Mr 5 2022 solutions

Junior problems J601 Let a, b, c, d be real numbers such that 7 > a ≥ b + 1 ≥ c + 3 ≥ d + 4 Prove that 1 7 − a + 4 6 − b + 9 4 − c + 16 3 − d ≥ a + b + c + d Proposed by Adrian Andreescu, University o.

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

Solution by Arkady Alt, San Jose, CA, USA

Let s ∶= a + b + c + d Since a < 7, b < 6, c < 4, d < 3 then s < 20 By Cauchy-Schwarz Inequality

Equality iff (7 − a, 6 − b, 4 − c, 3 − d) = (1, 2, 3, 4) ⇐⇒ (a, b, c, d) = (6, 4, 1, −1)

Also solved by Polyahedra, Polk State College, USA; Ivan Hadinata, Jember, Indonesia; Alex Grigoryan,Quantum College, Yerevan, Armenia; Corneliu Mănescu-Avram, Ploieşti, Romania; Daniel Văcaru, Pites,ti,Romania; Hyunbin Yoo, South Korea; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy;Henry Ricardo, Westchester Area Math Circle, USA; Sundaresh H R, Shivamogga, India; Theo Koupelis,Cape Coral, FL, USA; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Muhammad Thoriq,Yogyakarta, Indonesia; Israel Castillo Pilco, Huaral, Peru; Adam John Frederickson, Utah Valley University,

UT, USA; Dao Van Nam, MAY High School, Hoáng Mai, Ha Noi, Vietnam

J602 Prove that for any positive real number x,

Proposed by An Zhenping, Xianyang Normal University, China

Solution by Polyahedra, Polk State College, USA

Rationalizing a numerator, we have

√((√x)2+ 12) (12+ (√x+ 1)2) ≥√x+√x+ 1,with equality if and only if 1/√x=√x+ 1, that is, if and only if x = (√5− 1)/2

Also solved by Ivan Hadinata, Jember, Indonesia; Alex Grigoryan, Quantum College, Yerevan, Armenia;Muhammad Thoriq, Yogyakarta, Indonesia; Israel Castillo Pilco, Huaral, Peru; Adam John Frederickson,Utah Valley University, UT, USA; Corneliu Mănescu-Avram, Ploieşti, Romania; Daniel Văcaru, Pites,ti,Romania; G C Greubel, Newport News, VA, USA; Marin Chirciu,Colegiul Nat,ional Zinca Golescu, Pites,ti,Romania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Soham Dutta, DPS, Ruby Park,West Bengal, India; Sundaresh H R, Shivamogga, India; Theo Koupelis, Cape Coral, FL, USA; Nicuşor Zlota,Traian Vuia Technical College, Focşani, Romania; Prajnanaswaroopa S, Bangalore, India; Daniel Pascuas,Barcelona, Spain; Arkady Alt, San Jose, CA, USA

J603 Let ABC be a triangle with centroid G and M, N, P, Q be the midpoints of the segments AB, BC, CA, AG,respectively Prove that if

sin(A − B) sin C = sin(C − A) sin B,then points M, N, P, Q lie on a circle

Proposed by Mihaela Berindeanu, Bucharest, RomâniaSolution by Kousik Sett, India

Draw AD ⊥ BC H is the orthocenter AD cuts the nine-point circle at E which is the midpoint of AH.The nine-point circle pass through the points M, N, P , Q, and E

We have

AN = 12

√2b2+ 2c2− a2, AQ=1

3AN, AE =1

2AH = R cos A, and AD = bc

2R.

As sin C = sin(A + B) and sin B = sin(A + C), the given condition

sin(A − B) sin C = sin(C − A) sin B,

1

3×1

4(2b2+ 2c2− a2) = b2+ c2− a2

4 Ô⇒ b2+ c2= 2a2,which is true by (1)

Also solved by Alex Grigoryan, Quantum College, Yerevan, Armenia; Polyahedra, Polk State College,USA; Corneliu Mănescu-Avram, Ploieşti, Romania; Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;Telemachus Baltsavias, Kerameies Junior High School, Kefalonia, Greece; Theo Koupelis, Cape Coral, FL,USA

J604 Let m, n, p be odd positive integers such that (m − n)(n − p)(p − m) = 0 and

