Junior problems - AwesomeMath...Nowsupposethat x y z: (3) Then,wehave 3x3 52975 ,x d 52975 3 e= 26...

Junior problems

J313. Solve in real numbers the system of equations

x(y + z − x3

)= y

(z + x− y3

)= z

(x+ y − z3

)= 1.

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

Solution by Ashley Colopy, College at Brockport, SUNYWe can write the equations as

y + z − x3 = 1x

z + x− y3 = 1y

x+ y − z3 = 1z

⇐⇒

y + z = 1

x + x3

z + x = 1y + y3

x+ y = 1z + z3

From the first equation we can see that y+ z has the same sign as x. Therefore there is another variablehaving the same sign as x. Repeating this for the other two equations we conclude that for each variablethere is another one having the same sign as it. Thus either all three are negative or all three are positive.

We will solve the system assuming that all three variables are positive.Using the AM - GM inequality we get

y + z = 1x + x3 ≥ 2

√x3 1

x = 2x

z + x = 1y + y3 ≥ 2

√y3 1

y = 2y

x+ y = 1z + z3 ≥ 2

√z3 1z = 2z

Adding all three inequalities we get

2x+ 2y + 2z =1

x+ x3 +

1

y+ y3 +

1

z+ z3 ≥ 2x+ 2y + 2y.

Therefore all inequalities must be equalities and we get

1

x+ x3 = 2x

and similar equations for y and z.The equation is equivalent to

x4 − 2x2 + 1 = 0 ⇐⇒ (x2 − 1)2 = 0 ⇐⇒ x = 1.

Thus the solution of the system is x = y = z = 1.The negative case produces x = y = z = −1.

Also solved by Daniel Lasaosa, Pamplona, Spain; Peter C. Shim, The Pingry School, Basking Ridge, NJ;Seonmin Chung, Stuyvesant High School, NY; Adnan Ali, student at A.E.C.S-4, Mumbai, India; Albert Sta-dler, Herrliberg, Switzerland; Alok Kumar, Delhi, India; Arber Avdullahu,Mehmet Akif College,Kosovo; Ar-kady Alt, San Jose, California, USA; Francesc Gispert Sánchez, CFIS, Universitat Politécnica de Catalunya,Barcelona, Spain; Joshua Noah Benabou, Manhasset High School, NY; Farrukh Mukhammadiev, AcademicLyceum Nr.1 under the SamIES, Samarkand, Uzbekistan; Evgenidis Nikolaos, M. N. Raptou High School,Larissa, Greece; Prithwijit De, HBCSE, Mumbai, India; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Ta-shkent, Uzbekistan; Shatlyk Mamedov, School Nr. 21, Dashoguz, Turkmenistan; Titu Zvonaru, Comănes, ti,Romania and Neculai Stanciu, Buzău, Romania; Jhiseung Daniel Hahn, Phillips Exeter Academy, Exeter,NH, USA; William Kang, Bergen County Academies, Hackensack, NJ, USA; Jongyeob Lee, StuyvesantHigh School, NY, USA; Michael Tang, Edina High School, MN, USA; Jaesung Son, Ridgewood, NJ, USA;Polyahedra, Polk State College, FL, USA; Kwon Il Ko, Cushing Academy, MA, USA.

Mathematical Reflections 5 (2014) 1

J314. Alice was dreaming. In her dream, she thought that primes of the form 3k + 1 are weird. Then shethought it would be interesting to find a sequence of consecutive integers all of which are greater than1 and which are not divisible by weird primes. She quickly found five consecutive numbers with thisproperty:

8 = 23, 9 = 32, 10 = 2 · 5, 11 = 11, 12 = 22 · 3.

What is the length of the longest sequence she can find?

Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA

Solution by Adnan Ali, Student at A.E.C.S-4, Mumbai, IndiaFirst of all we note that we can have a sequence of maximum of 6 integers, as, in the sequence of any 7integers, there is obviously one which is divisible by the prime 7 = 3(2) + 1. Therefore, we show that therequired number is 6. It suffices to see that

953 = 1 · 953, 954 = 2 · 32 · 53, 955 = 5 · 191, 956 = 22 · 239, 957 = 3 · 11 · 29, 958 = 2 · 479.

Also solved by Daniel Lasaosa, Pamplona, Spain; Peter C. Shim, The Pingry School, Basking Ridge,NJ; Seonmin Chung, Stuyvesant High School, NY; Jhiseung Daniel Hahn, Phillips Exeter Academy, Exeter,NH, USA; William Kang, Bergen County Academies, Hackensack, NJ, USA; Rebecca Buranich, College atBrockport, SUNY; Jongyeob Lee, Stuyvesant High School, NY, USA; Michael Tang, Edina High School, MN,USA; Jaesung Son, Ridgewood, NJ, USA; Polyahedra, Polk State College, FL, USA; Kwon Il Ko, CushingAcademy, MA, USA.


J315. Let a, b, c be non-negative real numbers such that a+ b+ c = 1. Prove that√

4a+ 1 +√

4b+ 1 +√

4c+ 1 ≥√

5 + 2.

Proposed by Cosmin Pohoata, Columbia University, USA

Solution by Daniel Lasaosa, Pamplona, SpainDenote x =

√4a+ 1− 1, y =

√4b+ 1− 1, z =

√4c+ 1− 1, and s = x+ y + z. Note therefore that

s2 ≥ x2 + y2 + z2 = 4(a+ b+ c) + 3− 2(x+ 1 + y + 1 + z + 1) + 3 = 4− 2s,

or0 ≤

(s+ 1 +

√5)(

s+ 1−√

5),

for s ≥√

5− 1, with equality iff xy + yz + zx = 0, ie iff two of x, y, z are zero, or iff two of a, b, c are zero.It follows that √

4a+ 1 +√

4b+ 1 +√

4c+ 1 = s+ 3 ≥√

5 + 2,

with equality iff (a, b, c) is a permutation of (1, 0, 0).

Also solved by Ángel Plaza, Department of Mathematics, University of Las Palmas de Gran Canaria,Spain; Peter C. Shim, The Pingry School, Basking Ridge, NJ; Seonmin Chung, Stuyvesant High School, NY;Jhiseung Daniel Hahn, Phillips Exeter Academy, Exeter, NH, USA; William Kang, Bergen County Acade-mies, Hackensack, NJ, USA; Albert Stadler, Herrliberg, Switzerland; Arkady Alt, San Jose, California, USA;Debojyoti Biswas, Kolkata, India; Erdenebayar Bayarmagnai, School Nr.11, Ulaanbaatar, Mongolia; HenryRicardo, New York Math Circle; Marius Stanean, Zalau, Romania; Farrukh Mukhammadiev, Academic Ly-ceum Nr.1 under the SamIES, Samarkand, Uzbekistan; Nicusor Zlota, “Traian Vuia” Technical College,Focsani, Romania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Salem Malikic,Simon Fraser University, Burnaby, BC, Canada; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent,Uzbekistan; Shatlyk Mamedov, School Nr. 21, Dashoguz, Turkmenistan; Titu Zvonaru, Comănes, ti, Romaniaand Neculai Stanciu, Buzău, Romania; Utsab Sarkar, Chennai Mathematical Institute, India; Vincent Hu-ang, Schimelpfenig Middle School, Plano, TX; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia;George Gavrilopoulos, High School of Nea Makri, Athens, Greece; An Zhen-ping, Mathematics Department,Xianyang Normal University, Xianyang, Shaanxi, China; Jongyeob Lee, Stuyvesant High School, NY, USA;Michael Tang, Edina High School, MN, USA; Jaesung Son, Ridgewood, NJ, USA; Polyahedra, Polk StateCollege, FL, USA; Kwon Il Ko, Cushing Academy, MA, USA.


J316. Solve in prime numbers the equation

x3 + y3 + z3 + u3 + v3 + w3 = 53353.


