mr 2 2013 solutions

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Junior problems

J259. Among all triples of real numbers (x,y ,z) which lie on a unit sphere x2 + y2 + z2 = 1 find a triplewhich maximizes min(|x y|, |y z|, |z x|).

Proposed by Arkady Alt, San Jose, California, USA

Solution by Polyahedra, Polk State College, USA

Without loss of generality, assume that x y z. Then

m = min(|x y|, |y z|, |z x|) = min(y x, z y) z x2

.

Now x2 + y2 + z2 = 1 can be written as (z x)2 = 2 (z + x)2 2y2. Thus (z x)2 2, with equality ifand only if z + x = 0 = y. Hence m

22 , with the maximum of

22 attained at (x,y,z) =

22 , 0,

22

.

Also solved by Antonio Trusiani, Universit di Roma Tor Vergata, Roma, Italy; Daniel Lasaosa, Uni-

versidad Pblica de Navarra, Spain; Alessandro Ventullo, Milan, Italy; Radouan Boukharfane, Polytechniquede Montreal, Canada.

Mathematical Reflections 2 (2013) 1

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J260. Solve in integers the equationx4 y3 = 111.

Proposed by Jos Hernndez Santiago, Oaxaca, Mxico

Solution by Daniel Lasaosa, Universidad Pblica de Navarra, Spain

It can be easily checked that any perfect cube leaves, modulus 13, a remainder of

5,

1, 0, 1, 5, and any

perfect fourth power leaves, modulus 13, a remainder of 4, 0, 1, 3. But 111 leaves a remainder of 7 6modulus 13, which can be easily checked that cannot be obtained as the sum of one out of 4, 0, 1, 3 andone out of5, 1, 0, 1, 5. It follows that the proposed equation can have no solutions in integers (x, y).

Also solved by Kwan Chung Hang, Sir Ellis Kadoorie Secondary School, Hong Kong; Andrea Fiacco, and

Antonio Trusiani, Universit di Roma Tor Vergata, Roma, Italy; Corneliu Mnescu-Avram, Ploieti, Ro-

mania; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Alessandro Ventullo, Milan, Italy; Sayan Das,

Kolkata, India; Radouan Boukharfane, Polytechnique de Montreal, Canada; Polyahedra, Polk State College,

USA; George Batzolis, Mandoulides High School, Greece; AN-anduud Problem Solving Group, Ulaanbaatar,

Mongolia; Robin Park, Thomas Jefferson High School For Science and Technology; Ioan Viorel Codreanu,

Satulung, Maramures, Romania.


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J261. Let A1 . . . An be a polygon inscribed in a circle with center O and radius R. Find the locus of pointsM on the circumference such that

A1M2 + + AnM2 = 2nR2.

Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA

Solution by Daniel Lasaosa, Universidad Pblica de Navarra, SpainLet G be the center of mass of the n-gon, ie, for a coordinate system such that wlog its origin is located atthe circumcenter O of the n-gon, we have

OG =

OA1 +

OA2 + + OAn

n.

Note now that

AiM2 =

OM OAi

2= 2R2 2OM OAi,

where we have used that |OM| = |OAi| = R because the vertices of the n-gon, as well as M, are on thecircle with center O and radius R. Adding these equations for i = 1, 2, . . . , n, we find

A1M2 + + AnM2 = 2nR2 2nOM OG,

ie, the condition given in the problem statement is equivalent toOM OG = 0. We have now two cases:

Case 1: If G = O, the condition is satisfied for all M on the circumference, ie, the locus of points Msatisfying the condition given in the problem statement is the entire circumference.

Case 2: If G = O, the condition is equivalent to OG and OM being perpendicular, ie, the locus ofpoints M satisfying the condition given in the problem statement is the two ends of the diameter of thecircumference which is perpendicular to OG, G being the center

Also solved by Kwan Chung Hang, Sir Ellis Kadoorie Secondary School, Hong Kong; G.R.A.20 Problem

Solving Group, Roma, Italy; Radouan Boukharfane, Polytechnique de Montreal, Canada; George Batzolis,Mandoulides High School, Greece; Polyahedra, Polk State College, USA.


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J262. Find all positive integers m, n such thatm+1

n

= mn+1

.

Proposed by Roberto Bosch Cabrera, Havana, Cuba

Solution by Daniele Fakhoury, and Andrea Fiacco, Universit di Roma Tor Vergata, Roma, Italy

We note that if m + 1 > n 0 then

m + 1n

> 0 and nm + 1

= 0.On the other hand if 0 m + 1 < n

m + 1

n

= 0 and

n

m + 1

> 0.

Finally, if n = m + 1 0 then1 =

m + 1

n

=

n

m + 1

.

So equality holds iff n = m + 1 0.

