Tran, University of Medicine and Pharmacy, Ho ChiMinh, Vietnam; Lukas Seier, Charters Sixth Form, Sunningdale, England, UK; Albert Stadler, Herrliberg,Switzerland; Arkady Alt, San Jose, CA, USA; Arleo Robles, Charters Sixth Form, Sunningdale, England,UK; Corneliu Mănescu-Avram, Ploies,ti, Romania; Daniel Văcaru, Pites,ti, Romania; Dumitru Barac, Sibiu,Romania; Henry Ricardo, Westchester Area Math Circle, NY, USA; Joel Schlosberg, Bayside, NY, USA;Ioannis D. Sfikas, Athens, Greece; Joonsoo Lee, Dwight Englewood School, NJ, USA; Marin Chirciu, Co-legiul Nat,ional Zinca Golescu, Pites,ti, Romania; Nicuşor Zlota, Traian Vuia Technical College, Focşani,Romania; Oscar Brown, Charters School, Sunningdale, UK; Sumanth Ravipati, George Mason University,VA, USA; Titu Zvonaru, Comănes,ti, Romania; Jiho Lee, Canterbury School, New Milford, CT, USA.


J486. Let a, b, c be positive numbers. Prove that

bc

(2a + b)(2a + c) +ca

(2b + c)(2b + a) +ab

(2c + a)(2c + b) ≥ 1

3.

Proposed by An Zhenping, Xianyang Normal University, China

Solution by Henry Ricardo, Westchester Area Math Circle, NY, USAThe Cauchy-Schwarz inequality gives us

∑cyclic

bc

(2a + b)(2a + c) = ∑cyclic

(bc)2bc (2a + b)(2a + c)

≥(∑cyclic bc)

2

∑cyclic bc (2a + b)(2a + c)

= ∑cyclicb2c2 + 2abc∑cyclic a

∑cyclic b2c2 + 8abc∑cyclic a

= X + 2Y

X + 8Y≥ 1

3⇐⇒ X ≥ Y.

But X = ∑cyclic(bc)2 ≥ (bc)(ca) + (ca)(ab) + (ab)(bc) = abc(a + b + c) = Y is a consequence of the AGMinequality. Equality holds if and only if a = b = c.

Also solved by Daniel Lasaosa, Pamplona, Spain; Polyahedra, Polk State College, USA; Takuji Grigo-rovich Imaiida, Fujisawa, Kanagawa, Japan; Chirita Andrei-Giovani, The Greek Catholic Timotei CipariuHigh School, Bucharest, Romania; Brian Bradie, Christopher Newport University, Newport News, VA , USA;Nikos Kalapodis, Patras, Greece; Albert Stadler, Herrliberg, Switzerland; Arkady Alt, San Jose, CA, USA;Corneliu Mănescu-Avram, Ploies,ti, Romania; Ioan Viorel Codreanu, Satulung, Maramureş, Romania; Da-niel Văcaru, Pites,ti, Romania; Dumitru Barac, Sibiu, Romania; Ioannis D. Sfikas, Athens, Greece; MarinChirciu, Colegiul Nat,ional Zinca Golescu, Pites,ti, Romania; Nguyen Viet Hung, Hanoi University of Science,Vietnam; Nick Iliopoulos, 8th General Highschool of Trikala, Greece; Nicuşor Zlota, Traian Vuia TechnicalCollege, Focşani, Romania; Titu Zvonaru, Comănes,ti, Romania.


Senior problems

S481. Let n be a positive integer. Evaluaten

∑k=1

(n + k)4n3 + k3 .


Solution by Li Zhou, Polk State College, USABy synthetic division,

(n + k)4n3 + k3 = k

3 + 3nk2 + 3n2k + n3k2 − nk + n2 = k + 4n + 3n2rk,

where rk = 2k−nk2−nk+n2 . Notice that

rn−i =2(n − i) − n

(n − i)2 − n(n − i) + n2 = n − 2i

i2 − ni + n2 = −ri.

Thusn

∑k=1

rkk=n−i===

n−1

∑i=0

rn−i = rn − r0 +n

∑i=1

rn−i =2

n−

n

∑i=1

ri,

so ∑nk=1 rk = 1n . Therefore,

n

∑k=1

(n + k)4n3 + k3 =

n

∑k=1

(k + 4n + 3n2rk) =n(n + 1)

2+ 4n2 + 3n = n(9n + 7)

2.

Also solved by Daniel Lasaosa, Pamplona, Spain; Jeffrey Roh, St. Andrew’s School, DE, USA; An-drew Yang, Hotchkiss School, Lakeville, CT, USA; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan;Sebastian Foulger, Charters School, Sunningdale, England, UK; Jeewoo Lee, Townsend Harris HS, NY,USA; Jenna Park, Blair Academy, Blairstown, NJ, USA; Kelvin Kim, Bergen Catholic High School, NJ,USA; Chirita Andrei-Giovani, The Greek Catholic Timotei Cipariu High School, Bucharest, Romania; BrianBradie, Christopher Newport University, Newport News, VA, USA; Lukas Seier, Charters Sixth Form, Sun-ningdale, England, UK; Albert Stadler, Herrliberg, Switzerland; Arkady Alt, San Jose, CA, USA; CorneliuMănescu-Avram, Ploies,ti, Romania; Dimoulios Ioannis, Thessaloniki, Greece; Dumitru Barac, Sibiu, Roma-nia; Joonsoo Lee, Dwight Englewood School, NJ, USA; Moubinool Omarjee, Lycée Henri IV, Paris, France;Sushanth Sathish Kumar, PHS, SoCal, USA; Jiho Lee, Canterbury School, New Milford, CT, USA.