Solution by Evgenidis Nikolaos, M.N. Raptou High School, Larissa, GreeceLemma 1: Every cube of an integer is conguent with 1, 0,−1 (mod 9).Lemma 2: Every cube of an integer is congruent with 1, 0,−1 (mod 7).

Both of these lemmas can be easily proved by examining all forms of an integer (mod 9) and (mod 7),correspondingly.

First, we observe that if all of x, y, z, u, w, v are odd primes, then their cubes’ summary would be aneven number, contradiction. So at least one of them will be equal to 2.

• if five of x, y, z, u, v, w are equal to 2 then the initial equation has no solutions.

• if three of x, y, z, u, v, w are equal to 2, let us say w = v = u = 2, then we have

x3 + y3 + z3 = 53329,

which by Lemma 1 does not hold.

Suppose that only w = 2. Then it will be

x3 + y3 + z3 + u3 + v3 = 53345 (1)

By Lemma 1, we have that if any of the remaining x, y, z, u, v is not equal to 3, (1) does not hold, since53345 ≡ 2 (mod 9), while LHS cannot be conguent with it.

Hence, let v = 3.

• if it is also u = 3, we obtain the equation

x3 + y3 + z3 = 53291.

WLOG, suppose that x ≥ y ≥ z. This way, we get to the conclusion that x = 29, x = 31 and x = 37,for none of which we have a solution.

Back to the case only v = 3. This gives

x3 + y3 + z3 + u3 = 53318. (2)

By Lemma 2, we can similarly conclude that if any of x, y, z, u is not equivalent to 7, we get to acontradiction since 53318 ≡ 6 (mod 7) and LHS of (2) cannot be congruent with it.

• if any of x, y, z equals 7 (apart from u that is taken), for instance z = 7⇔ z3 = 343, then we have

y3 + z3 = 52632.

Now supposing x ≥ y, we can find that either x = 31 or x = 37. In each case there is no solution inprime integers.

So, let u = 7 and we take thatx3 + y3 + z3 = 52975.


Now suppose thatx ≥ y ≥ z. (3)

Then, we have

3x3 ≥ 52975⇔ x ≥ d52975

3e = 26

andx ≤ 38

since if x ≥ 38⇔ x3 ≥ 54872, that is impossible.We have taken that 26 ≤ x ≤ 38. Consequently, x = 29, x = 31 and x = 37. By (3), we can drop out

cases x = 29 and x = 31. For x = 37, we have

y3 + z3 = 2322.

Relation (3) implies that y3 ≥ 1311 and because y is an odd prime and 1311 ≤ y3 ≤ 2322, it must bey = 13 or y = 11 . For y = 11 the given equation has no solution, while for y = 13 we get z = 5. Hence, thegiven equation has a solution (x, y, z, u, v, w) = (37, 13, 5, 7, 3, 2) and all its permutations in prime integers.

Also solved by Daniel Lasaosa, Pamplona, Spain; Théo Lenoir, Institut Saint-Lô, Agneaux, France; PeterC. Shim, The Pingry School, Basking Ridge, NJ; Seonmin Chung, Stuyvesant High School, NY; WilliamKang, Bergen County Academies, Hackensack, NJ, USA; Albert Stadler, Herrliberg, Switzerland; AlessandroVentullo, Milan, Italy; Evgenidis Nikolaos, M. N. Raptou High School, Larissa, Greece; Samantha Paradis,College at Brockport, SUNY; Titu Zvonaru, Comănes, ti, Romania and Neculai Stanciu, Buzău, Romania;Jhiseung Daniel Hahn, Phillips Exeter Academy, Exeter, NH, USA; Michael Tang, Edina High School, MN,USA; Jaesung Son, Ridgewood, NJ, USA; Polyahedra, Polk State College, FL, USA; Kwon Il Ko, CushingAcademy, MA, USA.


J317. In triangle ABC, the angle-bisector of angle A intersects line BC atD and the circumference of triangleABC at E. The external angle-bisector of angle A intersects line BC at F and the circumference oftriangle ABC at G. Prove that DG ⊥ EF .


Solution by Adnan Ali, Student at A.E.C.S-4, Mumbai, IndiaSince ∠BAC + ∠CAY = 180◦,∠DAC + ∠CAF = 90◦ = ∠GAE. Hence, GE is the diameter of thecircumcircle of 4ABC. Consider 4FGE and let BC ∩ GE = Z. Then, since ∠BAE = ∠EAC, E is themidpoint of BC. Similarly G is the midpoint of BAC. Hence GE ⊥ BC ⇒ ∠CZE = 90◦. Thus FZ ⊥ GE,or that FZ is an altitude of 4FGE through F . Similarly, we know that ∠EAC + ∠CAF = 90◦. So,EA ⊥ FG. And so EA is the altitude of 4FGE through E. One can see that AE ∩ FZ = D. Hence toprove that DG ⊥ EF , it suffices to show that GX is an altitude of 4FGE which passes through D whereX = GD ∩EF . Since the altitudes of a triangle concur at a point, we conclude that the altitude through Galso passes through D. And that forces DG ⊥ EF and thus the result.

Also solved by Daniel Lasaosa, Pamplona, Spain; Ricardo Barroso Campos, Sevilla, Spain; Peter C. Shim,The Pingry School, Basking Ridge, NJ; Seonmin Chung, Stuyvesant High School, NY; William Kang, BergenCounty Academies, Hackensack, NJ, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia;George Gavrilopoulos, High School of Nea Makri, Athens, Greece; Andrea Fanchini, Cantú, Italy; GeorgiosBatzolis, Mandoulides High School, Thessaloniki, Greece; Yassine Hamdi, Lycée du Parc, Lyon, France;Farrukh Mukhammadiev, Academic Lyceum Nr.1 under the SamIES, Samarkand, Uzbekistan; PrithwijitDe, HBCSE, Mumbai, India; Jongyeob Lee, Stuyvesant High School, NY, USA; Jaesung Son, Ridgewood,NJ, USA; Titu Zvonaru, Comănes, ti, Romania and Neculai Stanciu, Buzău, Romania; Evgenidis Nikolaos,M. N. Raptou High School, Larissa, Greece; Polyahedra, Polk State College, FL, USA; Kwon Il Ko, CushingAcademy, MA, USA.


J318. Determine the functions f : R→ R satisfying f(x− y)− xf(y) ≤ 1− x for all real numbers x and y.

Proposed by Marcel Chirita, Bucharest, Romania

Solution by Polyahedra, Polk State College, USAClearly, f(x) ≡ 1 is a solution. We show that it is the only one.First, f (0− (−x))− 0f(−x) ≤ 1− 0, so f(x) ≤ 1 for all x. Next, f (2x− x)− 2xf(x) ≤ 1− 2x, that is,

(1− 2x) [f(x)− 1] ≤ 0.

So f(x) ≥ 1 for all x > 12 . Thus f(x) = 1 for all x > 1

2 .Finally, if x ≤ 1

2 then 2− x > 1, so f(2− x) = 1. Since f(2− x)− 2f(x) ≤ 1− 2, f(x) ≥ 1.Hence f(x) = 1 for all x ≤ 1

2 as well.

Also solved by Solution by Alessandro Ventullo, Milan, Italy; Daniel Lasaosa, Pamplona, Spain; PeterC. Shim, The Pingry School, Basking Ridge, NJ; Seonmin Chung, Stuyvesant High School, NY; WilliamKang, Bergen County Academies, Hackensack, NJ, USA; AN-anduud Problem Solving Group, Ulaanbaatar,Mongolia; Jhiseung Daniel Hahn, Phillips Exeter Academy, Exeter, NH, USA; Arber Avdullahu, MehmetAkif College,Kosovo; Arber Igrishta, Eqrem Qabej, Vushtrri, Kosovo; Arkady Alt, San Jose, California,USA; Paul Revenant, Lycée Champollion, Grenoble, France; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov,Tashkent, Uzbekistan; Titu Zvonaru, Comănes, ti, Romania and Neculai Stanciu, Buzău, Romania; JongyeobLee, Stuyvesant High School, NY, USA; Michael Tang, Edina High School, MN, USA; Jaesung Son, Rid-gewood, NJ, USA; Kwon Il Ko, Cushing Academy, MA, USA.