Also solved by Kwan Chung Hang, Sir El lis Kadoorie Secondary School, Hong Kong; Daniel Lasaosa,

Universidad Pblica de Navarra, Spain; Vlad Petrache, Mihaela Petric and Daniel Vcaru, Colegiul Eco-

nomic Maria Teiuleanu, Piteti, Romania; Angel Plaza, University of Las Palmas de Gran Canaria, Spain;

Polyahedra, Polk State College, USA; Corneliu Mnescu-Avram, Ploieti, Romania; Radouan Boukharfane,

Polytechnique de Montreal, Canada; Alessandro Ventullo, Milan, Italy; Robin Park, Thomas Jefferson High

School For Science and Technology.


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J263. The n-th pentagonal number is given by the formula pn =n(3n1)

2 . Prove that there are infinitelymany pentagonal numbers that can be written as a sum of two perfect squares of positive integers.

Proposed by Jos Hernndez Santiago, Oaxaca, Mxico

Solution by Robin Park, Thomas Jefferson High School For Science and Technology

We claim that there exist infinitely many pentagonal numbers pn that can be written as m2 + n2, where m

and n are positive integers. For this to be true, there must exist infinitely many perfect squares m2 that canbe expressed as n(n1)2 .

Let m2 =n(n1)

2 . This rearranges to (2n 1)2 8m2 = 1, which is a Pell Equation a2 kb2 = 1, wherea = 2n 1, k = 8, b = m, and is known to have an infinite number of solutions. Since k is even, a must beodd, so there exist infinitely many n that satisfy this equation.

Therefore there exist infinitely many pentagonal numbers that can be expressed as m2 + n2.

Also solved by Daniel Lasaosa, Universidad Pblica de Navarra, Spain; Andrea Fiacco, and Antonio Tru-

siani, Universit di Roma Tor Vergata, Roma, Italy; Alessandro Ventullo, Milan, Italy; Radouan Boukhar-

fane, Polytechnique de Montreal, Canada; Corneliu Mnescu-Avram, Ploieti, Romania; Polyahedra, PolkState College, FL, USA; George Batzolis, Mandoulides High School, Thessaloniki, Greece; AN-anduud Prob-

lem Solving Group, Ulaanbaatar, Mongolia.


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J264. In triangle ABC, 2A = 3B. Prove thata2 b2 a2 + ac b2 = b2c2.Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution by Giulia Giovannotti, Universit di Roma Tor Vergata, Roma, Italy

We note that A = 3B/2, C = A B = 5B/2, anda = 2R sin(3B/2), b = 2R sin(B), c = 2R sin(5B/2),

where R is the radius of the circumcircle. So it suffices to show that

(sin(3B/2)2 sin(B)2)(sin(3B/2)2 + sin(3B/2) sin(5B/2) sin(B)2) = sin(B)2 sin(5B/2)2.

Since sin(x) sin(y) = (cos(x y) cos(x + y))/2 this is equivalent to

(cos(2B) cos(3B))(cos(B) + cos(2B) cos(3B) cos(4B)) = (1 cos(2B))(1 cos(5B)).

Now cos(x)cos(y) = (cos(x y) + cos(x + y))/2, and by expanding the two sides of the above equation weobtain

2LHS = 2 2 cos(2B) 2cos(5B) + cos(7B) and 2RHS = 2 2 cos(2B) 2 cos(5B) + cos(7B).

So the required equation holds for any B.

Also solved by Polyahedra, Polk State College, FL, USA; Nicuor Zlota, Traian Vuia Technical Col-

lege, Focani, Romania; Daniel Lasaosa, Universidad Pblica de Navarra, Spain; Arkady Alt , San Jose

,California, USA; Radouan Boukharfane, Polytechnique de Montreal, Canada.


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Senior problems

S259. Let a,b,c,d,e be integers such that

a(b + c) + b(c + d) + c(d + e) + d(e + a) + e(a + b) = 0.

Prove that a + b + c + d + e divides a5

+ b5

+ c5

+ d5

+ e5

5abcde.Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution by Alessandro Ventullo, Milan, Italy

Suppose that a,b,c,d,e are the five roots 1, 2, 3, 4, 5 of a fifth degree polynomial P(x). Let

k =5

i=1

ki , sk =

1j1

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S260. Let m < n be positive integers and let x1, x2, . . . , xn be positive real numbers. If A is a subset of{1, 2, . . . , n}, define sA =

iA xi and A

c = {i {1, 2, . . . , n}|i / A}. Prove that|A|=m

sAsAc

mn m

n

m

,

where the sum is taken over all m-element subsets A of{1, 2, . . . , n}.

Proposed by Mircea Becheanu, University of Bucharest, Romania

Solution by Carlo Pagano, and Antonio Trusiani, Universit di Roma Tor Vergata, Roma, Italy

Let A, B {1, 2, . . . , n}. Since

s(A) + s(Ac) = s(B) + s(Bc) =ni=1

xi

it follows that s(A) s(B) iff 1/s(Ac) 1/s(Bc) and by Chebycevs inequality

|A|=m

sAsAc

1nm

|A|=m

sA |A|=m

1

sAc .