S482. Prove that in any regular 31-gon A0A1 . . .A30 the following equality holds:

1

A0A1= 1

A0A2+ 1

A0A4+ 1

A0A8+ 1

A0A15.

Proposed by Dragoljub Milošević, Gornji Milanovac, Serbia

Solution by Andrea Fanchini, Cantù, Italy

We denote with α = 2π31 and with R the circumradius from the center of the regular 31-gon to one of the

vertices.We apply the cosine theorem to the triangle OA0A1

A0A21 = 2R2(1 − cosα) = 4R2 sin2 α

2⇒ A0A1 = 2R sin

α

2

in the same way from the triangles OA0A2, OA0A4, OA0A8 we obtain

A0A2 = 2R sinα, A0A4 = 2R sin 2α, A0A8 = 2R sin 4α

and from the triangle OA0A15 we have

A0A215 = 2R2 (1 − cos(15 ⋅ 2π

31)) = 2R2 (1 − cos(16 ⋅ 2π

31)) = 4R2 sin2 8α ⇒ 2R sin 8α

therefore we have to prove that

cscα

2= cscα + csc 2α + csc 4α + csc 8α (1)


Now we havecscα + cotα = 1

sinα+ cosα

sinα= 1 + cosα

sinα

but cos2 α2 = 1+cosα2 then

cscα + cotα =2 cos2 α2

2 sin α2 cos α2

= cotα

2⇒ cscα = cot

α

2− cotα

and in the same way we have that

cscα

2= cot

α

4− cot

α

2, csc 2α = cotα − cot 2α, csc 4α = cot 2α − cot 4α, csc 8α = cot 4α − cot 8α

so the (1) is also equal to

2 cotα

2= cot

α

4+ cot 8α ⇒ 2 cot

π

31= cot

π31

2+ cot

16π

31=

and finally the RHS is

= cotπ31

2− tan

π31

2=

1 + cos π31

sin π31

−1 − cos π

31

sin π31

= 2 cotπ

31

Q.e.d.

Also solved by Daniel Lasaosa, Pamplona, Spain; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa,Japan; Li Zhou, Polk State College, USA; Albert Stadler, Herrliberg, Switzerland; Dimoulios Ioannis, Thes-saloniki, Greece; Ioannis D. Sfikas, Athens, Greece; Nicuşor Zlota, Traian Vuia Technical College, Focşani,Romania; Titu Zvonaru, Comănes,ti, Romania.


S483. For any real number a let ⌊a⌋ and a be the greatest integer less than or equal to a and the fractionalpart of a, respectively. Solve the equation

16x⌊x⌋ − 10x = 2019.

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

Solution by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, JapanLet n = ⌊x⌋ and a = x.

16x⌊x⌋ − 10x = 2019

⇔ 16n(n + a) − 10a = 2019

⇔ a = 2019 − 16n2

16n − 10.

We obtain 0 ≤ a = 2019−16n2

16n−10 < 1. If n ≥ 1, then 2019 − 16n2 ≥ 0 and 2019 − 16n2 < 16n − 10. Therefore weobtain n = 11, a = 0.5 and x = 11.5. Similarly, if n ≤ 0, then 16n2 − 2019 ≥ 0 and 16n2 − 2019 < 10 − 16n, nosolution satisfying these inequalities. The only solution is x = 11.5.

Also solved by Daniel Lasaosa, Pamplona, Spain; Jeffrey Roh, St. Andrew’s School, DE, USA; AndrewYang, Hotchkiss School, Lakeville, CT, USA; Sebastian Foulger, Charters School, Sunningdale, England,UK; Jeewoo Lee, Townsend Harris HS, NY, USA; Min Jung Kim, Tabor Academy, MA, USA; Kelvin Kim,Bergen Catholic High School, NJ, USA; Chirita Andrei-Giovani, The Greek Catholic Timotei Cipariu HighSchool, Bucharest, Romania; Bryant Hwang, Korea International School; Brian Bradie, Christopher New-port University, Newport News, VA, USA; Li Zhou, Polk State College, USA; Albert Stadler, Herrliberg,Switzerland; Arkady Alt, San Jose, CA, USA; Corneliu Mănescu-Avram, Ploies,ti, Romania; Dumitru Ba-rac, Sibiu, Romania; Joe Simons, Utah Valley University, UT, USA; Ioannis D. Sfikas, Athens, Greece;Joonsoo Lee, Dwight Englewood School, NJ, USA; Nicuşor Zlota, Traian Vuia Technical College, Focşani,Romania; Sushanth Sathish Kumar, PHS, SoCal, USA; Sachin Kumar, Bhabha Atomic Research Center,Mumbai, India; Sumanth Ravipati, George Mason University, VA, USA; Titu Zvonaru, Comănes,ti, Romania;Jiho Lee, Canterbury School, New Milford, CT, USA.


S484. Let a, b, c be positive real numbers such that a + b + c = 2. Prove that

a2 (1

b− 1)(1

c− 1) + b2 (1

c− 1)(1

a− 1) + c2 (1

a− 1)(1

b− 1) ≥ 1

3.