Senior problems

S313. Let a, b, c be nonnegative real numbers such that√a+√b+√c = 3. Prove that√

(a+ b+ 1)(c+ 2) +√

(b+ c+ 1)(a+ 2) +√

(c+ a+ 1)(b+ 2) ≥ 9.


First solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia√a+√b+√c = 3⇔ 3−

√c =√a+√b

⇔ 9− 6√c+ c = a+ b+ 2

√ab

⇒ 9− 6√c+ c ≤ a+ b+ (a+ b)

⇔ 9

2− 3√c+

c

2≤ a+ b

⇔ 11

2− 3√c+

c

2≤ a+ b+ 1.

Hence

(a+ b+ 1)(c+ 2) ≥(

11

2− 3√c+

c

2

)(c+ 2)

= 9 +(√c− 1

)2(√c− 2)2 ≥ 9

⇒√

(a+ b+ 1)(c+ 2) ≥ 3

Similarly, √(b+ c+ 1)(a+ 2) ≥ 3√(c+ a+ 1)(b+ 2) ≥ 3.

Adding all above inequality the desired inequality is proved. The equality holds only when a = b = c = 1 .

Second solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia∑cyc

√(a+ b+ 1)(c+ 2) =

∑cyc

√(a+ b+ 1)(1 + 1 + c)

Applying Cauchy-Schwartz’s inequality,

≥∑cyc

(√a · 1 +

√b · 1 + 1 ·

√c) = 9.

Hence the inequality is proved.The equality holds only when a = b = c = 1.

Also solved by Li Zhou, Polk State College, USA; Daniel Lasaosa, Pamplona, Spain; William Kang, Ber-gen County Academies, Hackensack, NJ, USA; George Gavrilopoulos, High School of Nea Makri, Athens,Greece; An Zhen-ping, Mathematics Department, Xianyang Normal University, Xianyang, Shaanxi, China;Jongyeob Lee, Stuyvesant High School, NY, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; ArberIgrishta, Eqrem Qabej, Vushtrri, Kosovo; Arkady Alt, San Jose, California, USA; Bodhisattwa Bhowmik,RKMV, Agartala, Tripura, India; Erdenebayar Bayarmagnai, School Nr.11, Ulaanbaatar, Mongolia; Far-rukh Mukhammadiev, Academic Lyceum Nr.1 under the SamIES, Samarkand, Uzbekistan; Nicusor Zlota,“Traian Vuia” Technical College, Focsani, Romania; Evgenidis Nikolaos, M. N. Raptou High School, La-rissa, Greece; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, Italy; Sardor Bozorboyev,Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan; Shatlyk Mamedov, School Nr. 21, Dashoguz, Turkmenis-tan; Titu Zvonaru, Comănes, ti, Romania and Neculai Stanciu, Buzău, Romania; Utsab Sarkar, ChennaiMathematical Institute, India; Ioan Viorel Codreanu, Satulung, Maramures, Romania.


S314. Let p, q, x, y, z be real numbers satisfying

x2y + y2z + z2x = p and xy2 + yz2 + zx2 = q.

Evaluate(x3 − y3

) (y3 − z3

) (z3 − x3

)in terms of p and q.

Proposed by Marcel Chirita, Bucharest, Romania

Solution by Prithwijit De, HBCSE, Mumbai, IndiaLet ω be a non-real cube root of unity. We can write (x3 − y3)(y3 − z3)(z3 − x3) = UVW where

U = (x− y)(y − z)(z − x) = q − p (1)

V = (x− yω)(y − zω)(z − xω) = qω2 − pω (2)

andW = (x− yω2)(y − zω2)(z − xω2) = qω − pω2. (3)

Thus UVW = (q − p)(qω2 − pω)(qω − pω2) = q3 − p3. Therefore

(x3 − y3)(y3 − z3)(z3 − x3) = q3 − p3.

Also solved by Li Zhou, Polk State College, USA; Daniel Lasaosa, Pamplona, Spain; William Kang,Bergen County Academies, Hackensack, NJ, USA; Adnan Ali, student at A.E.C.S-4, Mumbai, India; Al-bert Stadler, Herrliberg, Switzerland; Alok Kumar, Delhi, India; Arber Avdullahu, Mehmet Akif College,Kosovo; Arkady Alt, San Jose, California, USA; Bunyod Boltayev, Khorezm, Uzbekistan; Debojyoti Biswas,Kolkata, India; Erdenebayar Bayarmagnai, School Nr.11, Ulaanbaatar, Mongolia; Farrukh Mukhammadiev,Academic Lyceum Nr.1 under the SamIES, Samarkand, Uzbekistan; Navid Safei, University of Technoogyin Policy Making of Science and Technology, Iran; Nicusor Zlota, “Traian Vuia” Technical College, Focsani,Romania; Evgenidis Nikolaos, M. N. Raptou High School, Larissa, Greece; Sardor Bozorboyev, LyceumS.H.Sirojjidinov, Tashkent, Uzbekistan; Shatlyk Mamedov, School Nr. 21, Dashoguz, Turkmenistan; TituZvonaru, Comănes, ti, Romania and Neculai Stanciu, Buzău, Romania.


S315. Consider triangle ABC with inradius r. Let M and M ′ be two points inside the triangle such that∠MAB = ∠M ′AC and ∠MBA = ∠M ′BC. Denote by da, db, dc and d′a, d

′b, d′c the distances from M

and M ′ to the sides BC,CA,AB, respectively. Prove that

dadbdcd′ad′bd′c ≤ r6.

Proposed by Nairi Sedrakyan, Yerevan, Armenia

Solution by Khakimboy Egamberganov, National University of Uzbekistan, Tashkent, UzbekistanWe have that the points M and M ′ are isogonal points in the triangle ABC with da · d′a = db · d′b = dc · d′c.Let

k = da · d′a = db · d′b = dc · d′cand

Sa = [BMC], Sb = [AMC], Sc = [AMB],

S′a = [BM ′C], S′b = [AM ′C], S′c = [AM ′B].

So we get that

k =4SaS

′a

a2=

4SbS′b

b2=

4ScS′c

c2=(2√SaS′a + 2

√SbS

′b + 2

√ScS′c

a+ b+ c

)2and by Cauchy-Schwartz inequality,

k ≤(2Sa + 2Sb + 2Sc)(2S

′a + 2S′b + 2S′c)

(a+ b+ c)2= r2.

Hence dadbdcd′ad′bd′c = k3 ≤ r6 and the equality holds if and only if M ≡M ′ ≡ I, where I is incenter of the

triangle ABC.

Also solved by Li Zhou, Polk State College, USA; Daniel Lasaosa, Pamplona, Spain; Titu Zvonaru,Comănes, ti, Romania and Neculai Stanciu, Buzău, Romania; Bodhisattwa Bhowmik, RKMV, Agartala, Tri-pura, India; Marius Stanean, Zalau, Romania; Farrukh Mukhammadiev, Academic Lyceum Nr.1 under theSamIES, Samarkand, Uzbekistan; Prithwijit De, HBCSE, Mumbai, India.


S316. Circles C1(O1, R1) and C2(O2, R2) intersect in points U and V . Points A1, A2, A3 lie on C1 and pointsB1, B2, B3 lie on C2 such that A1B1, A2B2, A3B3 are passing through U . Denote by M1,M2,M3 themidpoints of A1B1, A2B2, A3B3. Prove that M1M2M3V is a cyclic quadrilateral.