Moreover, by AM-HM inequality,1nm

|A|=m

1

sAc

nm

|A|=m sAc

.

Now we note that

|A|=m

sA =

n 1m 1

ni=1

xi and|A|=m

sAc =

n 1

n m 1 n

i=1

xi.

Hence |A|=m

sAsAc

nm

|A|=m sA

|A|=m sAc=

nm

n1m1

n1nm1

= mn m

n

m

.

Also solved by Kwan Chung Hang, Sir Ellis Kadoorie Secondary School, Hong Kong; AN-anduud Problem

Solving Group, Ulaanbaatar, Mongolia; Daniel Lasaosa, Universidad Pblica de Navarra, Spain; Robin Park,

Thomas Jefferson High School For Science and Technology; Li Zhou, Polk State College, Florida, USA;

George Batzolis, Mandoulides High School, Thessaloniki, Greece; Radouan Boukharfane, Polytechnique de

Montreal, Canada.


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S261. Let ABC be a triangle with circumcircle and let K be the circle simultaneously tangent to AB, ACand , internally. Let X be a point on the circumcircle of ABC and let Y, Z be the intersections of with the tangents from X with respect to K. As X varies on , what is the locus of the incenters oftriangles XY Z?

Proposed by Cosmin Pohoata, Princeton University, USA

Solution by Li Zhou, Polk State College, USALet i be the inversion with respect to the circle centered at A and with radius

bc, where b = AC and

c = AB. Let B be i(B), etc.. Then is the line BC, and thus K is the excircle ofABC opposite A.Let s and r be the semiperimeter and inradius ofABC = ACB. Then

AD =bc

AD=

2[ABC]

s sin A=

2r

sin A.

Hence, the distance from F to AC, being half of that from D to AC, is r. This completes the proof of thelemma.

The lemma implies that F and A are images of each other under the inversion j with respect to K.Therefore, the locus of the incenters of

XY Z is the circle j(). Of course, whether to include the tangency

point of and K depends on whether XY Z is allowed to degenerate into a point.Also solved by Daniel Lasaosa, Universidad Pblica de Navarra, Spain; Robin Park, Thomas Jefferson

High School For Science and Technology.


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S262. Let a,b,c be the sides of a triangle and let ma, mb, mc be the lengths of its medians. Prove that

a2 + b2 + c2 ab bc ca 4 ma2 + mb2 + mc2 mamb mbmc mcma .Proposed by Arkady Alt, San Jose, California, USA

Solution by Radouan Boukharfane, Polytechnique de Montreal, Canada

We know that for a triangle ABC, if ma,mb,mc are its medians on BC,CA,AB, so we have :

m2a + m2b + m

2c =

3

4

a2 + b2 + c2

Taking account this identity, we are left to prove that :

mbmc 1

2

a2 +

1

4

bc

We will prove that :

mbmc

a2

2

+bc

bBy squaring both sides of this inequality we get :

16m2am2b (2a2 + bc)2

which is equivalent to :

162(c2 + a2) b2

4 2(a

2 + b2) c24

(2a2 + bc)2 (b c)2(a + b + c)(a b c) 0Which is true. We used here the identities:

m2b =2(c2 + a2) b2

4and

m2c =2(a2 + b2) c2

4

We prove similary:

mamc b2

2+

ac

b

and

mamb c2

2 +

ab

b

By summing up the last inequalities we proved we get our result.

Also solved by Scott H. Brown, Auburn University, Montgomery, AL; Daniel Lasaosa, Universidad Pblica

de Navarra, Spain; Perfetti Paolo, Dipartimento di Matematica, Universit degli studi di Tor Vergata Roma,

Roma, Italy; Nicuor Zlota, Traian Vuia Technical College, Focani, Romania; Li Zhou, Polk State Col-

lege, USA; Robin Park, Thomas Jefferson High School For Science and Technology; Ioan Viorel Codreanu,

Satulung, Maramures, Romania; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Ioan Viorel

Codreanu, Satulung, Maramures, Romania.


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S263. Prove that for all n 2 and all 1 i n we haven

j=1

(1)njn+j

n

nnj

i +j

= 1.

Proposed by Marcel Chirita, Bucharest, Romania


It can be easily found, by taking particular cases with n = 2, 3, that the proposed relation is not necessarilytrue. It is however always true that

S(n, i) =

nj=0

(1)njn+j

n

nnj

i +j

= 0

for 1 i n. In fact, we will prove a more general result, namely that for all positive integer i, and allnon-negative integer n, we have

S(n, i) =

n

j=0

(

1)nj

n+jn

nnji +j =

(i

1)(i

2)

(i

n)

i(i + 1) (i + n),

which clearly results in S(n, i) = 0 for all integer 1 i n. In order to obtain this result, we will use thefollowing

Claim: For all positive integer n, we have

T(n) =n

j=0

(1)nj

n +j

n

n

n j

= 1.

