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OCTOGON MATHEMATICAL MAGAZINE

Vol. 20, No.2, October 2012, pp 625-652

Print: ISSN 1222-5657, Online: ISSN 2248-1893

http://www.uni-miskolc.hu/∼matsefi/Octogon/

625

Math Competitions Corner

Jose Luis Dıaz-Barrero 25

No. 2

This section of the Journal offers readers an opportunity to solve interestingmathematical problems appeared previously in High School MathematicalOlympiads and University Competitions or used by trainers and contestantsto prepare Math Competitions. Elegant solutions, generalizations of theproblems posed and new suitable proposals are always welcomed. Proposalsshould be accompanied by solutions. The origin of the problems appearedpreviously will be revealed when the solutions are published.Send submittals to: Jose Luis Dıaz-Barrero, Applied Mathematics III,Barcelona Tech, Jordi Girona 1-3, C2, 08034 Barcelona, Spain or by e-mail(preferred) to: <[email protected]>Solutions to the problem stated in this issue should be posted before May 15,2003

PROBLEMS

MC–21. Find all real solutions of the following system of equations

x1x2 + 1 = 4x2,

x2x3 + 1 = x3,

x3x4 + 1 = 4x4,

. . .

x2011 · x2012 + 1 = 4x2012,

x2012 · x1 + 1 = x1.

25Received: 21.09.20122010 Mathematics Subject Classification. 11-06.Key words and phrases. Contest.

626 Jose Luis Dıaz-Barrero

MC–22. Is there a polynomial A(x) with real coefficients such that

A(m) = 2012m

for all positive integer m?

MC–23. Find all the integers a > 1 such that any prime divisor of a6 − 1 isa divisor either of a3 − 1 or a2 − 1.

MC–24. Suppose N is the sum of the squares of three positive integers.For all positive integer n, show that N2n

is also the sum of the squares ofthree positive integers.

MC–25. Let a, b, c be the lengths of the sides of a triangle ABC. Provethat √

a+ b− c

a+ b+ c+

√b+ c− a

a+ b+ c+

√c+ a− b

a+ b+ c≤

√3

MC–26. Let k be a positive integer. Compute

1

k!

∞∑

n1=1

∞∑

n2=1

. . .

∞∑

nk=1

1

n1n2 . . . nk(n1 + n2 + . . .+ nk + 1)

MC–27. Let P0, P1, P2, . . . be a sequence of convex polygons such that, foreach k ≥ 0, the vertices of Pk+1 are the midpoints of all sides of Pk. Provethat there exists a unique point lying inside all these polygons.

MC–28. Calculate

limn→∞

1

2n

n∑

k=0

{(−1)k

(n

k

)( √k√

k +√n

)}

MC–29. Let a, b, c be three complex numbers such thata|bc| + b|ca| + c|ab| = 0. Prove that

|(a− b)(b− c)(c − a) ≥ 3√

3 |abc|

Math Competitions Corner 627

MC–30. Let n be a positive integer. Compute

S(n) =

=n∑

k=1

[arctg

(k2 + k − 1

(k2 + k + 2)(k2 + k + 1)

)arctg

(k4 + 2k3 + 2k2 + k + 2

(2k + 1)(k2 + k + 1)

)]

MC–31. Determine whether the real numberln(11 + 5

√2)

ln(5 + 11√

2)is rational or

not.

MC–32. Prove that for all A ∈ M2(R) there exist X,Y ∈ M2(R) suchthat A = X3 + Y 3 and XY = Y X.

MC–33. Let a1, a2, . . . , an, and b1, b2, . . . , bn be positive real numbers.Prove that

(n∑

i=1

bi

)Pni=1 bi n∏

i=1

(ai + bi

)bi ≤(

n∑

i=1

ai +

n∑

i=1

bi

)Pni=1 bi n∏

i=1

bbii

When does equality occur?

MC–34. Let A1, A2, . . . , An be finite sets. Prove that

∣∣∣∣∣

n⋂

i=1

Ai

∣∣∣∣∣ =∑

i

|Ai|−∑

i<j

|Ai ∪Aj|+∑

i<j<k

|Ai ∪Aj ∪Ak|− . . .+(−1)n+1

∣∣∣∣∣

n⋃

i=1

Ai

∣∣∣∣∣

MC–35. Compute

limn→∞

n∏

i=1

(i∑

k=1

3k2 + 9k + 7

(k + 1)3(k + 2)3

)


SOLUTIONS

No problem is ever permanently closed. We will be very pleased consideringfor publication new solutions or comments on the past problems.