Proposed by An Zhenping, Xianyang Normal University, China

Solution by Dimoulios Ioannis, Thessaloniki, GreeceNote that all ∑ notations refer to ∑cyclic, unless otherwise mentioned.Let x = a/2, y = b/2, z = c/2. Then x + y + z = 1. Plugging in it suffices to prove that

∑a2 (1

b− 1)(1

c− 1) ≥ 1

3⇐⇒

∑4x2 ( 1

2y− 1)( 1

2z− 1) ≥ 1

3⇐⇒

3∑x3(x − y + z)(x + y − z) ≥ xyz ⇐⇒3∑(x5 + 2x3yz − x3y2 − x3z2) ≥ xyz(x + y + z)2 ⇐⇒

3∑x5 + 6∑x3yz ≥∑x3yz + 2∑x2y2z + 3∑x3y2 + 3∑x3z2 ⇐⇒3∑x5 + 5∑x3yz ≥ 2∑x2y2z + 3 ∑

sym

x3y2

By Muirhead’s Inequality∑sym

x3yz ≥ ∑sym

x2y2z

and by Shur’s Inequality of 5th degree and Muirhead’s Inequality

∑x5 +∑x3yz ≥ ∑sym

x4y ≥ ∑sym

x3y2

Adding the second to the first inequality three times gives the desired result.

Also solved by Daniel Lasaosa, Pamplona, Spain; Nikos Kalapodis, Patras, Greece; Jeffrey Roh, St. An-drew’s School, DE, USA; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Chirita Andrei-Giovani,The Greek Catholic Timotei Cipariu High School, Bucharest, Romania; Li Zhou, Polk State College, USA;Albert Stadler, Herrliberg, Switzerland; Arkady Alt, San Jose, CA, USA; Dumitru Barac, Sibiu, Romania;Ioannis D. Sfikas, Athens, Greece; Nguyen Viet Hung, Hanoi University of Science, Vietnam; Nicuşor Zlota,Traian Vuia Technical College, Focşani, Romania; Titu Zvonaru, Comănes,ti, Romania.


S485. Find all positive integers n for which there is a real constant c such that

(c + 1) (sin2n x + cos2n x) − c (sin2(n+1) x + cos2(n+1) x) = 1,

for all real numbers x.


Solution by Li Zhou, Polk State College, USALet fn(c, x) be the left side of the given equation. Then f1(0, x) = sin2 x + cos2 x = 1 for all real x. Also,

f2(2, x) = 3 (sin4 x + cos4 x) − 2 (sin6 x + cos6 x)

= 3 (sin2 x + cos2 x)2 − 2 (sin2 x + cos2 x)3 = 1

for all real x. Now suppose that n ≥ 3. We have fn (c, π4 ) =c+22n = 1, so c = 2n − 2. Meanwhile,

fn (c, π3) = (c + 1) (3n + 1)

4n−c (3n+1 + 1)

4n+1= 1,

so c = 4n+1−4(3n+1)3n+3 . Equating these two expressions of c we get 4n+1 + 2 = 6n + 2 ⋅ 3n + 3 ⋅ 2n, which is false for

n = 3. Moreover, since 45 < 64, 4n+1 < 6n for all n ≥ 4. Therefore, n = 1 and 2 are the only such integers.

Also solved by Daniel Lasaosa, Pamplona, Spain; Jeffrey Roh, St. Andrew’s School, DE, USA; Takuji Gri-gorovich Imaiida, Fujisawa, Kanagawa, Japan; Sebastian Foulger, Charters School, Sunningdale, England,UK; Jeewoo Lee, Townsend Harris HS, NY, USA; Chirita Andrei-Giovani, The Greek Catholic Timotei Cipa-riu High School, Bucharest, Romania; Jiho Lee, Canterbury School, New Milford, CT, USA; Albert Stadler,Herrliberg, Switzerland; Corneliu Mănescu-Avram, Ploies,ti, Romania; Ioannis D. Sfikas, Athens, Greece;Joonsoo Lee, Dwight Englewood School, NJ, USA; Dumitru Barac, Sibiu, Romania.


S486. Let ABC be an acute triangle. Let B1,C1 be the midpoint of AC and AB, respectively and B2,C2

be the foot of altitude from B,C, respectively. Let B3,C3 be the reflection of B2,C2 across the lineB1C1. The lines BB3 and CC3 intersect in X. Prove that XB =XC.

Proposed by Mihaela Berindeanu, Bucharest, Romania

Solution by Li Zhou, Polk State College, USA

Since C1 is the center of the circumcircle Ω of ABB2, B3 is on Ω. Since C1B1 ∥ BC, B2B3 ⊥ BC. Hence,∠CBB3 = 90 − A, which implies that the circumcenter O of ABC is on BB3. Likewise, O is on CC3 aswell. Therefore, X = O, completing the proof.

Also solved by Andrea Fanchini, Cantù, Italy; Daniel Lasaosa, Pamplona, Spain; Takuji GrigorovichImaiida, Fujisawa, Kanagawa, Japan; Chirita Andrei-Giovani, The Greek Catholic Timotei Cipariu HighSchool, Bucharest, Romania; Corneliu Mănescu-Avram, Ploies,ti, Romania; Dimoulios Ioannis, Thessaloniki,Greece; Albert Stadler, Herrliberg, Switzerland; Sushanth Sathish Kumar, PHS, SoCal, USA; Titu Zvonaru,Comănes,ti, Romania.


Undergraduate problems

U481. Evaluatelimn→∞

1

n(⌊e

1n ⌋ + ⌊e

2n ⌋ +⋯ + ⌊e

nn ⌋) .


Solution by Daniel Lasaosa, Pamplona, SpainSince for all 1 ≤ k ≤ n we have 0 < k

n ≤ 1, then 3 > e ≥ e kn > 1, or ⌊e kn ⌋ takes only values 1 or 2, taking value 2

iff k ≥ n ln 2. For each n, denote Kn = ⌊n ln 2⌋, where since ln 2 is irrational, the value is 1 for k = 1,2, . . . ,Kn

and 2 for k =Kn + 1,Kn + 2, . . . , n, resulting in

⌊e1n ⌋ + ⌊e

2n ⌋ +⋯ + ⌊e

nn ⌋ =Kn + 2(n −Kn) = 2n −Kn.