First solution by Li Zhou, Polk State College, USA

CC2

C1

YZ

X

O

V

M1

B1O1

U

O2

A1

Let O,X, Y, Z be the midpoints of O1O2, A1U,UB1, UM1, respectively. Then

X + Y

2=

1

2

(A1 + U

2+U +B1

2

)=

1

2

(A1 +B1

2+ U

)=M1 + U

2= Z.

Hence OZ ‖ O1X, and thus OZ ⊥ UM1. Therefore, M1 is on the circle C centered at O and of radiusOU = OV . Likewise M2 and M3 are on C as well.


Second solution by Daniel Lasaosa, Pamplona, SpainLemma: Two secant circles (O1, R1) and (O2, R2) intersect at U, V . Let A,B be the respective points where aline through either U or V meets again both circles. Then, the midpointM of AB satisfies OM = OU = OV ,where O is the midpoint of O1O2.

Proof: Consider a cartesian coordinate system with center at O, such that O1 ≡ (−d, 0), O2 ≡ (d, 0),where 2d < R1 +R2 because the circles are secant. The equations of the circles are then

(x+ d)2 + y2 = R12, (x− d)2 + y2 = R2

2.

Let U ≡ (∆, h) and V ≡ (∆,−h), where (∆ + d)2 + h2 = R12 and (∆ − d)2 + h2 = R2

2. Any line throughU has equation y = h+m(x−∆), or its intersections with (O1, R1) satisfy(

m2 + 1)x2 + 2dx− 2m2∆x+ 2mhx+

(m2 − 1

)∆2 − 2d∆− 2mh∆ = 0.

Since one of the roots, corresponding to U , is ∆, the other root equals the independent term divided by(m2 + 1)∆, and substitution in the equation for the line through U yields

A ≡

((m2 − 1

)∆− 2d− 2mh

m2 + 1,−(m2 − 1

)h+ 2m∆ + 2md

m2 + 1

).

Similarly,

B ≡

((m2 − 1

)∆ + 2d− 2mh

m2 + 1,−(m2 − 1

)h+ 2m∆− 2md

m2 + 1

),

yielding

M ≡

((m2 − 1

)∆− 2mh

m2 + 1,−(m2 − 1

)h+ 2m∆

m2 + 1

).

It follows that

OM2 =

((m2 − 1

)2+ 4m2

) (∆2 + h2

)(m2 + 1)2

= ∆2 + h2 = OU2 = OV 2.

The Lemma follows, also when the line passes through V because of the symmetry with respect to lineO1O2.

From the Lemma, it follows that the circle through U, V with center in the midpoint O of O1O2 passesalso through M1,M2,M3. The conclusion follows.

Also solved by Adnan Ali, A.E.C.S-4, Mumbai, India; Marius Stanean, Zalau, Romania; Prithwijit De,HBCSE, Mumbai, India; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan.


S317. Let ABC be an acute triangle inscribed in a circle of radius 1. Prove that

tanA

tan3B+

tanB

tan3C+

tanC

tan3A≥ 4

(1

a2+

1

b2+

1

c2

)− 3.


Solution by Arkady Alt, San Jose, California, USASince a = 2 sinA, b = 2 sinB, c = 2 sinC and

(tanA− tanB)

(1

tan3A− 1

tan3B

)= −

(tanA− tanB)2(tan2A+ tanA tanB + tan2B

)tan3A tan3B

≤ 0

then by Rearrangement Inequality∑cyc tanA· 1

tan3B≥∑

cyc tanA · 1

tan3A=∑

cyc

1

tan2A=∑

cyc cot2A =∑

cyc

(1

sin2A− 1

)=∑

cyc

1

sin2A− 3 = 4

(1

a2+

1

b2+

1

c2

)− 3.

Also solved by Daniel Lasaosa, Pamplona, Spain; AN-anduud Problem Solving Group, Ulaanbaatar,Mongolia; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Bodhisattwa Bhowmik, RKMV, Agartala,Tripura, India; Marius Stanean, Zalau, Romania; Farrukh Mukhammadiev, Academic Lyceum Nr.1 underthe SamIES, Samarkand, Uzbekistan; Prithwijit De, HBCSE, Mumbai, India; Sardor Bozorboyev, LyceumS.H.Sirojjidinov, Tashkent, Uzbekistan; Titu Zvonaru, Comănes, ti, Romania and Neculai Stanciu, Buzău,Romania; Li Zhou, Polk State College, USA.


S318. Points A1, B1, C1, D1, E1, F1 are lying on the sides of AB,BC,CD,DE,EF, FA of a convex hexagonABCDEF such that

AA1

AB=AF1

AF=CC1

CD=CB1

BC=ED1

ED=EE1

EF= λ.

Prove that A1D1, B1E1, C1F1 are concurrent if and only if [ACE][BDF ] =

(λ

1−λ

)2.


Solution by Khakimboy Egamberganov, National University of Uzbekistan, Tashkent, UzbekistanWe have that

(1− λ)2[ACE]− λ2[BDF ] = (1− λ)2([OAE] + [OEC] + [OCA])− λ2([OBD] + [ODF ] + [OFB]) (1).

Suppose that A1D1, B1E1, C1F1 are concurrent at one point O. Then, we get that

0 = [OA1D1] = (1−λ)[OA1E]−λ[OA1D] = (1−λ)(

(1−λ)[OAE]−λ[OBE])−λ(λ[OBD]−(1−λ)[OAD]

)=

= (1− λ)2[OAE]− λ2[OBD] + λ(1− λ)(

[OAD]− [OBE]),

so(1− λ)2[OAE]− λ2[OBD] = λ(1− λ)

([OBE]− [OAD]

).

Analoguosly,(1− λ)2[OEC]− λ2[OFB] = λ(1− λ)

([OFC]− [OBE]

)(1− λ)2[OCA]− λ2[ODF ] = λ(1− λ)

([OAD]− [OFC]

).

Thus, since (1) we have that [ACE][BDF ] =

(λ

1−λ

)2.

Suppose that [ACE][BDF ] =

(λ

1−λ

)2and we will prove that A1D1, B1E1, C1F1 are concurrent. Since (1) we

can find easily that there exists a point O such that

0 = (1− λ)2[ACE]− λ2[BDF ] = [OA1D1] + [OB1E1] + [OC1F1].

So[OA1D1] = [OB1E1] = [OC1F1]

and O be the intersection point of the lines A1D1, B1E1, C1F1.

Also solved by Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan; Li Zhou, Polk StateCollege, USA.


Undergraduate problems

U313. Let X and Y be nonnegative definite Hermitian matrices such that X−Y is also nonnegative definite.Prove that tr

(X2)≥ tr

(Y 2).

Proposed by Radouan Boukharfane, Sidislimane, Morocco

Solution by Henry Ricardo, New York Math CircleFirst we show that tr(XY ) ≥ 0 : For nonnegative definite Hermitian matrices X and Y , there exist ma-trices A and B such that X = AA∗ and Y = BB∗, giving us tr(XY ) = tr(AA∗BB∗) = tr(A∗BB∗A) =tr ((A∗B)(A∗B)∗) ≥ 0.

Now we writetr(X2)− tr(Y 2) = tr ((X − Y )X) + tr (Y (X − Y )) ≥ 0,

which follows from the result proved above since X − Y is nonnegative definite.

Also solved by Daniel Lasaosa, Pamplona, Spain; AN-anduud Problem Solving Group, Ulaanbaatar,Mongolia; Radouan Boukharfane, Sidislimane, Morocco; Alessandro Ventullo, Milan, Italy; Li Zhou, PolkState College, USA.


U314. Prove that for any positive integer k,

limn→∞

(1 + n√

2 + · · ·+ n√k

k

)n>k

e,

where e is Euler constant.