Proof: Consider the following way to generate n-element subsets C of set A = A1 A2 with A1 ={1, 2, . . . , n}, A2 = {n + 1, n + 2, . . . , 2n}: we first remove n j elements out of A2 (which can be done in nnj ways), and then choose n elements out of the remaining j elements of A2, plus the n elements of A1(which can be done in

n+jn

ways). Given j, we generate

nnj

n+jn

sets, not necessarily distinct. Consider

a set C = C1 C2, where C1 = C A1 and C2 = C A2, and C2 has exactly k elements. As long as j k,C can be generated. Out of the

n

nj

ways in which we remove n j elements, there are exactly nknj waysin which we can choose to remove n j elements out of A2 that will leave the k elements of C2 availablefor selection into C. Therefore, any given n-subset element C such that C2 has exactly k elements, can bechosen in

nknj

ways. Let us count, for each C, the number of ways in which we can choose it when n j

is even, minus the number of ways in which we can choose it when n j is odd, which clearly equals T(n)when added over all possible subsets C. For any particular set with n 1 k 0, this number of ways is

n

j=k(1)nj

n k

n j =nk

m=0(1)m

n k

m = (1 1)nk = 0,

where we have defined m = nj. Therefore, the contribution to T(n) from all sets C= A2 is zero. Moreover,C = A2 can be chosen in exactly one way, when j = n, ie when no elements of A2 are removed prior tothe selection of the n-element subset. Since n j = 0 is even, this contributes +1 to T(n), being the onlynonzero term. The Claim follows.

Returning to the proposed problem, note that, since n + i = (n j) + (i +j), n +j = (i +j) (i n),while

n

nj

= nnj

n1n1j

and

n+jn

= n+jn

n1+jn1

, we have

(n + i)

n+jn

nnj

i +j

= (i n)n1+j

n1 n1

n1j

i +j+

n +j

n n

n

j

+


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+

n 1 +j

n 1

n 1n 1 j

,

or after substitution of this relation, and using the Claim, we have

(i + n)S(n, i) = (i n)S(n 1, i) + T(n) T(n 1) = (i n)S(n 1, i).

Together with

S(1, i) =

1j=0

(1)1j1 +j

i +j = 1

i +2

i + 1 =i

1

i(i + 1) ,

the improved, more general result follows after straightforward induction.

Also solved by Anastasios Kotronis, Athens, Greece; Radouan Boukharfane, Polytechnique de Montreal,

Canada; Perfetti Paolo, Dipartimento di Matematica, Universit degli studi di Tor Vergata Roma, Roma,

Italy; Li Zhou, Polk State College, USA; G.R.A.20 Problem Solving Group, Roma, Italy; Albert Stadler,

Switzerland; Tsouvalas Konstantinos, University of Athens, Athens, Greece.


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S264. Let a,b,c,x,y,z be positive real numbers such that ab + bc + ca = xy + yz + zx = 1. Prove that

a(y + z) + b(z + x) + c(x + y) 2.

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

Solution by Perfetti Paolo, Dipartimento di Matematica, Universit degli studi di Tor Vergata Roma,

Roma, ItalyLet A = a2 + b2 + c2, X = x2 + y2 + z2, P = ab + bc + ca, Q = xy + yz + zx.

The inequality is cyc

a(y + z) = (a + b + c)(x + y + z) cyc

ax + 2

Squaring and employing P = Q = 1, we get

(A + 2)(X+ 2)

2 +cyc

ax

2

CauchySchwarz yields cyc

ax

AX

thus we come to

(A + 2)(X+ 2)

2 +

AX2

or

AX+ 2A + 2X+ 4 4 +

AX+ AX A + X 2

AX

which is trivial.

Also solved by Kwan Chung Hang, Sir Ellis Kadoorie Secondary School, Hong Kong; Semchankau Ali-

aksei, Minsk, Belarus; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Li Zhou, Polk State College,

USA; Daniel Lasaosa, Universidad Pblica de Navarra, Spain; Nicuor Zlota, Traian Vuia Technical Col-

lege, Focani, Romania; Arkady Alt , San Jose ,California, USA; Radouan Boukharfane, Polytechnique de

Montreal, Canada; Ioan Viorel Codreanu, Satulung, Maramures, Romania.


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Undergraduate problems

U259. Compute

limn

1 + 1n(n+a)

n3

1 + 1n(n+b)n3

.


Solution by Giulio Calimici, and Emiliano Torti, Universit di Roma Tor Vergata, Roma, Italy

We note that

n3 ln

1 +

1

n(n + a)

= n3

1

n(n + a)+ O(1/n4)

= n

1

1 + a/n+ O(1/n2)

= n

1 a

n+ O(1/n2)

= n a + O(1/n).

Hence 1 + 1n(n+a)

n3

1 + 1n(n+b)

n3 = exp

n3 ln

1 +

1

n(n + a)

n3 ln

1 +

1

n(n + b)

= exp (n a + O(1/n) n + b + O(1/n)) exp(b a) = eba.