7

m+ 8

n+9

p = 2

Find the least possible value of mnp

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

Solution by the author

We get m = 7, n = p = 17 or m = 63, n = p = 9, with the least mnp being 2023 So, the least possible value ofmnp is 2023

Also solved by Theo Koupelis, Cape Coral, FL, USA; Daniel Văcaru, Pites,ti, Romania; Ivan Hadinata,Jember, Indonesia; Alex Grigoryan, Quantum College, Yerevan, Armenia; Muhammad Thoriq, Yogyakarta,Indonesia; Polyahedra, Polk State College, USA; Israel Castillo Pilco, Huaral, Peru; Prajnanaswaroopa

S, Bangalore, India; Emon Suin, Ramakrishna Mission Vidyalaya, Narendrapur, India; Sundaresh H R,Shivamogga, India

Solution by Theo Koupelis, Cape Coral, FL, USA

(a) Let s, E be the triangle’s semiperimeter and area, respectively Then we have 4m2

a = 2(b2+ c2) − a2 ≥(b + c)2− a2 = (a + b + c)(b + c − a) = 4s(s − a), and similarly 4m2

abc⋅ (2s2− a2− b2− c2)

= 2sabc⋅ (ab + bc + ca − s2) = 2 +2(s − a)(s − b)(s − c)

abc

= 2 + E2s⋅ 4Eabc = 2 + r

2R.Equality occurs when a = b = c

(b) For an acute triangle the quantities cos A, cos B, cos C are positive We have 4m2

a = 2(b2+ c2) − a2 =

b2+c2+2bc cos A ≤ (b2+c2)(1+cos A), and similarly 4m2

b ≤ (a2+c2)(1+cos B), and 4m2

c≤ (a2+b2)(1+cos C).Thus,

cos A+ cos B + cos C − 1 = b2+ c2− a2

Also solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Alex Grigoryan, Quantum lege, Yerevan, Armenia; Polyahedra, Polk State College, USA; Israel Castillo Pilco, Huaral, Peru; CorneliuMănescu-Avram, Ploieşti, Romania; Daniel Văcaru, Pites,ti, Romania; Marin Chirciu,Colegiul Nat,ional Zin-

Col-ca Golescu, Pites,ti, Romania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Scott H.Brown, Auburn University Montgomery, AL, USA; Nicuşor Zlota, Traian Vuia Technical College, Focşani,Romania; Dao Van Nam, MAY High School, Hoáng Mai, Ha Noi, Vietnam; Arkady Alt, San Jose, CA,USA; Titu Zvonaru, Comănes,ti, România

J606 Let a, b, c be real numbers in the interval [0, 1] Prove that

Proposed by Marius Stănean, Zalău, România

Solution by Polyahedra, Polk State College, USA

The inequalities are equivalent to

2 +2b2

√ca

2 +2c2

√ab

2b +c2(a + b)

2c = ab + bc + ca

Since 1 + bc − (b + c) = (1 − b)(1 − c) ≥ 0, 1 + bc ≥ b + c Therefore,

1+ 2abc = a(1 + bc) + b(1 + ca) + c(1 + ab) + (1 − a)(1 − b)(1 − c) − (ab + bc + ca)

≥ a(b + c) + b(c + a) + c(a + b) − (ab + bc + ca) = ab + bc + ca,completing the proof

Also solved by Alex Grigoryan, Quantum College, Yerevan, Armenia; Paolo Perfetti, Università deglistudi di Tor Vergata Roma, Italy; Theo Koupelis, Cape Coral, FL, USA

Senior problems

S601 Solve in integers the equation

16x2y2(x2+ 1)(y2− 1) + 4(x4+ y4+ x2− y2) = 20232− 1

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

Solution by the author

The equation can be rewritten as

(4x4+ 4x2)(4y4− 4y2) + (4x4+ 4x2) + (4y4− 4y2) + 1 = 20232,which is equivalent to

implies x2= 144 and y2 = 4 The solutions (x, y) are (12, 2), (12, −2), (−12, 2), (−12, −2)

Also solved by Ivan Hadinata, Jember, Indonesia; Israel Castillo Pilco, Huaral, Peru; Adam John derickson, Utah Valley University, UT, USA; Le Hoang Bao, Tien Giang, Vietnam; Sundaresh H R, Shiva-mogga, India; Theo Koupelis, Cape Coral, FL, USA

Fre-S602 For every integer n > 1 let

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

Solution by Theo Koupelis, Cape Coral, FL, USA

When n is odd, the number of terms inside the parentheses for both An+1/An and Bn+1/Bn is even For

An+1/An,the average value of the sum of two terms that are symmetrically positioned in the summation isgiven by

1

2((m + 1)/21 + 1

n+ 2 − (m + 1)/2) =

2(n + 2)(m + 1)(2n + 3 − m).