Solution by Daniel Lasaosa, Pamplona, SpainLet f(x) = n

√x. This function is clearly increasing (with first derivative f(x) = x

nn√x) and concave (with

second derivative f ′′(x) = − n−1n2x2

n√x). It follows that, for x ∈ [k − 1, k], we have

f(x) ≤ n√k − n− 1

n2k2n√k(k − x),

with equality iff x = k, or

nk n√k

n+ 1=

∫ k

0f(x)dx =

k∑j=1

∫ j

j−1f(x)dx <

k∑j=1

n√j −

n∑j=1

n− 1

n2j2n√j

∫ j

j−1(j − x)dx =

=k∑j=1

n√j −

k∑j=1

n− 1

2n2j2n√j,

or equivalently, (1 + n√

2 + · · ·+ n√k

k

)n>

n n√k

n+ 1+n− 1

2n2k

k∑j=1

n√j

j2

n

>

> k

(n

n+ 1

)n+n− 1

n+ 1

n√k

2k

k∑j=1

n√j

j2> k

(n

n+ 1

)n+n− 1

n+ 1

n√k

2k

k∑j=1

1

j2.

The limit of the second term when n → ∞ is clearly positive, since the factors outside the sum have limit12k , and the sum is positive and does not depend on n. It follows that

limn→∞

(1 + n√

2 + · · ·+ n√k

k

)n> k lim

n→∞

(n

n+ 1

)n= k lim

n→∞

1(1 + 1

n

)n =k

e.

The conclusion follows.

Also solved by Daniel Lasaosa, Pamplona, Spain; Henry Ricardo, New York Math Circle; Arkady Alt ,San Jose,California, USA; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Alessandro Ventullo, Mi-lan, Italy; Jaesung Son, Ridgewood, NJ, USA; Ángel Plaza, Department of Mathematics, University of LasPalmas de Gran Canaria, Spain; Albert Stadler, Herrliberg, Switzerland; Utsab Sarkar, Chennai Mathemati-cal Institute, India; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Corneliu Mănescu-Avram,Transportation High School, Ploiesti, Romania; Khakimboy Egamberganov, National University of Uzbekis-tan, Tashkent, Uzbekistan; William Kang, Bergen County Academies, Hackensack, NJ, USA; Nicusor Zlota,“Traian Vuia” Technical College, Focsani, Romania; Paolo Perfetti, Università degli studi di Tor VergataRoma, Roma, Italy; Li Zhou, Polk State College, USA.


U315. Let X and Y be complex matrices of the same order with XY 2−Y 2X = Y . Prove that Y is nilpotent.

Proposed by Radouan Boukharfane, Sidislimane, Morocco

Solution by Li Zhou, Polk State College, USALet J be the Jordan canonical form of Yn×n. Then there is P such that PY P−1 = J . Let Z = PXP−1, thenthe given condition becomes ZJ2− J2Z = J . Let J1, . . . , Jk be the Jordan blocks of J corresponding to theeigenvalues λ1, . . . , λk, not necessarily distinct. Then

J =

J1 0 · · · 0

0 J2 · · · 0...

.... . .

...0 0 · · · Jk

, Z =

Z1 ∗ · · · ∗∗ Z2 · · · ∗...

.... . .

...∗ ∗ · · · Zk

,where Zi and Ji have the same order for each i. By block multiplication, the given condition implies ZiJ2

i −J2i Zi = Ji for each i. Now let ni be the order of Ji, then

niλi = tr(Ji) = tr(ZiJ2i )− tr(J2

i Zi) = tr(J2i Zi)− tr(J2

i Zi) = 0,

so λi = 0 for 1 ≤ i ≤ k. Hence Jnii = 0, thus Jm = 0, where m = max{n1, . . . , nk}. Therefore, Y m =

P−1JmP = 0 as well.

Also solved by Khakimboy Egamberganov, National University of Uzbekistan, Tashkent, Uzbekistan; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Alessandro Ventullo, Milan, Italy.


U316. The sequence {Fn} is defined by F1 = F2 = 1, Fn+2 = Fn+1 + Fn for n ≥ 1. For any natural numberm define v2(m) as v2(m) = n if 2n|m and 2n+1|m. Prove that there is exactly one positive number µsuch that the equation

v2([µn]!) = v2(F1F2 · · ·Fn)

is satisfied by the infinitely many positive integers n. determine that number µ.

Proposed by Albert Stadler, Herrliberg, Switzerland

First solution by by the proposerWe claim that the equation

v2([µn]!) = v2(F1F2 · · ·Fn) (1)

has infinitely many solutions in n if and only if µ = 56 . We start with the following:

Lemma 1:Let p be a prime. Let n be a natural number whose base p representation equals n =

∑kj=0 njp

j , where0 ≤ nj ≤ p − 1, 0 ≤ j ≤ k. For any natural number m define vp(m) = n if pn|m and pn+1 6 |m. Then

vp(n!) =∑

j≥1

[n

pj

]=n− n0 − n1 − n2 − · · · − nk

p− 1.

Proof:Among the integers 1, 2, . . . , n there are exactly

[np

]which are divisible by p, namely

p, 2p, ...,

[n

p

]p. (2)

The integers between 1 and n which are divisible by p2 (a subset of (2)) are

p2, 2p2, . . . ,

[n

p2

]p2,

which are[np2

]in number, and so on. The number of integers between 1 and n, which are divisible by

pj but not by pj+1 is exactly[npj

]−[

npj+1

]. Hence p divides n! exactly

∑j≥1

j

([n

pj

]−[n

pj+1

])=∑j≥1

[n

pj

]times. We have ∑

j≥1

[n

pj

]=∑j≥1

n

pj−∑j≥1

{n

pj

}

=n

p− 1−∑j≥1

{n0 + n1p+ n2p

2 + · · ·+ nkpk

pj

}=

=n

p− 1−

k∑j=1

n0 + n1p+ n2p2 + · · ·+ nj−1p

j−1

pj−∑j≥k+1

n0 + n1p+ n2p2 + · · ·+ nkp

k

pj=

n

p− 1−∑k

j=1

1

pj∑j

i=1 ni−1pi−1− 1

pk(p− 1)

∑ki=0 nip

i =n

p− 1−∑k

i=1 ni−1∑k

j=i pi−j−1− 1

pk(p− 1)

∑ki=0 nip

i =

n

p− 1− 1

p− 1

k∑i=1

ni−1(1− pi−1−k)−1

pk(p− 1)

k∑i=0

nipi =


=n− n0 − n1 − n2 − · · · − nk

p− 1.

This concludes the proof of Lemma 1.Lemma 2:

v2(Fn) =

0, if n ≡ 1, 2 mod 3

1, if n ≡ 3 mod 6

3, if n ≡ 6 mod 12

v2(n) + 2, if n ≡ 0 mod 12

Proof:See Lemma 2 of Lengyel’s paper"The order of the Fibonacci and Lucas numbers"

Let [µn] =∑

i ei2i and

[ n12

]=∑

i fi2i be the binary representations of [µn] and

[n12

]respectively. Then,

by Lemma 1 and Lemma 2:

v2([µn]!) =∑k≥1

[[µn]

2k

]=∑i

ei(2i − 1) = [µn]−

∑i

ei,

v2(F1F2 · · ·Fn) =n∑k=1

v2(Fk) =∑

0<6k+3≤n1 +

∑0<12k+6≤n

3 +∑

0<12k≤n2 +

∑0<12k≤n

v2(12k) =

=∑

0<6k+3≤n1 +

∑0<12k+6≤n

3 +∑

0<12k≤n4 +

∑0<k≤ n

12

v2(k) =

[n+ 3

6

]+ 3

[n+ 6

12

]+ 4

[ n12

]+ v2

([ n12

]!)

=

[n

6+

1

2

]+ 3

[n

12+

1

2

]+ 5

[ n12

]−∑i

fi.

We have∑

i ei = O(log n) and∑

i fi = O(log n). So, as n tends to infinity,

v2([µn]!) = [µn]−∑i

ei = µn+O(log n)

and

v2(F1F2 · · ·Fn) =

[n

6+

1

2

]+ 3

[n

12+

1

2

]+ 5

[ n12

]−∑i

fi =5n

6+O(log n).