Also solved by Albert Stadler, Switzerland; Stanescu Florin, Serban Cioculescu school, Gaesti, Romania;

Daniel Lasaosa, Universidad Pblica de Navarra, Spain; Moubinool Omarjee Lyce Henri IV , Paris , France;

AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Alessandro Ventullo, Milan, Italy; Anastasios

Kotronis, Athens, Greece; Robin Park, Thomas Jefferson High School For Science and Technology; PerfettiPaolo, Dipartimento di Matematica, Universit degli studi di Tor Vergata Roma, Roma, Italy; Radouan

Boukharfane, Polytechnique de Montreal, Canada; Li Zhou, Polk State College, USA.


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U260. Determine all functions f : R R which are derivatives and satisfy

f2

f(x) dx.

Proposed by Mihai Piticari, "Drago Voda" National College, Romania

Solution by Daniele Fakhoury, and Andrea Fiacco, Universit di Roma Tor Vergata, Roma, Italy

If f exists in an open interval I R then, by hypothesis,

f(x)(2f(x) 1) = 0 x I.

By continuity of f, the above equation implies that

f(x) = 0 x I or f(x) = x2

+ c x I.

In the general case, since f is a derivative, by Darbouxs Theorem it follows that f has intermediate valueproperty. This implies that our f is continuous and it is non-differentiable in at most two points a

b:

f(x) =

12(x b) x (b, +),0 x [a, b],12(x a) x (, a).

Also solved by Stanescu Florin, Serban Cioculescu school, Gaesti, Romania; Li Zhou, Polk State College,

USA.


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U261. Let Tn(x) be the sequence of Chebyshev polynomials of the first kind, defined by T0(x) = 0, T1(x) = x,and

Tn+1(x) = 2xTn(x) Tn1(x)for n 1. Prove that for all x 1 and all positive integers n

x n

Tn(x) 1 + n(x 1).



The formal definition of Chebyshev polynomials is as in the problem statement, but with T0(x) = 1. We willactually use this definition in the following solution, since among other things, the definition with T0(x) = 0results after induction in Tn(1) = n for all n, hence the second inequality would not hold. With the definitionT0(x) = 1, as we will soon see, both inequalities hold.

Assume that x is a fixed value, and for said x, consider the x-dependent sequence (sn)n0 defined bys0 = 1, s1 = x, and for all n 2, sn = 2xsn1 sn2. Using standard techniques, it follows that

sn =x + x2 1n + x x2 1n

2 x + x2 1+ x x2 1

2

n

= xn,

where we have used the power-mean inequality, which is valid since x x2 1 are positive reals because0 x2 1 < x for all x 1. Note that equality holds iff equality in the power mean inequality holds, ieiff

x2 1 = 0, for x = 1, or iffn = 1 for all x.

The second inequality can be rewritten as

Tn(x) (1 + n(x 1))n .

For n = 0, both sides are identically 1, while for n = 1, both sides are identically x, or the inequality holds

with equality for all x 1 in these cases. For n = 2, the inequality rewrites as2x2 1 (2x 1)2, 2(x 1)2 0,

clearly true, with equality iff x = 1. Now, we will show by induction that, for all n 2 and all x, we have

Tn(x) = 2(x 1)n1k=1

(n k)Tk(x) + n(x 1)T0(x) + 1.

For n = 2 and n = 3, this result is respectively equivalent to

T2(x) = 2(x

1)T1(x) + 2(x

1) + 1 = 2x2

1,

T3(x) = 2(x 1)T2(x) + 4(x 1)T1(x) + 3(x 1) + 1 = 2xT2(x) x,clearly true in both cases. These are our base cases, and if the result holds for n, n 1, then

Tn+1(x) = 2(x 1)Tn(x) + 2Tn(x) Tn1(x) =

= 2(x 1)Tn(x) + 4(x 1)Tn1(x) + 2(x 1)n2k=1

(n k + 1)Tk(x) + (n + 1)(x 1)T0(x) + 1,

where we have used the hypothesis of induction for n, n 1, and the result clearly holds for n + 1 too. Henceit holds for all positive integer n.


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Now, this means that, if for some n 3 the inequality holds for 1, 2, . . . , n 1, then

Tn(x) 2(x 1)n1k=1

(n k) (1 + k(x 1))k + n(x 1) + 1,

with equality iff x = 1, since T2(x) = (1 + 2(x 1))2 iff x = 1, and if x = 1 then both sides are identi-cally 1. Using the expression for the sum of the geometric progression with ratio 1 + n(x 1) from 1 to(1 + n(x 1))

n

1

, we obtain

(1 + n(x 1))n 1 = n(x 1)n1k=0

(1 + n(x 1))k ,

or it suffices to show that

n1k=1

(1 + n(x 1))k n1k=1

2(n k)n

(1 + k(x 1))k .