We now have

n+ 1

n+ 2⋅

n+ 1m(2n + 2 − m) ≥

1

2⋅(m + 1)(2n + 3 − m)2(n + 2) ⇐⇒ (2n + 3)(n − m + 1)2≥ 0,which is obvious

When n is even, the number of terms inside the parentheses for both An+1/An and Bn+1/Bn is odd In thiscase, the middle term in the summation for An+1/Anis 1/(n + 1), and the middle term in the summation for

Also solved by G C Greubel, Newport News, VA, USA; Dao Van Nam, MAY High School, Hoáng Mai,

Ha Noi, Vietnam; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Nicuşor Zlota, TraianVuia Technical College, Focşani, Romania; Arkady Alt, San Jose, CA, USA

S603 Let ABC be a triangle with incenter I and centroid G Line AG intersects BC in E and line AIintersects BC in D and the circumcircle in M Point P is the orthogonal projection of E onto AM.Prove that if (AB + AC)2= 16AP ⋅ DM then GI is parallel to BC.

Proposed by Mihaela Berindeanu, Bucharest, România

Solution by Kousik Sett, India

M is the midpoint of minor arc BC Join M, E Then ∠MED = 90○

Again, since EP ⊥ DM, we get

M P2− DP2= ME2− DE2

Ô⇒ MD ⋅ (MP − DP ) = ME2−a4(b + c)2(b − c)22

Ô⇒ MD ⋅ DP = a2(b − c)2

By (1) and (2), we get

AP⋅ DM = a4(b + c)2(b − c)22 +(b + c)a2bc2 = a2

4 .The given condition can be written as

Also solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Telemachus Baltsavias, KerameiesJunior High School, Kefalonia, Greece; Theo Koupelis, Cape Coral, FL, USA

S604 Let x, y, z be real numbers such that −1 ≤ x, y ≤ 1 and x + y + z + xyz = 0 Prove that

√

x+ 1 +√y+ 1 +√z+ 1 ≤√9+ xyz

Proposed by Marius Stănean, Zalău, Romania

Solution by the author

Without loss of generality, we may assume that

z= min{x, y, z}

We have two cases:

Case 1: If x + y + z ≤ 0, then by the Cauchy-Schwarz Inequality, we have

√

x+ 1 +√y+ 1 +√z+ 1 ≤√3(1 + x + 1 + y + 1 + z)

=√9+ 3(x + y + z)

≤√9− x − y − z =√9+ xyz,with equality when x = y = z = 0

Case 2: If x + y + z > 0, then since xyz = −(x + y + z) < 0 it follows that z < 0 < x, y Setting t = x+ y

2 ≤ 1,

by the Cauchy-Schwarz Inequality, we have

√

x+ 1 +√y+ 1 ≤√2(x + y + 2) = 2√1+ t,and using the AM-GM Inequality, we have

2√(1 + t)(1 + t2) + 1 − t ≤√9+ 9t2− 2t3.Squaring both sides of the inequality, expanding and reducing terms, it becomes

2(1 − t)√(1 + t)(1 + t2) ≤ (1 − t)(3t2+ t + 2),or

t2(1 − t)(9t2+ 2t + 9)3t2+ t + 2 + 2√(1 + t)(1 + t2) ≥ 0,clearly true The equality holds when t = 1 which means x = y = 1 and z = −1

Also solved by Daniel Văcaru, Pites,ti, Romania; Paolo Perfetti, Università degli studi di Tor VergataRoma, Italy; Theo Koupelis, Cape Coral, FL, USA; Nicuşor Zlota, Traian Vuia Technical College, Focşani,Romania

S605 Let ABC be an acute triangle with circumcenter O and circumradius R Let Ra; Rb; Rc be thecircumradii of triangles OBC, OCA, OAB, respectively Prove that triangle ABC is equilateral if andonly if