MC–1. Find all functions f : N → [0,+∞) such that f(1000) = 10 and

f(n+ 1) =

n∑

k=0

1

f2(k) + f(k)f(k + 1) + f2(k + 1)

for all integer n ≥ 0. (Here, f2(i) means (f(i))2.)

(IMAC – 2011)

Solution 1. Note that

f(n+ 1) =

n−1∑

k=0

1

f2(k) + f(k)f(k + 1) + f2(k + 1)

+1

f2(n) + f(n)f(n+ 1) + f2(n+ 1)

= f(n) +1

f2(n) + f(n)f(n+ 1) + f2(n+ 1)

From where

(f(n+ 1) − f(n))(f2(n) + f(n)f(n+ 1) + f2(n+ 1)

)= 1

Implying that

f3(n+ 1) − f3(n) = 1.

That is, f3(n) = f3(n+ 1) − 1. Since f(1000) = 10 = 3√

1000, then byinduction it is easy to obtain that the function f is f(n) = 3

√n for n ∈ N: if

f(m) = 3√m, for m > 1, then it also holds that

f(m− 1) = 3√f3(m) − 1 = 3

√m− 1 and f(m+ 1) = 3

√f3(m) + 1 = 3

√m+ 1.

Jose Gabriel Alonso (student), 4◦ E.S.O., Colegio Garoe and Angel Plaza,Universidad de Las Palmas de Gran Canaria, Spain.


Solution 2. We have

f(n+ 1) − f(n) =

n∑

k=0

1

f2(k) + f(k)f(k + 1) + f2(k + 1)

−n−1∑

k=0

1

f2(k) + f(k)f(k + 1) + f2(k + 1)

=1

f2(n) + f(n)f(n+ 1) + f2(n+ 1)

Rearranging terms, we get f3(n+ 1) − f3(n) = 1 from which follows

f3(n+ 1) = 1 + f3(n) = 2 + f3(n− 1) = . . . = (n+ 1) + f3(0)

Putting n = 999, we get f3(1000) = 1000 + f3(0) from which followsf3(0) = 0 and f(0) = 0. So, f3(n+ 1) = n+ 1 and f(n) = 3

√n.

Mihaly Bencze, Brasov, Romania

Also solved by Mohammad W. Alomari, Jerash University, Jordan, ArnauMessegue, Barcelona Tech, Barcelona, Spain and Jose Luis Dıaz-Barrero,Barcelona Tech, Barcelona, Spain.

MC–2. Let ABDC be a cyclic quadrilateral inscribed in a circle C. Let Mand N be the midpoints of the arcs AB and CD which do not contain C andA respectively. If MN meets side AB at P, then show that

AP

BP=AC +AD

BC +BD

(IMAC – 2011)

Solution. Applying Ptolemy’s theorem to the inscribed quadrilateralACND, we have

AD · CN +AC ·ND = AN · CD


Since N is the midpoint of the arc CD, then we have CN = ND = x, andAN · CD = (AC +AD)x. Likewise, considering the inscribed quadrilateralCNDB we have BN · CD = (BD +BC)x. Dividing the precedingexpressions, yields

AN

BN=AC +AD

BD +BC

Applying the bisector angle theorem, we have

AN

BN=AP

BP

This completes the proof.

Ivan Geffner, Barcelona Tech, Barcelona, Spain

Also solved by Jose Luis Dıaz-Barrero, Barcelona Tech, Barcelona, Spain.

MC–3. Let f : R → R be a continuous function on R. A point x is called ashadow point of f if and only if there exists y ∈ R, y > x such thatf(y) > f(x). Suppose that all the points of the open interval I = (a, b),(a < b) are shadow points of f and a and b, are not shadow points of f.Prove that

(i) f(x) ≤ f(b) for all a < x < b.

(ii) f(a) = f(b).

(IMC – 2011)

Solution. (i) We argue by contradiction. Indeed, suppose that existsx1 ∈ (a, b) such that f(x1) > f(b). Since b it is not a shadow point of f , thenfor all y > b is f(y) ≤ f(b). That is, we have for all y > b,


f(x1) > f(b) ≥ f(y) (1)

Since x1 is a shadow point of f , then exists a point x2 > x1 such thatf(x2) > f(x1). By (1) is x1 < x2 < b. Since x2 is also a shadow point of f ,then exists x3 such that x2 < x3 < b with f(x3) > f(x2). Carrying out thisprocedure we can build up an increasing sequence {xn}n≥1 bounded by b. So,{xn}n≥1 is convergent. Let limn→+∞ xn = x. We claim that x = b. Indeed,for all n is xn < b and limn→+∞ xn ≤ b. If were x < b, then must exits xn

such that x < xn < b which contradicts the way in what {xn}n≥1 was builtup. So, x = b.