Ifn is even, when k = n2 the term in the LHS is (1 + n(x 1))n2 , and in the RHS is

1 + n2 (x 1)

n2 , clearly

not larger, and equal iff x = 1. Whether n is odd or even, every k other than

n

2 can be grouped in pairs ofsum n, or it suffices to show that, for all integer k such that 1 k < n2 , we have

(1 + n(x 1))k + (1 + n(x 1))nk 2(n k)n

(1 + k(x 1))k + 2kn

(1 + (n k)(x 1))nk .

Now, since k < n k < n for all such n, we have 1 + k(x 1) 1 + (n k)(x 1) 1 + n(x 1), withequality iff x = 1, or it suffices to show that

n 2kn

(1 + n(x 1))nk n 2kn

(1 + n(x 1))k ,

clearly true since n 2k > 0 and 1 + n(x 1) 1, with equality again iff x = 1.The conclusion follows, equality holds in both equalities iff either n = 1 and for all x 1, or x = 1 for

all n.

Also solved by Daniel Vcaru, Colegiul Economic Maria Teiuleanu, Piteti, Romania; Radouan Boukhar-

fane, Polytechnique de Montreal, Canada.


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U262. Let a and b be positive real numbers. Find limn nn

i=1

a + bi

.


Solution by G.R.A.20 Problem Solving Group, Roma, Italy

Let xn =

ni=1

a + bi

. Then

limnxn+1

xn = limna + bn + 1 = a,

which implies that the sequence n

xn has the same limit.

Also solved by Daniel Lasaosa, Universidad Pblica de Navarra, Spain; Robinson Higuita Diego Harvin,

Universidad de Antioquia, Colombia; Radouan Boukharfane, Polytechnique de Montreal, Canada; Perfetti

Paolo, Dipartimento di Matematica, Universit degli studi di Tor Vergata Roma, Roma, Italy; Moubinool

Omarjee ,Lyce Henri IV , Paris , France; Stanescu Florin Serban Cioculescu school, Gaesti, Romania;

Angel Plaza, Department of Mathematics, University of Las Palmas de Gran Canaria, Spain; Konstantinos

Tsou- valas, University of Athens, Athens, Greece; Anastasios Kotronis, Athens, Greece; AN-anduud Problem

Solving Group, Ulaanbaatar, Mongolia; Li Zhou, Polk State College, USA.


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U263. Let n 2 be an integer. A general n n magic square is a matrix A Mn(R) such that the sum ofthe elements in each row of A is the same. Prove that the set of n n general magic squares is anR-vector space and find its dimension.

Proposed by Cosmin Pohoata, Princeton University, USA


Denote by Sn(R) the set of general n n magic squares. Clearly Sn(R) Mn(R), and Mn(R) is well-known to be an R-vector space. Let A, B Sn(R) and , R, then A + B Sn(R) because the sumsof elements in each row of A + B are all equal to s + t, where s, t are the respective values of the sumsof elements in each row for A, B. We conclude that Sn(R) is an R-vector subspace ofMn(R), and it remainsonly to find its dimension.

Consider any n (n 1) matrix A, and any real value s. Clearly, we can generate a matrix A Sn(R)as follows: the first n 1 elements in row i of A are the elements in row i of A, and the n-th element is sminus the sum of the previous n 1 elements. Reciprocally, given a matrix A Sn(R) such that the sum ofthe elements of its rows is s, there is exactly one n (n 1) matrix A resulting of erasing the last column inA. Note moreover that two distinct matrices A Sn(R), either they differ in one of the values of the n 1first columns, or they differ in the sum of each row, or they are the same, whereas two distinct n

(n

1)

matrices, and/or two distinct value of the sum s of each row, generate distinct matrices in Sn(R). It followsthat we can establish a bijection between the set Sn(R), and the set of pairs formed by one real s, and onen (n 1) matrix, or the dimension of Sn(R) is n(n 1) + 1 = n2 n + 1.

Also solved by Harun Immanuel, ITS Surabaya; Andrea Fiacco, and Antonio Trusiani, Universit di

Roma Tor Vergata, Roma, Italy; Radouan Boukharfane, Polytechnique de Montreal, Canada; Li Zhou,

Polk State College, USA.


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U264. Let A be a finite ring such that 1 + 1 = 0. Prove that the equations x2 = 0 and x2 = 1 have the samenumber of solutions in A.

Proposed by Mihai Piticari, "Dragos Voda" National College, Romania

Solution by Daniele Fakhoury, Andrea Fiacco, Emiliano Torti, and Antonio Trusiani, Universit di Roma

Tor Vergata, Roma, Italy

Let Z = x A : x2 = 0 and U = x A : x2 = 1 and let us define the map f : A A as f(x) = x + 1for any x A. We claim that f is a bijective map from Z onto U. Indeed, if x Z then f(x) U because

f(x)2 = (x + 1)2 = x2 + (1 + 1)x + 1 = x2 + 1 = 1.