R3+ R2(Ra+ Rb+ Rc) = 4RaRbRc

Proposed by Marian Ursărescu, Roman, România

Solution by Kousik Sett, India

If triangle ABC is equilateral then

cos A cos B+ cos B cos C + cos C cos A = 1 − 2 cos A cos B cos C (2)Using the following identity

cos2A+ cos2B+ cos2C= 1 − cos A cos B cos Cfor any triangle ABC, (2) can be written as

cos2A+ cos2B+ cos2C= cos A cos B + cos B cos C + cos C cos A,

which implies

∑(cos A − cos B)2= 0,which gives

cos A= cos B = cos C Ô⇒ A = B = C

as triangle ABC is acute and hence triangle ABC is equilateral

Also solved by Daniel Văcaru, Pites,ti, Romania; Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;Telemachus Baltsavias, Kerameies Junior High School, Kefalonia, Greece; Theo Koupelis, Cape Coral, FL,USA; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania

S606 Let a, b, c be positive real numbers Prove that

a2

b2 +b2

c2 +c2

a2 + abc ≥ a + b + c + 1

Proposed by An Zhenping, Xianyang Normal University, China

Solution by Marius Stănean, Zalău, România

Using Weighted AM-GM Inequality, we have

b43

⋅b

2 6

c26

⋅c

2 6

a26

=aband from AM-GM Inequality

3(c

a+c

b) +1

3abc≥ c,and all these 3 inequalities summed up lead us to the desired result

Also solved by Theo Koupelis, Cape Coral, FL, USA; Israel Castillo Pilco, Huaral, Peru; Paolo Perfetti,Università degli studi di Tor Vergata Roma, Italy; Prodromos Fotiadis, Nikiforos High School, Drama, Greece;Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania

Undergraduate problems

U601 Let a, b, c, d be real numbers such that all roots of the polynomial P (x) = x5− 10x4+ ax3+ bx2+ cx + dare greater than 1 Find the minimum possible value of a + b + c + d

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

First solution by the author

By the AM-GM Inequality we have

Second solution by Theo Koupelis, Cape Coral, FL, USA

Let xi, i= 1, , 5 be the real roots of the polynomial, with xi> 1 Using Vieta’s formulas we get

xixj+ ∑5

i,j,k=1 i≠j≠k≠i

xixjxk− ∑5

i,j,k,`=1 i≠j≠k≠i

i=1(xi− 1) ≤ 1, and thus a + b + c + d ≥ 8

Also solved by Daniel Pascuas, Barcelona, Spain; Ivan Hadinata, Jember, Indonesia

√

ax2n+ bdx.

Proposed by Ovidiu Furdui and Alina Sîntămărian, Cluj-Napoca, România

Solution by the authors

The limit equals to √1

a We have, since √ax2n+ b >√axn, that

1

n ∫

n 0

xn

√

x2n+b a

x

√

x2+b a

dx

= √1

a⋅1n

≥√ x

x2+ b a

, which holds for all x ≥ 1

It follows, based on (1) and (2) that

1

√

a⋅1n

R, Shivamogga, India; Theo Koupelis, Cape Coral, FL, USA; Yunyong Zhang, Chinaunicom; Nicuşor Zlota,Traian Vuia Technical College, Focşani, Romania

U603 Find all functions f ∶ R Ð→ R which satisfy the conditions:

(a) f(x + 1)f(y) − f(xy) = (2x + 1)f(y), for all x, y ∈ R,

(b) f(x) > 2x for all x > 2

Proposed by Mircea Becheanu, Canada

Solution by the author

It is easy to see that f = 0 and f(x) = x2 satisfy condition (a) The constant function f = 0 does not satisfycondition (b) and we will prove that only f(x) = x2 is a solution for our problem

For the pair (x, 0) we have f(x + 1)f(0) − f(0) = (2x + 1)f(0) If f(0) /= 0 we get f(x + 1) = 2(x + 1),for all x This means f(x) = 2x for all x But f(x) = 2x does not verify the equation Therefore we have

f(0) = 0

For the pair (0, 1) we have f(1)2− f(0) = f(1) ⇔ f(1)2= f(1) Hence, f(1) = 1 or f(1) = 0

Assume that f(1) = 0 For that pair (x, 1) we obtain f(x + 1)f(1) − f(x) = (2x + 1)f(1), giving thesolution f(x) = 0, for all x