Now, clearly a real 0 < δn < 1 exists such that Kn = n ln 2 − δn, or the proposed limit equals

limn→∞

2n −Kn

n= 2 − ln 2 + lim

n→∞

δnn

= 2 − ln 2.

The conclusion follows.

Also solved by Jeffrey Roh, St. Andrew’s School, DE, USA; Andrew Yang, Hotchkiss School, Lakevil-le, CT, USA; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Sebastian Foulger, Charters School,Sunningdale, England, UK; Jeewoo Lee, Townsend Harris HS, NY, USA; Jenna Park, Blair Academy, Blair-stown, NJ, USA; Min Jung Kim, Tabor Academy, MA, USA; Brian Bradie, Christopher Newport University,Newport News, VA, USA; Li Zhou, Polk State College, USA; Jiho Lee, Canterbury School, New Milford, CT,USA; Albert Stadler, Herrliberg, Switzerland; Corneliu Mănescu-Avram, Ploies,ti, Romania; Dumitru Ba-rac, Sibiu, Romania; Henry Ricardo, Westchester Area Math Circle, NY, USA; Joel Schlosberg, Bayside,NY, USA; Ioannis D. Sfikas, Athens, Greece; Joonsoo Lee, Dwight Englewood School, NJ, USA; Khakim-boy Egamberganov, ICTP; Moubinool Omarjee, Lycée Henri IV, Paris, France; Nicuşor Zlota, Traian VuiaTechnical College, Focşani, Romania; Sumanth Ravipati, George Mason University, VA, USA; Sushanth Sa-thish Kumar, PHS, SoCal, USA; Tamoghno Kandar, IIT Bombay, India; Thiago Landim de Souza Leão,Federal University of Pernambuco, Recife, Brazil; Akash Singha Roy, Chennai Mathematical Institute, India;Prajnanaswaroopa S, Amrita University, Coimbatore, India; Arkady Alt, San Jose, CA, USA.


U482. For any positive integer n consider the polynomial fn = x2n+xn+1. Prove that for any positive integerm there is a positive integer n such that fn has exactly m irreducible factors in Z[X].

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

First solution by Daniel Lasaosa, Pamplona, SpainNote first that all n ≥ 1, we have

x3⋅2n+1 − 1 = (x3⋅2n−1 − 1) (x3⋅2n−1 + 1) (x2n + 1) (x2n+1 − x2n + 1) ,

or since the cyclotomic polynomial Φ3⋅2n+1(x) has degree ϕ (3 ⋅ 2n+1) = 2 ⋅ 2n = 2n+1, where ϕ is the Eulertotient function, and it is irreducible in Z[X], then

Φ3⋅2n+1(x) = x2n+1 − x2n + 1,

and it is irreducible.

Lemma: For every non negative integer k, the polynomial

pk(x) = x2k+1 + x2k + 1

is the product of exactly k + 1 irreducible factors in Z[X].Proof: For k = 0, we have p0(x) = x2 +x+ 1 is irreducible in Z[X], since otherwise either −1 or +1 would

be a root, but p0(−1) = 1 and p0(1) = 3. If the Lemma holds for k, note that for k + 1 we have

pk+1(x) = (x2k+1 + x2k + 1) (x2k+1 − x2k + 1) = pk(x) ⋅Φ3⋅2n+1(x)

or since pk has exactly k + 1 irreducible factors and Φ3⋅2n+1(x) is irreducible, then pk+1 has exactly k + 2irreducible factors in Z[X]. The Lemma follows.

By the Lemma, for every positive integer m, pm−1(x) = f2m−1(x) has exactly m irreducible factors inZ[X]. The conclusion follows.


Second solution by Li Zhou, Polk State College, USAActually, for each m, there are infinitely many such n. Indeed, take any prime p different from 3, and letn = pm−1. Denote by Φd(x) the d-th cyclotomic polynomial.Then it is well known (http://mathworld.wolfram.com/CyclotomicPolynomial.html) that

fn(x) =x3n − 1

xn − 1=∏d∣3nΦd(x)∏d∣nΦd(x)

=m

∏k=1

Φ3pk−1(x)

and each Φ3pk−1(x) is irreducible in Z[X].

Also solved by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Akash Singha Roy, Chennai Ma-thematical Institute, India; Joel Schlosberg, Bayside, NY, USA; Ioannis D. Sfikas, Athens, Greece; MoubinoolOmarjee, Lycée Henri IV, Paris, France; Albert Stadler, Herrliberg, Switzerland.


U483. Evaluatelimn→∞

1

n3∑

1≤i<j<k≤n

cot−1 ( in) cot−1 ( j

n) cot−1 (k

n)

Proposed by Nicuşor Zlota, Focşani, Romania

Solution by Brian Bradie, Christopher Newport University, Newport News, VA, USARecognize that

limn→∞

1

n3∑

1≤i<j<k≤n


n) cot−1 (k

n)

= ∫1

0∫

x

0∫

y

0cot−1 z cot−1 y cot−1 xdz dy dx.

By integration by parts,

∫ cot−1 z dz = z cot−1 z + ∫z

1 + z2 dz = z cot−1 z + 1

2ln(1 + z2) +C,

so

∫y

0cot−1 z dz = y cot−1 y + 1

2ln(1 + y2).