If x1, x2 Z and f(x1) = f(x2) then

x1 = x1 + 1 + 1 = f(x1) + 1 = f(x2) + 1 = x2 + 1 + 1 = x2

and it follows that f is injective on Z. If y U then y2 = 1. Let x = y + 1 and we have that

x2 = (y + 1)2 = y2 + (1 + 1)y + 1 = y2 + 1 = 1 + 1 = 0,

so x Z and f(x) = y + 1 + 1 = y, that is F(Z) = U.

Also solved by Maria Miliaresi and Anastasios Kotronis, Athens, Greece; AN-anduud Problem Solving

Group, Ulaanbaatar, Mongolia; Daniel Lasaosa, Universidad Pblica de Navarra, Spain; Harun Immanuel,

ITS Surabaya; Corneliu Mnescu-Avram, Ploieti, Romania.


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Olympiad problems

O259. Solve in integers the equation x5 + 15xy + y5 = 1.

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

Solution by Arkady Alt , San Jose ,California, USA

Due to symmetry of equation we may assume that x y.Since in the case x = y we get equation 2x5 + 15x2 = 1 which obviously have no solution inintegers then further we also can assume x > y.Consider three cases.1. xy = 0.Then we get solution (x, y) = (1, 0) ;2. xy < 0.Then due to supposition x > y we have x > 0, y < 0 and by replacing y in equationwith y we obtain equation x5 15xy y5 = 1, where x, y 1(because now x, y are positive integers).Since x5 y5 = 15xy + 1 > 0 then x y > 0 x y + 1 yieldsx5

y5 = (x

y) x4 + y4 + xy x2 + y2 + x2y2 5 (x y) x2y2 and xy 2.Therefore,

15xy + 1 5 (x y) x2y2 3xy + 13

(x y) x2y2 = 3 (x y) xy 3xy

x y andsince 1 x y then x y = 1 and xy 3.But system

x y = 1

xy = 3have no integer solutions. Then remains system

x y = 1

xy = 2

which give us solution x = 2, y = 1.Since (x, y) = (2, 1) satisfy x5 15xy y5 = 1then (x, y) = (2, 1) , (1, 2) are solutions of original equation in the case xy < 0.3. Let xy > 0. It is possible if x, y < 0.Then by replacing (x, y) in original equation with(x, y) we obtain equation x5 + y5 = 15xy 1 where 1 x < y.Hence, xy 2 and x + y 2x + 1 3.Since x5 + y5 = (x + y) x4 + y4 xy x2 + y2 + x2y2 =(x + y)xy + x2 + y2 (x y)

2 + x2y2 (x + y) xy + x2 + y2+ x2y2 (x + y) 3xy + x

2y2 =(x + y) (3 + xy) xy 3 (3 + 2) xy = 15xy then 15xy 1 = x5 + y5 15xy,that is the contradiction.Thus, all solutions of original equation are (x, y) = (1, 0) , (0, 1) , (2, 1) , (1, 2) .

Also solved by Li Zhou, Polk State College, USA; George Batzolis, Mandoulides High School, Thessaloniki,

Greece; Daniel Lasaosa, Universidad Pblica de Navarra, Spain; Semchankau Aliaksei, Minsk, Belarus;

AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Antonio Trusiani, Universit di Roma Tor

Vergata, Roma, Italy.


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O260. Let p be a positive real number. Define a sequence (an)n1 by a1 = 0 and

an =

n + 1

2

p+ an

2

for n 2. Find the minimum of annp1 over all positive integers n.


Solution by Radouan Boukharfane, Polytechnique de Montreal, Canada

We observe that :

a2n =

2n + 1

2

p+ a 2n2 = n

p + an

a2n+1 =

2n + 2

2

p+ a 2n+12 = (n + 1)

p + an

therefore we conclude from this two expressions that:

a2n+1 a2n = (n + 1)p

np

that can be rewriten :

an+1 an =n

2+ 1

pn

2

pfrom where we deduce that an is increasing and that for n 3:

an = 1 +n1k=2

k

2+ 1

p

k

2

p

The sequence sn =an

np1 is decreasing and the minimum of sn over all positive integers n is attained for

limn sn = limn

1 +n1

k=2

k2 + 1

p k2pnp 1

or:

n1k=2

k2 + 1

p k2pnp 1 =

1

2p

n1k=2 ((k + 2)

p kp)np 1

or:

n1

k=2 [(k + 2)p kp] =

n1

k=2 [(k + 2)p (k + 1)p + (k + 1)p kp]

=n1k=2

[(k + 2)p (k + 1)p] +n1k=2

[(k + 1)p kp] = (n + 1)p 3p + np 2p

limn

n1

k=2[(k+2)pkp]np1 = limn

(n+1)p3p+np2pnp1

= limn

(n+1)p+np

np1 + limn

Cste 3p 2p

np1 = limn(n+1)p+np

np1 + 0

= limn

(n+1)p+np

np1 = limn(n+1)p

np1 + limnnp

np1

= limn

(n+1)p

np

np1np

+ limn

np

np1 =limn

(n+1n )p

1 + 1 =11 + 1 = 2


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which means that :

limn sn =

1

2p1

which is the minimum we look for. We are done.