So a necessary condition that (1) has infinitely many solutions is µ =5

6.

Consider the numbers n = 81 ·2k, k ≥ 2 Then[

5n

6

]= 135 ·2k−1 = (27+22+21+20)2k−1 , v2

([5n

6

]!

)=

135 · 2k−1 − 4,[ n

12

]= 27 · 2k−2 = (24 + 23 + 21 + 20)2k−2 , v2

([ n12

]!)

= 27 · 2k−1 − 4, v2(F1F2 · · ·Fn) =[n

6+

1

2

]+ 3

[n

12+

1

2

]+ 5

[ n12

]−∑

i fi = 27 · 2k−1 + 3 · 27 · 2k−2 + 5 · 27 · 2k−2 − 4 =

135 · 2k−1 − 4 = v2

([5n

6

]!

).

This concludes the proof.


http://www.fq.math.ca/Scanned/33-3/lengyel.pdf

Second solution by Li Zhou, Polk State College, USAWe show that µ = 5

6 and the equation is satisfied by n = 3 · 2m + 17 for all m ≥ 5.From the recurrence we see that Fn is odd for all n not divisible by 3. Note that F3 = 2, F6 = 23, and

F12 = 24 · 9. Assume q is an odd integer below. Then (F3q, F6) = F(3q,6) = F3 and (F6q, F12) = F(6q,12) = F6,so v2 (F3q) = 1 and v2 (F6q) = 3. Now F2n = FnLn, where L0 = 2, L1 = 1, and Ln+2 = Ln+1 +Ln for n ≥ 0.By the well-known fact that L3n =

∑ni=0

(ni

)2iLi, we get L6k ≡ L0 (mod 4), that is, v2 (L6k) = 1 for all

k ≥ 0. It is then easy to see by induction on m that v2 (F6·2mq) = m+ 3 for all m ≥ 0. Therefore,

v2 (F1 · · ·Fn) =

⌈1

2

⌊n3

⌋⌉+ 3

⌊n6

⌋+ v2

(⌊n6

⌋!)

=

⌈1

2

⌊n3

⌋⌉+ 3

⌊n6

⌋+

∞∑i=1

⌊bn/6c

2i

⌋∼ n

6+n

2+n

6=

5n

6, n→∞.

On the other hand,

v2 (bµnc!) =∞∑i=1

⌊bµnc

2i

⌋∼ µn, n→∞.

Hence, to satisfy the given equation for infinitely many n, µ must necessarily be 56 .

Finally, for all n = 3 · 2m + 17 with m ≥ 5, the two formulas above become

v2 (F1 · · ·Fn) = 2m−1 + 3 + 3(2m−1 + 2

)+ 2m−1 = 2m+1 + 2m−1 + 9,

v2

(⌊5n

6

⌋!

)=

∞∑i=1

⌊5 · 2m−1 + 14

2i

⌋=∞∑i=1

⌊2m+1 + 2m−1 + 14

2i

⌋

=

⌊14

2

⌋+

⌊14

4

⌋+

⌊14

8

⌋+

m∑i=0

2i +m−2∑i=0

2i

= 11 +(2m+1 − 1

)+(2m−1 − 1

)= 2m+1 + 2m−1 + 9.


U317. For any natural numbers s, t and p, prove that there is a numberM(s, t, p) such that every graph witha matching of size at least M(s, t, p) contains either a clique Ks, an induced complete bipartite graphKt,t or an induced matching Mp. Does the result remain true if we replace the word "matching" by"path"?

Proposed by Cosmin Pohoata, Columbia University, USA

Solution by the proposerThe result is true for both "matching" and "path" cases. We’ll only include a proof of it for "matching";for the "path" proof, we refer to A. Atminas, V.V. Lozin, I. Razgon, Linear time algorithm for computinga small complete biclique in graphs without long induced paths, SWAT, 2012: 142-152.

The key part of the proof is the following analogue for bipartite graphs, which is easier to prove.

Lemma:.For any natural numbers t and p, there is a natural number N(t, p) such that every bipartite graph with amatching of size at least N(t, p) contains either a complete bipartite graph Kt,t or an induced matching Mp.

Proof:For p = 1 and arbitrary t, we can define N(t, p) = 1. Now, for each fixed t, we prove the result by inductionon p. Without loss of generality, we prove it only for values of the form p = 2s. Suppose we have shown thelemma for p = 2s for some s ≥ 0. Let us now show that it is sufficient to set N(t, 2p) = RB(t, RB(t,N(t, p))),where RB is the bipartite Ramsey number.

Consider a graph G with a matching of size at least RB(t, RB(t,N(t, p))). Without loss of generality,we may assume that G contains no vertices outside of this matching. We also assume that G does notcontain an induced Kt,t, since otherwise we are done. Then G must contain the bipartite complement ofKRB(t,N(t,p)),RB(t,N(t,p)) with vertex classes, say A and B. Now let C and D consist of the vertices matchedto vertices in A and B respectively in the original matching in G.

Note that A, B, C, D are pairwise disjoint. Graphs G[A∪C] and G[B ∪D] (induced on sets A∪C andB ∪D) now each contain a matching of size RB(t,N(t, p)). There are no edges between A and B, yet theremight be edges between C and D. By our assumption, G[C ∪D] is however Kt,t-free, therefore, by Ramsey’stheorem, it must contain the bipartite complement of KN(t,p),N(t,p), with vertex sets C ′ ⊂ C, D′ ⊂ D. LetA′ ⊂ A and B′ ⊂ B be the set of vertices matched to C ′ and D′ respectively in the original matching in G.Now there are no edges in G[A′ ∪ B′] and none in G[C ′ ∪D′], but G[A′ ∪ C ′] and G[B′ ∪D′] both containa matching of size N(t, p). Since G is Kt,t-free, by the induction hypothesis, we conclude that they bothcontain an induced matching Mp. Putting these together we find that G contains an induced M2p. Thisproves the Lemma.

Returning to the problem, define M(s, t, p) := R(s,R(s,N(t, p))), where R is the classical Ramseynumber and N(t, p) is the N from the Lemma. Suppose that G is a (Ks,Kt,t)-free graph with a matchingof size R(s,R(s,N(t, p))). Since G is Ks-free, by Ramsey’s theorem, it must contain an independent set Aof size R(s,N(t, p)). Let B be the set of vertices matched to A. Since G[B] is Ks-free, it must contain anindependent set B′ of size N(t, p). Let A′ be the set of vertices matched to B′; the graph H = G[A′ ∪B′] isa bipartite graph with a matching of size N(t, p). By the above, since H is Kt,t-free, it thus follows that Hcontains an induced matching Mp. This completes the proof.


U318. Determine all possible values of∑∞

k=1(−1)q(k)

k2, where q(x) is a quadratic polynomial that assumes only

integer values at integer places.

Proposed by Albert Stadler, Herrliberg, Switzerland

Solution by Li Zhou, Polk State College, USAIt is well known that

q(x) = a

(x

2

)+ b

(x

1

)+ c

(x

0

)=

1

2ax(x− 1) + bx+ c

for some a, b, c ∈ Z with a 6= 0. Note that for k ∈ N,

q(k + 4) = q(k) + 4ak + 6a+ 4b ≡ q(k) (mod 2).