Next,

∫x

0cot−1 y (y cot−1 y + 1

2ln(1 + y2)) dy = 1

2(y cot−1 y + 1

2ln(1 + y2))

2 RRRRRRRRRRR

x

0

= 1

2(x cot−1 x + 1

2ln(1 + x2))

2

and

∫1

0cot−1 x ⋅ 1

2(x cot−1 x + 1

2ln(1 + x2))

2

dx = 1

6(x cot−1 x + 1

2ln(1 + x2))

3 RRRRRRRRRRR

1

0

= 1

6(π

4+ 1

2ln 2)

3

= 1

384(π + 2 ln 2)3.

Thus,

limn→∞

1

n3∑

1≤i<j<k≤n


n) cot−1 (k

n) = 1

384(π + 2 ln 2)3.

Also solved by Daniel Lasaosa, Pamplona, Spain; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa,Japan; Li Zhou, Polk State College, USA; Akash Singha Roy, Chennai Mathematical Institute, India; AlbertStadler, Herrliberg, Switzerland; Dumitru Barac, Sibiu, Romania; Ioannis D. Sfikas, Athens, Greece.


U484. Find all polynomials P (x) for which:

P (a + b) = 6 (P (a) + P (b)) + 15a2b2(a + b),

for all complex numbers a and b such that a2 + b2 = ab.

Proposed by Titu Andreescu, USA and Mircea Becheanu, Canada

Solution by Li Zhou, Polk State College, USALetting a = b = 0 yields P (0) = 12P (0), so P (0) = 0. Clearly, P /≡ 0. Thus P (x) = cnxn +⋯ + c1x, with n ≥ 1and cn ≠ 0.

Suppose that a2+b2 = ab. We claim that for each k ≥ 1, there is a constant fk such that ak+bk = fk(a+b)k.Indeed, f1 = 1, and since a2 + b2 = (a + b)2 − 2ab = (a + b)2 − 2 (a2 + b2), f2 = 1

3 . Inductively, for k ≥ 2,

ak+1 + bk+1 = (ak + bk) (a + b) − ab (ak−1 + bk−1) = fk(a + b)k+1 −1

3fk−1(a + b)k+1,

that is, fk+1 = fk − 13fk−1. In particular, f3 = 0 and f4 = −1

9 = f5. Using a generating function we can also

obtain fk = (√3+i

2√3)k+ (√3−i

2√3)k. Therefore, for k ≥ 6,

∣6fk∣ ≤ 12 ∣√

3 + i2√

3∣k

= 12( 1√3)k

≤ 12

27< 1.

Equating coefficients of (a+ b)k in P (a+ b) = 6(P (a)+P (b))+ 53(a+ b)

5, we get (1− 6fk)ck = 0 for all k ≠ 5,thus ck = 0 for all k ≠ 5. Also, c5 = 6f5c5 + 5

3 = −23c5 +

53 , so c5 = 1. In conclusion, P (x) = x5 is the only

solution.

Also solved by Daniel Lasaosa, Pamplona, Spain; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa,Japan; Jeffrey Roh, St. Andrew’s School, DE, USA; Akash Singha Roy, Chennai Mathematical Institute,India; Albert Stadler, Herrliberg, Switzerland; Corneliu Mănescu-Avram, Ploies,ti, Romania; Ioannis D.Sfikas, Athens, Greece.


U485. Let f ∶ [0,1]→ (0,∞) be a continuous function and let A be the set of all positive integers n for whichthere is a real number xn such that

∫1

xnf(t)dt = 1

n.

Prove that the set xnn∈A is an infinite sequence and find

limn→∞

n (xn − 1) .

Proposed by Florin Rotaru, Focşani, Romania

Solution by Thiago Landim de Souza Leão, Federal University of Pernambuco, Recife, BrazilSince f(t) > 0 ∀t ∈ [0,1], then ∫ 1

0 f(t)t. = ε > 0 and ∃n0 such that ε > 1n0. Moreover,

F (x) = ∫1

xf(t)t.

is a decreasing differentiable function F ∶ [0,1]→ [0, ε] and satisfying F ′(x) = −f(x).Hence, ∀n ≥ n0 ∃xn such that F (xn) = 1

n .Even more, by the Mean Value Theorem, ∃ξ ∈ [xn,1] such that

F (1) − F (xn) = (1 − xn)F ′(ξ) = (1 − xn)f(ξ)

and n(1 − xn) = − 1f(ξ) . Taking n→∞, we get

limn→∞

n(1 − xn) = −1

f(1) .

Also solved by Daniel Lasaosa, Pamplona, Spain; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Ja-pan; Li Zhou, Polk State College, USA; Akash Singha Roy, Chennai Mathematical Institute, India; Prajnana-swaroopa S, Amrita University, Coimbatore, India; Khakimboy Egamberganov, ICTP; Moubinool Omarjee,Lycée Henri IV, Paris, France; Sergio Esteban Muñoz, Universidad de Buenos Aires, Argentina; AlbertStadler, Herrliberg, Switzerland.


U486. Let ⌊x⌋ be the floor function and let k ≥ 3 be a positive integer. Evaluate

∫∞

0

⌊x⌋xk

dx.

Proposed by Metin Can Aydemir, Ankara, Turkey

Solution by Joel Schlosberg, Bayside, NY, USA

∫∞

0

⌊x⌋xk

dx

=∞

∑n=0∫

n+1

n

⌊x⌋xk

dx

=∞

∑n=0∫

n+1

n

n

xkdx

= 0 +∞

∑n=1

n

(1 − k)xk−1 ∣n+1

n

= 1

k − 1

∞

∑n=1

( n

nk−1− n

(n + 1)k−1)

= 1

k − 1

∞

∑n=1

( 1

nk−2− 1

(n + 1)k−2 +1

(n + 1)k−1)

= 1

k − 1[ζ(k − 2) − (ζ(k − 2) − 1) + (ζ(k − 1) − 1)]

= ζ(k − 1)k − 1

.