Also solved by Daniel Lasaosa, Universidad Pblica de Navarra, Spain; Li Zhou, Polk State College, USA


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O261. Find all positive integers n for which

(n) (n) 4n,

where (n) is the sum of positive divisors of n and is Eulers totient function.

Proposed by Albert Stadler Buchenrain, Herrliberg, Switzerland

Solution by Andrea Fiacco, Emiliano Torti, and Antonio Trusiani, Universit di Roma Tor Vergata,

Roma, Italy

We will show that the inequality holds iff n {1, 8, p , p2 : p is a prime}.For n = 1 this is trivial. If n = p then

(p) (p) = 1 +p (p 1) = 2 4p

which holds for any prime p. If n = p2 then

(p) (p) = 1 +p +p2 (p2 p) = 1 + 2p 4

p2 = 4p

which holds for any prime p. Now, we note that

(n) (n) = n

d|n

1

dd|n

(d)

d

2n

p|n

1

p.

If n = pk with k 3 then, by the above inequality,

4pk/2 (pk) (pk) 2pk1

that is 4 pk2 which holds iff n {23, 24, 33}. It is easy to verify that the required inequality holds onlyfor n = 8.

Finally, if n is divisible by r 2 distinct primes then, by AM-GM inequality,(n) (n) 2n

p|n

1

p>

2nr

(p|np)1/r

2rn1 1r 4n1 12 = 4n,

and the required inequality does not hold.

Also solved by Semchankau Aliaksei, Minsk, Belarus; George Batzolis, Mandoulides High School, Thes-

saloniki, Greece; Radouan Boukharfane, Polytechnique de Montreal, Canada; Daniel Lasaosa, Universidad

Pblica de Navarra, Spain; Kwan Chung Hang, Sir Ellis Kadoorie Secondary School, Hong Kong; Li Zhou,

Polk State College, USA.


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O262. Let n 3 be an integer. Consider a convex n-gon A1 . . . An for which there is a point P in its interiorsuch that AiP Ai+1 =

2n for all i [1, n 1]. Prove that P is the point which minimizes the sum of

distances to the vertices of the n-gon.



Consider a coordinate system with center in P and such that the positive horizontal semiaxis is directedalong ray OAn, or denoting by ri the distances OAi, we have Ai (ri cos(i), ri sin(i)) for i = 1, 2, . . . , n.Let now Q be any point in the plane, denoted by Q (r cos , r sin ). Clearly,

QAi =

r2 + ri2 2rri cos( i) =

(ri r cos( i))2 + r2 sin2( i)

ri r cos( i),with equality iff either r = 0, or i is an integral multiple of . Adding all such inequalities, andsubtracting the sum of all distances to P, we find

ni=1

QAi ni=1

P Ai ri r cos( i) = rni=1

cos( i) = 0,

where we have used that, as it is well-known, the last sum is zero (it is equivalent to the statement "a regularn-gon exists such that one of its sides forms an angle with one of the axes"). Note that, since i cannotbe simultaneously a multiple of for i = 1, 2, . . . , n (the polygon could then have at most two vertices,absurd), we conclude that equality holds iff r = 0, ie iffQ = P.

Also solved by Semchankau Aliaksei, Minsk, Belarus; Radouan Boukharfane, Polytechnique de Montreal,

Canada; Li Zhou, Polk State College, USA.


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O263. A tournament T with a linear order < on its vertices is called an ordered tournament and is denoted by(T,

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O264. Let p > 3 be a prime. Prove that 2p1 1 (mod p2) if and only if the numerator of(p1)/2k=2

1

k

1 +

1

2+ + 1

k 1

is a multiple of p.

Proposed by Gabriel Dospinescu, Lyon, France

Solution by G.R.A.20 Problem Solving Group, Roma, Italy

So it suffices to show that

(p1)/2k=2

Hk1(1)k

=1

2

H(p1)/2(1)

2 H(p1)/2(2) 2q2 (mod p)where Hk1(d) = 1 + 12d + + 1(k1)d .Since by Wolstenholmes Theorem Hp1(1) Hp1(2) 0 (mod p) for any prime p > 3, it follows that

2q =2p 2

p=

1

p

p1k=1

pk

p1k=1

(1)k1k

(mod p)

=

p1k=1

1

k 2

(p1)/2k=1

1

2k= Hp1(1) H(p1)/2(1) H(p1)/2(1) (mod p),

and

H(p1)/2(2) =1

2

(p1)/2k=1

1

k2+

1

2

p1k=(p+1)/2

1

(p k)2 1

2Hp1(2) 0 (mod p).

Hence,

12

H(p1)/2(1)2 H(p1)/2(2) 12 (2q)2 + 0 2q2.

Also solved by Semchankau Aliaksei, Minsk, Belarus; Radouan Boukharfane, Polytechnique de Montreal,

Canada.

mr 2 2013 solutions

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