Now denote 12ax(x− 1) + bx+ c by qa,b,c(x), with a, b, c ≡ 0 or 1 (mod 2). Then

S0,0,0 =∞∑k=1

(−1)q0,0,0(k)

k2=

1

12+

1

22+

1

32+

1

42+ · · · = π2

6,

S0,1,0 =∞∑k=1

(−1)q0,1,0(k)

k2= − 1

12+

1

22− 1

32+

1

42− · · · = −π

2

12,

S1,0,0 =∞∑k=1

(−1)q1,0,0(k)

k2=

1

12− 1

22− 1

32+

1

42+

1

52− 1

62− 1

72+

1

82+ · · ·

=

(1

12− 1

32+

1

52− 1

72+ · · ·

)− 1

4

(1

12− 1

22+

1

32− 1

42+ · · ·

)= G− π2

48,

where G is Catalan’s constant, and

S1,1,0 =

∞∑k=1

(−1)q1,1,0(k)

k2= − 1

12− 1

22+

1

32+

1

42− 1

52− 1

62+

1

72+

1

82− · · ·

= −(

1

12− 1

32+

1

52− 1

72+ · · ·

)− 1

4

(1

12− 1

22+

1

32− 1

42+ · · ·

)= −G− π2

48.

Finally, S0,0,1 = −S0,0,0, S0,1,1 = −S0,1,0, S1,0,1 = −S1,0,0, and S1,1,1 = −S1,1,0.

Also solved by Roberto Mastropietro and Emiliano Torti, University of Rome Tor Vergata", Italy; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Daniel Lasaosa, Pamplona, Spain.


Olympiad problems

O313. Find all positive integers n for which there are positive integers a0, a1, . . . , an such that a0 +a1 + · · ·+an = 5(n− 1) and

1

a0+

1

a1+ · · ·+ 1

an= 2.


Solution by Adnan Ali, Student in A.E.C.S-4, Mumbai, IndiaFrom the AM-HM Inequality,

5(n− 1)

n+ 1≥n+ 1

2.

So, n2 − 8n+ 11 ≤ 0⇒ 1 < 4−√

5 ≤ n ≤ 4 +√

5 < 7. Hence 2 ≤ n ≤ 6. But it is easy to see that

2 + 2 + 1 = 5,1

2+

1

2+

1

1= 2, (for n = 2)

3 + 3 + 3 + 1 = 10,1

3+

1

3+

1

3+

1

1= 2, (for n = 3)

2 + 2 + 2 + 3 + 6 = 15,1

2+

1

2+

1

2+

1

3+

1

6= 2, (for n = 4)

3 + 3 + 3 + 2 + 3 + 6 = 20,1

3+

1

3+

1

3+

1

2+

1

3+

1

6= 2, (for n = 5)

3 + 3 + 3 + 4 + 4 + 4 + 4 = 25,1

3+

1

3+

1

3+

1

4+

1

4+

1

4+

1

4= 2, (for n = 6)

Hence n ∈ {2, 3, 4, 5, 6}.

Also solved by Daniel Lasaosa, Pamplona, Spain; William Kang, Bergen County Academies, Hackensack,NJ, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Michael Tang, Edina High School,MN, USA; Ángel Plaza, Department of Mathematics, University of Las Palmas de Gran Canaria, Spain;Albert Stadler, Herrliberg, Switzerland; Alessandro Ventullo, Milan, Italy; Arber Avdullahu, Mehmet AkifCollege, Kosovo; Arber Igrishta, Eqrem Qabej, Vushtrri, Kosovo; Bodhisattwa Bhowmik, RKMV, Agartala,Tripura, India; Erdenebayar Bayarmagnai, School Nr.11, Ulaanbaatar, Mongolia; Francesc Gispert Sán-chez, CFIS, Universitat Politécnica de Catalunya, Barcelona, Spain; Henry Ricardo, New York Math Circle;Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan; Shatlyk Mamedov, School Nr. 21, Da-shoguz, Turkmenistan; Titu Zvonaru, Comănes, ti, Romania and Neculai Stanciu, Buzău, Romania Li Zhou,Polk State College, USA.


O314. Prove that every polynomial p(x) with integer coefficients can be represented as a sum of cubes ofseveral polynomials that return integer values for any integer x.


Solution by Navid Safei, University of Technology in Policy Making of Science and Technology, IranAt first thanks to the following identity one can find that the polynomial x, could be written as sum ofcubes of five integer valued polynomials as follows:

x = x3 +

(x3 − x

6

)3

+

(x3 − x

6

)3

+

(−1− x3 − x

6

)3

+

(1− x3 − x

6

)3

It is obvious that the polynomialx3 − x

6is an integer-valued polynomial. By use of the following we can

find that the polynomial x2 , could be written as sum of cubes of integer-valued polynomials( just take placein the above identity x through x2) . Now consider the polynomial P (x) with integer coefficients then wecan write the polynomial by the form P (x) = P1(x

3) +xP2(x3) +x2P3(x

3) where P1, P2, P3 are polynomialswith integer coefficients . it is obvious that these polynomials (i.e. P1, P2, P3) could be written as the sumof cubes of polynomials with integer coefficients and thus integer-valued. Since polynomials x, x2 could bewritten so , the statement was proved.

Also solved by Henri Godefroy, Stanislas Secondary School, Paris, France; Li Zhou, Polk State College,USA.


O315. Let a, b, c be positive real numbers. Prove that(a3 + 3b2 + 5

) (b3 + 3c2 + 5

) (c3 + 3a2 + 5

)≥ 27(a+ b+ c)3.


Solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Roma, ItalyThe AGM yields a3 + 1 + 1 ≥ 3a hence

(a3 + 3b2 + 5)(b3 + 3c2 + 5)(c3 + 3a2 + 5) ≥≥ (3a+ 3b2 + 3)(3b+ 3c2 + 3)(3c+ 3a2 + 3) =

= 27(a+ b2 + 1)(b+ c2 + 1)(c+ a2 + 1)

we come to

(a+ b2 + 1)(b+ c2 + 1)(c+ a2 + 1) ≥ (a+ b+ c)3

Now Holder comes in

(a+ b2 + 1)(1 + b+ c2)(a2 + 1 + c) ≥

(∑cyc

a13 1

13a

23

)3

whence the result.

Also solved by George Gavrilopoulos, High School of Nea Makri, Athens, Greece; An Zhen-ping, Ma-thematics Department, Xianyang Normal University, Xianyang, Shaanxi, China; Albert Stadler, Herrliberg,Switzerland; Erdenebayar Bayarmagnai, School Nr.11, Ulaanbaatar, Mongolia; Khakimboy Egamberganov,National University of Uzbekistan, Tashkent, Uzbekistan; Daniel Lasaosa, Pamplona, Spain; Marius Sta-nean, Zalau, Romania; Farrukh Mukhammadiev, Academic Lyceum Nr.1 under the SamIES, Samarkand,Uzbekistan; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan; Titu Zvonaru, Comănes, ti,Romania and Neculai Stanciu, Buzău, Romania; Li Zhou, Polk State College, USA.


O316. Prove that for all integers k ≥ 2 there exists a power of 2 such that at least half of the last k digitsare nines. For example, for k = 2 and k = 3 we have 212 = . . . 96 and 253 = . . . 992.

Proposed by Roberto Bosch Cabrera, Havana, Cuba

Solution by Daniel Lasaosa, Pamplona, SpainClaim: If x ≡ −1 (mod 5k) for some integers x and k ≥ 1, then x5 ≡ −1 (mod 5k+1).Corollary: For every positive integer k, we have 22·5

k−1 ≡ −1 (mod 5k).Proof: If x ≡ −1 (mod 5k), an integer u exists such that x = u · 5k − 1, or

x5 =(u5 · 54k−1 − u4 · 53k + 2u3 · 52k − 2u2 · 5k + u

)5k+1 − 1.

The Claim follows. Since 22 = 5 · 1− 1, the result holds for k = 1, serving as the base case for induction overk, the step being guaranteed by the Claim. The Corollary follows.

By the Corollary, for every positive integer k, a positive integer u exists such that

22·5k−1

= u5k − 1, 0 < u10k − 22·5k−1+k = 2k.