Also solved by Daniel Lasaosa, Pamplona, Spain; Jeffrey Roh, St. Andrew’s School, DE, USA; AndrewYang, Hotchkiss School, Lakeville, CT, USA; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Seba-stian Foulger, Charters School, Sunningdale, England, UK; Brian Bradie, Christopher Newport University,Newport News, VA, USA; Li Zhou, Polk State College, USA; Akash Singha Roy, Chennai MathematicalInstitute, India; Prajnanaswaroopa S, Amrita University, Coimbatore, India; Albert Stadler, Herrliberg, Swi-tzerland; Corneliu Mănescu-Avram, Ploies,ti, Romania; Henry Ricardo, Westchester Area Math Circle, NY,USA; Ioannis D. Sfikas, Athens, Greece; Khakimboy Egamberganov, ICTP; Daniel López-Aguayo, InstitutoTecnológico y de Estudios Superiores de Monterrey, Mexico; Moubinool Omarjee, Lycée Henri IV, Paris,France; Sumanth Ravipati, George Mason University, VA, USA; Sushanth Sathish Kumar, PHS, SoCal,USA; Thiago Landim de Souza Leão, Federal University of Pernambuco, Recife, Brazil.


Olympiad problems

O481. Prove thatn

∏k=1

(1 − 4 sinπ

5ksin

3π

5k) = − sec

π

5n.


Solution by Li Zhou, Polk State College, USABy the double-angle and product-to-sum formulas,

cosπ

5k(1 − 4 sin

π

5ksin

3π

5k) = cos

π

5k− 2 sin

2π

5ksin

3π

5k= cos

π

5k+ cos

π

5k−1− cos

π

5k= cos

π

5k−1.

By telescoping,

n

∏k=1

(1 − 4 sinπ

5ksin

3π

5k) =

n

∏k=1

cosπ

5k−1sec

π

5k= cos

π

50sec

π

5n= − sec

π

5n.

Also solved by Daniel Lasaosa, Pamplona, Spain; Andrew Yang, Hotchkiss School, Lakeville, CT, USA;Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Sebastian Foulger, Charters School, Sunningdale,England, UK; Jeewoo Lee, Townsend Harris HS, NY, USA; Min Jung Kim, Tabor Academy, MA, USA; Chi-rita Andrei-Giovani, The Greek Catholic Timotei Cipariu High School, Bucharest, Romania; Brian Bradie,Christopher Newport University, Newport News, VA, USA; Albert Stadler, Herrliberg, Switzerland; DumitruBarac, Sibiu, Romania; Joonsoo Lee, Dwight Englewood School, NJ, USA; M.A.Prasad, India; MoubinoolOmarjee, Lycée Henri IV, Paris, France; Sumanth Ravipati, George Mason University, VA, USA; Titu Zvo-naru, Comănes,ti, Romania; Akash Singha Roy, Chennai Mathematical Institute, India; Jiho Lee, CanterburySchool, New Milford, CT, USA.


O482. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 1. Prove that

a2

c3+ b

2

a3+ c

2

b3≥ (a + b + c)3.

Proposed by Dragoljub Miloşević, Gornji Milanovac, Serbia

Solution by Adamopoulos Dionysios, 3rd High School, Pyrgos, GreeceBy Hölder inequality we have that

(a2

c3+ b

2

a3+ c

2

b3)(a2 + b2 + c2)(c2 + a2 + b2) ≥

⎛⎝

3

√a4

c+ 3

√b4

a+ 3

√c4

b

⎞⎠

3

Since a2 + b2 + c2 = 1, we obtain that

a2

c3+ b

2

a3+ c

2

b3≥⎛⎝

3

√a4

c+ 3

√b4

a+ 3

√c4

b

⎞⎠

3

As a result it suffices to prove that

3

√a4

c+ 3

√b4

a+ 3

√c4

b≥ a + b + c

Setting a = x3, b = y3, c = z3 our inequality becomes

x4

z+ y

4

x+ z

4

y≥ x3 + y3 + z3

which is obvious due to the rearrangement inequality for triples (x4, y4, z4) and ( 1x ,

1y ,

1z).

Also solved by Daniel Lasaosa, Pamplona, Spain; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa,Japan; Nikos Kalapodis, Patras, Greece; Albert Stadler, Herrliberg, Switzerland; Ioan Viorel Codreanu, Sa-tulung, Maramureş, Romania; Dimoulios Ioannis, Thessaloniki, Greece; Dumitru Barac, Sibiu, Romania;Erfan Amouzad Khalili, Iran; Jamal Gadirov, Istanbul University, Turkey; Ioannis D. Sfikas, Athens, Gree-ce; Khakimboy Egamberganov, ICTP; Leung Hei Chun, SKH Tang Shiu Kin Secondary School, Hong Kong;M.A.Prasad, India; Marin Chirciu, Colegiul Nat,ional Zinca Golescu, Pites,ti, Romania; Nguyen Viet Hung,Hanoi University of Science, Vietnam; Nick Iliopoulos, 8th General Highschool of Trikala, Greece; NicuşorZlota, Traian Vuia Technical College, Focşani, Romania; Sarah B. Seales, Prescott, AZ, USA; SumanthRavipati, George Mason University, VA, USA; Sushanth Sathish Kumar, PHS, SoCal, USA; Titu Zvonaru,Comănes,ti, Romania; Li Zhou, Li Zhou, Polk State College, USA; Akash Singha Roy, Chennai Mathema-tical Institute, India; Chirita Andrei-Giovani, The Greek Catholic Timotei Cipariu High School, Bucharest,Romania.