So, we are in the case f(0) = 0 and f(1) = 1 Plugging the pair (−1, y) in the condition (a) one obtains

f(0)f(y) − f(−y) = −f(y), which shows that f(−y) = f(y) Hence, the function f is even and we will study

it for x > 0 For the pair (1, 1) we obtain f(2)f(1) − f(1) = 3f(1), then f(2) = 4 For the pair (2, 1) weobtain f(3)f(1) − f(2) = 5f(1), showing that f(3) = 9

We prove by induction that f(n) = n2 for all positive integers n Assume that f(n) = n2 and plug thepair (n, 1) We obtain f(n + 1)f(1) = f(n + 1) = f(n) + (2n + 1)f(1) = n2+ (2n + 1) = (n + 1)2

For the pair (n,1

f(n+ 1

n ) = (n+ 1

Plugging the pair (1

n, m) and using (1) and (2) one obtains

From condition (a) we obtain

Taking x > 1 we have xy > y and every positive real number z > y cand be written under the form z = xy forsome x > 1 If x > 1 we have x+1 > 2 and by condition (b), f(x+1) > 2(x+1) giving that f(x+1)−(2x+1) > 1.Going back in the equation (3) we obtain

showing that the function f is monotonic increasing on [0, +∞).

We will show now that that f(x) = x2 for all x > 0 Let (an)n≥1 and (bn)n≥1 be two sequences of positiverational numbers converging to x and such that

Also solved by Theo Koupelis, Cape Coral, FL, USA

U604 If

f(m) ∶= ∫01(xm− 1) ln(1 + x)x ln(x) dxthen evaluate,

lim

m→0(2π29m −8f(m)

2j+1 can factor out of the even terms:

3m2 ) = lim

m→0(2π29m − 83m2(mζ(2)

2 −3m2ζ(3)

8 + o(m2)))

= lim

m→0(2π29m −4ζ(2)

3m + ζ(3) − o(1))

= lim

m→0(2π29m −2π29m + ζ(3) − o(1))

= ζ(3)

Also solved by Daniel Pascuas, Barcelona, Spain; Seán M Stewart, Thuwal, Saudi Arabia; MatthewToo, Brockport, NY, USA; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Brian Bradie,Christopher Newport University, Newport News, VA, USA; G C Greubel, Newport News, VA, USA; TheoKoupelis, Cape Coral, FL, USA; Yunyong Zhang, Chinaunicom

U605 Let a, b, c be nonzero complex numbers having the same modulus for which

a3+ b3+ c3= rabc,where r is a real number

(i) Prove that −1 ≤ r ≤ 3

(ii) Prove that if r < 3 then one and only one of the equations ax2+ bx + c = 0, bx2 + cx + a = 0,

cx2+ ax + b = 0 has a root of modulus 1

Proposed by Florin Stănescu, Găeşti, România

Solution by Theo Koupelis, Cape Coral, FL, USA

(i) Let a = ρeiθ, b= ρeiφ,and c = ρeiω.Then the condition a3+b3+c3= rabc leads to ei3θ+ei3φ+ei3ω= rei(θ+φ+ω)

or r = ei(2θ−φ−ω)+ ei(2φ−θ−ω)+ ei(2ω−θ−φ).Setting α = 2θ − φ − ω, β = 2φ − θ − ω, and γ = 2ω − θ − φ we get

cos α+ cos β + cos γ = r,sin α+ sin β + sin γ = 0,

If α + β = 2kπ, where k ∈ Z, we get r = 2 cos(kπ) cos(kπ − β) + 1 = 1 + 2 cos β If α = 2kπ we get r =

1+ cos β + cos(2kπ + β) = 1 + 2 cos β Similarly, if β = 2kπ, we get r = 1 + 2 cos α In all three cases we have

−1 ≤ r ≤ 3 Also, r = 3 only when α, β, γ are even multiples of π

(ii) Without loss of generality, let the equation ax2+ bx + c = 0 have a root of modulo 1, that is, x = eiλ.Then

α and r would be equal to 3, which contradicts the condition r < 3 Thus, when r < 3, one and only one ofthe equations ax2+ bx + c = 0, bx2+ cx + a = 0, cx2+ ax + b = 0 has a root of modulus 1

Also solved by Theo Koupelis, Cape Coral, FL, USA