Note that if k = 2` ≥ 2 is an even positive integer, then 2k < 10`, clearly true since it is equivalent to(52

)`> 1, whereas if k = 2` + 1 ≥ 3 is an odd integer larger than 1, then 2k < 10`, which is equivalent to

2`+1 < 5`, true for ` = 1 (equivalent to k = 3), and clearly true for any ` ≥ 1 since the RHS is multiplied by5 every time ` grows in one unit, the LHS being multiplied only by 2. It follows that, whatever the last `digits are, the previous k−` digits (which are clearly at least half of the last k digits) are 9’s. The conclusionfollows.

Also solved by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Albert Stadler, Herrliberg,Switzerland; Li Zhou, Polk State College, USA.


O317. Twelve scientists met at a math conference. It is known that every two scientists have a common friendamong the rest of the people. Prove that there is a scientist who knows at least five people from theattendees of the conference.


First solution by Li Zhou, Polk State College, USANumber the scientists 1 through 12 and let S = {1, . . . , 12}. For each i ∈ S, denote by Fi the set of friendsof i. Assume that |Fi| ≤ 4 for all i.(a) Claim 1: |Fi| = 4 for all i.

Indeed, if |Fi| ≤ 3 for some i, then∣∣∣∣∣∣⋃k∈Fi

(Fk \ {i})

∣∣∣∣∣∣ ≤∑k∈Fi

|Fk \ {i}| ≤ 3|Fi| ≤ 9 < 11,

so there is j ∈ S \ {i} such that i and j do not have a friend in common, a contradiction.(b) Claim 2: If i and j are friends then |Fi ∩ Fj | = 1.

Indeed, suppose that Fi = {j, k, l,m} and k, l ∈ Fj as well. Note that i and m must have a friendin common, so |Fm ∩ {j, k, l}| ≥ 1. Hence, |(Fj ∪ Fk ∪ Fl ∪ Fm) \ {i, j, k, l,m}| ≤ 6 < 7, so there is n ∈S \ {i, j, k, l,m} such that i and n do not have a friend in common, a contradiction.(c) Without loss of generality, assume now that F1 = {2, 3, 4, 5} with F1 ∩ F2 = {3} and F1 ∩ F4 = {5}, asin the left graph below. For 1 to have a friend in common with each scientist in {6, . . . , 12}, we must have{6, . . . , 12} ⊆ (F2 ∪ F3 ∪ F4 ∪ F5). Now each of 2, 3, 4, 5 needs to have two friends in {6, . . . , 12}, so exactlyone scientist in {6, . . . , 12} has two friends in {2, 3, 4, 5} who are not friends. Without loss of generality, say6 is a friend of 2 and 4, then we have the complete left graph below. However, for 2 to have a friend incommon with each of 11 and 12, 7 has to be a friend of both 11 and 12, which contradicts Claim 2 since{5, 7} ⊆ F11 ∩ F12. This contradiction completes the proof.

6

2

1

4

3

5

7

8

9

10

12

11

1

2

3

4

5

1086

79

11

(d) Finally we notice that it does require twelve scientists to draw the conclusion, as shown by the rightgraph above with eleven scientists.


Second solution by Michael Ma, Plano, TXLet each of the scientists represent a vertex on a graph G. Now draw an edge between two vertices if the twocorresponding scientists are friends. Now we know that every two vertices share a common neighbor. We willproceed now by contradiction. So assume that no vertex is of degree at least 5. Now say there exists a vertexwith degree 3 or less. Then in G there are 11 other vertices that share a common neighbor. This commonneighbor must be one of the 3 or less neighbors. Then by the pigeonhole principle there is a neighbor thathas at least 44 edges to the other 11 vertices. Since this neighbor also has an edge to the original vertex,this vertex has degree at least 5. Contradiction.

So now every vertex has degree at least and at most and therefore exactly 4. Now take a vertex A. Sayit’s neighbors are B,C,D,E. Now the other 7 points have to share one of B,C,D,E with A. Say one ofB,C,D,E is adjacent to 3 or more of these 7 points. WLOG say B. Then there are only 3 since the degreeof B is 4. Then B must share a vertex with A. But none of these 3 points are neighbors with A and A isnot a neighbor of itself. Contradiction.

So one of B,C,D,E is adjacent to at least 1 of the seven points and the other 3 are neighbors of exactly2 of the 7 points. WLOG say E is adjacent to at least 1 and E is adjacent to F,D is adjacent to G,H,C isadjacent to I, L, and B is adjacent to J,K. Now E has to be adjacent with on of B,C,D. WLOG say D.Therefore since B and C cannot both be adjacent to E,B and C must be adjacent. Therefore E is adjacentto one of G,H, I, L, J,K. Assume E is connected to one of G,H. WLOG let it be H. Then J,K, I, L mustshare a neighbor with D. Since they are not adjacent to A or E, they must be adjacent to H or G. But Hcan only have one more neighbor and G only 2. Contradiction.

Hence E is connected to one of I, J,K,L. WLOG let it be L. Now since H shares a neighbor with D,Gand H are adjacent. Similarly I, L and K,J are adjacent.

Now since L and E must share a common neighbor and L cannot be adjacent to A or D, its mustbe adjacent to F . Similarly, since K and J cannot be neighbors of L, since it’s degree is 4,K and J areneighbors of F . Now since H and B share a common neighbor, H is connected to KorJ . WLOG assumeit’s K. Similarly we get that J is adjacent to G. Now since H and C share a common neighbor, H and Iare adjacent. Similarly, so are G and I. Now H and K share no common neighbors. Contradiction. So thereis a vertex of degree at least 5.

Also solved by Francesc Gispert Sánchez, CFIS, Universitat Politécnica de Catalunya, Barcelona, Spain;Philip Radoslavov Grozdanov, Yambol, Bulgaria.


O318. Find all polynomials f ∈ Z[X] with the property that for any distinct primes p and q, f(p) and f(q)are relatively prime.

Proposed by Marius Cavachi, Constanta, Romania

Solution by Li Zhou, Polk State College, USAFor each degree n ≥ 0, f(x) = ±xn clearly have this property. We show that they are the only ones.

Suppose that f(x) = anxn + · · · + a1x + a0 is such a polynomial, with n ≥ 0. Let p be any prime such

that p > |an|+ · · ·+ |a0|. If there is a prime p1 6= p such that p1|f(p), then by Dirichlet’s theorem, there isa prime q = p1u+ p for some positive integer u. Hence f(q) ≡ f(p) ≡ 0 (mod p1), that is, p1| (f(p), f(q)), acontradiction. Therefore |f(p)| must be a power of p. Observe that

|f(p)| ≤ |an|pn + · · ·+ |a1|p+ |a0| ≤ (|an|+ · · ·+ |a0|) pn < pn+1,

|f(p)| ≥ |an|pn − (|an−1|+ · · ·+ |a0|) pn−1 ≥ |an|pn − (p− 2)pn−1.

If |an| ≥ 2 then |f(p)| > pn, so f(p) cannot be a power of p. Thus |an| = 1. Then |f(p)| > pn−1, so|f(p)| = pn, which implies f(p)− anpn ∈ {0,±2pn}. But

|f(p)− anpn| = |an−1pn−1 + · · ·+ a1p+ a0| ≤ (|an−1|+ · · ·+ |a0|) pn−1 < pn,

so f(p)− anpn = 0. This means that f(x)− anxn has all primes greater than |an|+ · · ·+ |a0| as zeros, thusit must be identically 0, that is, f(x) = anx

n with |an| = 1.

Also solved by Daniel Lasaosa, Pamplona, Spain; Navid Safei, University of Technoogy in Policy Makingof Science and Technology, Iran; Khakimboy Egamberganov, National University of Uzbekistan, Tashkent,Uzbekistan.


Junior problems - AwesomeMath...Nowsupposethat x y z: (3) Then,wehave 3x3 52975 ,x d 52975 3 e= 26...

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Transcript of Junior problems - AwesomeMath...Nowsupposethat x y z: (3) Then,wehave 3x3 52975 ,x d 52975 3 e= 26...