O483. Find all integers n for which (4n2 − 1) (n2 + n) + 2019 is a perfect square.


Solution by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, JapanLet f(n) =

√(4n2 − 1)(n2 − n) + 2019, then

f(1) = 45, 45 < f(2) < 46, , 49 < f(3) < 50, 57 < f(4) < 58, 70 < f(5) < 71,

89 < f(6) < 90, 113 < f(7) < 114, 142 < f(8) < 143, 176 < f(9) < 177.

For n ≥ 10 we obtain 2n2 + n − 1 < f(n) < 2n2 + n + 5, that is,

f(n) = 2n2 + n, 2n2 + n + 1, 2n2 + n + 2, 2n2 + n + 3, 2n2 + n + 4.

If f(n) = 2n2 + n, no integer solution. If f(n) = 2n2 + n + 1, no integer solution. If f(n) = 2n2 + n + 2, nointeger solution. If f(n) = 2n2+n+3, no integer solution. If f(n) = 2n2+n+4, no integer solution. Thereforethe only solution is n = 1.

Also solved by Daniel Lasaosa, Pamplona, Spain; Jeffrey Roh, St. Andrew’s School, DE, USA; Seba-stian Foulger, Charters School, Sunningdale, England, UK; Jeewoo Lee, Townsend Harris HS, NY, USA;Chirita Andrei-Giovani, The Greek Catholic Timotei Cipariu High School, Bucharest, Romania; Albert Stad-ler, Herrliberg, Switzerland; Nick Iliopoulos, 8th General Highschool of Trikala, Greece; Corneliu Mănescu-Avram, Ploies,ti, Romania; Dumitru Barac, Sibiu, Romania; Ioannis D. Sfikas, Athens, Greece; KhakimboyEgamberganov, ICTP; M.A.Prasad, India; Nguyen Viet Hung, Hanoi University of Science, Vietnam; Su-manth Ravipati, George Mason University, VA, USA; Titu Zvonaru, Comănes,ti, Romania; Akash SinghaRoy, Chennai Mathematical Institute, India; Li Zhou, Polk State College, USA.


O484. Let ABC be a triangle with AB = AC. Points E and F lie on AB and AC, respectively so that EFpasses through the circumcenter of ABC. Let M be the midpoint of AB, let N be the midpoint ofAC and set P = FM ∩EN . Prove that the lines AP and EF are perpendicular.

Proposed by Tovi Wen, USA

First solution by by Andrea Fanchini, Cantù, Italy

We use barycentric coordinates with reference to the triangle ABC.

We consider a generic point E(t,1 − t,0) on AB where t is a parameter. Then the line that passes throughthe circumcenter of ABC is

EO ∶ (1 − t)b2x − b2ty + ((3SA + SB)t − 2SA)z = 0

so the point F isF = EO ∩AC = (2SA − (3SA + SB)t ∶ 0 ∶ (1 − t)b2)

and the line EFEF ∶ (1 − t)b2x − b2ty − (2SA − (3SA + SB)t)z = 0

now we have M(1 ∶ 1 ∶ 0) and N(1 ∶ 0 ∶ 1) therefore

FM ∶ −(1 − t)b2x + (1 − t)b2y + (2SA − (3SA + SB)t)z = 0

EN ∶ (1 − t)x − ty − (1 − t)z = 0


Then, the point P is

P = FM ∩EN = (−2(SB + 2SA)t2 + 2(SB + 2SA)t − b2 ∶ 2SAt2 + (SB − 3SA)t + (SA − SB) ∶ (1 − t)(2t − 1)b2)

and the line APAP ∶ (1 − t)(2t − 1)b2y − (2SAt2 + (SB − 3SA)t + (SA − SB))z = 0

Finally, we have

EF∞ = AP∞⊥ = (−2(2SA + SB)t + 2SA ∶ 2(2SA + SB)t − (3SA + SB) ∶ SA + SB)

Q.e.d.


Second solution by Li Zhou, Polk State College, USA

Let O be the circumcenter of ABC, Q be the orthocenter of AEF , and EV and FU be two altitudes ofAEF . It suffices to show that A,Q,P are collinear. By Desargues’ theorem, this will follow from the claimthat triangles ENV and FMU are perspective from a point X. Indeed,

AU

AV= AFAE

= FOOE

= FOFE

⋅ FEOE

= FNFV

⋅ UEME

,

thusAU

UE⋅ FVV A

= FN

ME= AMME

⋅ FNNA

.

By Menelaus’ theorem, the lines UV , MN , and EF are concurrent, completing the proof.

Also solved by Daniel Lasaosa, Pamplona, Spain; Chirita Andrei-Giovani, The Greek Catholic TimoteiCipariu High School, Bucharest, Romania; Adamopoulos Dionysios, 3rd High School, Pyrgos, Greece; Cor-neliu Mănescu-Avram, Ploies,ti, Romania; Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti,Romania.


O485. Prove that any infinite set of positive integers contains two numbers whose sum has a prime divisorgreater than 102020.

Proposed by Navid Safaei, Sharif University of Technology, Tehran, Iran

Solution by Daniel Lasaosa, Pamplona, SpainWe will prove a (slightly) more general statement, namely that for any infinite set A of positive integers,there are infinitely many primes p such that one number of the form m + n, where m ≠ n and m,n ∈ A, isdivisible by p.

Assume the contrary, ie that given an infinite set A, there is a finite set of distinct primes p1, p2, . . . , pusuch that any number of the form m + n, with m ≠ n and m,n ∈ A, can be written as a product of saidprimes. Define the sequence of sets A0 = A,A1,A2, . . . ,Au such that Au ⊂ Au−1 ⊂ ⋯ ⊂ A1 ⊂ A0 = A are allinfinite, and such that Ai is defined starting from Ai−1 for i = 1,2, . . . , u, depending on whether pi satisfiesone of the following two conditions:

Case 1: If there is a (possibly zero) multiplicity αi such that there are infinitely many elements n ∈ Adivisible by pi with multiplicity αi, then c = n

piαi

is a positive integer which is not a multiple of pi for infinitelymany elements n ∈ A; if this occurs for more than one multiplicity αi, choose one of them at random. Thereis therefore one (or possible more than one) of the pi − 1 possible nonzero remainders modulo pi, denoted byri, which repeats itself infinitely many times in the c’s (if more than one, then choose one at random). Ifpi is an odd prime, define Ai as the infinite set of such n’s. In the particular case where pi = 2, note thatthe c’s are odd, or each one leaves a remainder either 1 or 3 when divided by 4, one of these two remaindersappearing infinitely many times (if both, then choose one of them at random). Define in this case Ai as theset of the infinitely many n’s such that the corresponding c’s leave this remainder.

Case 2: If no nonnegative multiplicity αi is such that infinitely many elements n ∈ A are divisible bypi with multiplicity αi, then there must be infinitely many multiplicities αi such that at least one n ∈ A isdivided by pi with multiplicity αi. For each such multiplicity, chose at random exactly one of the n’s whichare divisible by pi with multiplicity αi, and let Ai be the set formed by the infinitely many such n’s.

Consider now any two distinct elements m,n ∈ Au, and let D be their greatest common divider. SinceD ≤m,n and m ≠ n, we have 2D <m + n.

Now, if pi is an odd prime which falls under Case 1, clearly pi divides D with multiplicity αi since itdivides both m,n with said multiplicity, and since m+n

piαi

leaves a remainder 2ri when divided by pi, where riis a nonzero remainder modulo pi, then pi also divides m+n with multiplicity αi. If pi = 2 falls under Case 1,then 2 divides D again with multiplicity αi, but m+n

piαi

yields a remainder of 2 when divided by 4, regardlessof whether m

piαi

and npiαi

yield both a remainder of 1, or both a remainder of 3, when divided by 4. Hencepi = 2 divides m + n with multiplicity αi + 1. On the other hand, if pi is any prime which falls under Case2, then m + n is divisible by pi with multiplicity equal to the least of the two multiplicities with which pidivides m and n, and likewise for D.

It follows that, except for a possible additional factor of 2 in m + n if prime pi = 2 falls under Case 1,m + n and D are divisible by exactly the same primes and with the same multiplicity, or m + n ∈ D,2D,for m + n ≤ 2D <m + n. The conclusion follows from this contradiction.

Also solved by Chirita Andrei-Giovani, The Greek Catholic Timotei Cipariu High School, Bucharest,Romania; M.A.Prasad, India; Akash Singha Roy, Chennai Mathematical Institute, India; Albert Stadler,Herrliberg, Switzerland.


O486. Let a, b, c be positive real numbers. Prove that

a2 + b2 + c2 ≥ a 3

√b3 + c3

2+ b 3

√c3 + a3

2+ c 3

√a3 + b3

2


Solution by Li Zhou, Polk State College, USAFirst,

(a2 + b2 + c2)2 − (a + b + c) [a (b2 − bc + c2) + b (c2 − ca + a2) + c (a2 − ab + b2)]= a2(a − b)(a − c) + b2(b − c)(b − a) + c2(c − a)(c − b) ≥ 0

by Schur’s inequality. Now we use Hölder’s inequality to get

(a2 + b2 + c2)3 = (a2 + b2 + c2) (a2 + b2 + c2)2

≥ (a2 + b2 + c2)(b + c2

+ c + a2

+ a + b2

) [a (b2 − bc + c2) + b (c2 − ca + a2) + c (a2 − ab + b2)]

≥⎛⎝

3

√a2 ⋅ b + c

2⋅ a (b2 − bc + c2) + 3

√b2 ⋅ c + a

2⋅ b (c2 − ca + a2) + 3

√c2 ⋅ a + b

2⋅ c (a2 − ab + b2)

⎞⎠

3

=⎛⎝a

3

√b3 + c3

2+ b 3

√c3 + a3

2+ c 3

√a3 + b3

2

⎞⎠

3

.

Also solved by Daniel Lasaosa, Pamplona, Spain; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa,Japan; Nikos Kalapodis, Patras, Greece; Albert Stadler, Herrliberg, Switzerland; Ioan Viorel Codreanu, Sa-tulung, Maramureş, Romania; Ioannis D. Sfikas, Athens, Greece; Khakimboy Egamberganov, ICTP; MarinChirciu, Colegiul Nat,ional Zinca Golescu, Pites,ti, Romania; Nicuşor Zlota, Traian Vuia Technical College,Focşani, Romania; Sarah B. Seales, Prescott, AZ, USA; Titu Zvonaru, Comănes,ti, Romania; Akash SinghaRoy, Chennai Mathematical Institute, India.


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Junior problems - AwesomeMath€¦ · J483. Leta;b;cberealnumberssuchthat13a+41b+13c=2019 and max‰V 41 13 a−bV;V 13 41 b−cV;Sc−aS’≤1: Provethat2019 ≤a2 +b2 +c2 ≤2020