Since f is continuous in R, then limn→+∞ f(xn) = f(b). But we havesupposed the f(xn) > f(x1) > f(b). Hence, we have

limn→+∞

f(xn) = f(b) ≥ f(x1) > f(b)

Contradiction. So, for all x ∈ (a, b) is f(x) ≤ f(b).

(ii) Let {yn}n≥1 be a sequence of points of (a, b) such that limn→+∞ yn = a,then

limn→+∞

f(yn) = f(a)

By (i) is f(yn) ≤ f(b). So, f(a) = limn→+∞ f(yn) ≤ f(b). Now, we will seethat f(a) = f(b). Indeed, if f(a) < f(b), then a will be a shadow point of f.But a it not a shadow point of f , so f(a) = f(b) and we are done.

Jose Luis Dıaz-Barrero, Barcelona Tech, Barcelona, Spain

Also solved by Ivan Geffner Fuenmayor, Barcelona Tech, Barcelona, Spain.

MC–4. Place n points on a circle and draw in all possible chord joiningthese points. If no three chord are concurrent, find (with proof) the numberof disjoint regions created.

(IMAC – 2011)

Solution. First, we prove that if a convex region crossed by L lines with Pinterior points of intersection, then the number of disjoint regions created isRL = L+ P + 1. To prove the preceding claim, we argue by MathematicalInduction on L. Let R be an arbitrary convex region in the plane. For each

L ≥ 0, let A(L) be the statement that for each P ∈{1, 2, . . . ,

(L2

)}, if L lines


that cross R, with P intersection points inside R, then the number ofdisjoint regions created inside R is RL = L+ P + 1.When no lines intersect R, then P = 0, and so, R0 = 0 + 0 + 1 = 1 and A(0)holds. Fix some K ≥ 0 and suppose that A(K) holds for K lines and someP ≥ 0 with RK = K + P + 1 regions. Consider a collection C of K + 1 lineseach crossing R (not just touching), choose some line ℓ ∈ C, and apply A(K)to C\{ℓ} with some P intersection points inside R and RK = K + P + 1regions. Let S be the number of lines intersecting ℓ inside R. Since onedraws a (K + 1)−st line ℓ, starting outside R, a new region is created when ℓfirst crosses the border of R, and whenever ℓ crosses a line inside of R.Hence the number of new regions is S + 1. Hence, the number of regionsdetermined by the K + 1 lines is, on account of A(K),

RK+1 = RK + S + 1 = (K + P + 1) + S + 1 = (K + 1) + (P + S) + 1,

where P + S is the total number of intersection points inside R. Therefore,A(K + 1) holds and by the PMI the claim is proven. Finally, since the circleis convex and any intersection point is determined by a unique 4−tuple ofpoints, then there are P =

(n4

)intersection points and L =

(n2

)chords and

the number of regions is R =(n4

)+(n2

)+ 1.


Also solved by Ivan Geffner Fuenmayor, Barcelona Tech, Barcelona, Spain.

MC–5. Let a, b, c ∈ (0,+∞) such that a+ b+ c = 1. Prove that

a

a3 + b2c+ c2b+

b

b3 + c2a+ a2c+

c

c3 + a2b+ b2a≤ 1 +

8

27abc

(IMAC – 2011)

Solution. To prove the the inequality claimed, we will apply CBSinequality. Indeed, we have

(∑

cyc

a

)2

=

(∑

cyc

√x

√a2

x

)2

≤(∑

cyc

x

)(∑

cyc

a2

x

)

Writing the preceding in the most convenient form

∑

cyc

a2

x≥(∑

cyc

a

)2/(∑

cyc

x

)


we have

a3 + b2c+ c2b =a2

1a

+b2

1c

+c2

1b

≥(∑

cyc

a

)2/(∑

cyc

1

a

)

from which immediately follows

a

a3 + b2c+ c2b= a

∑

cyc

1

a

/(∑

cyc

a

)2

So, on account of the preceding and the constrain, we have

L ≤(∑

cyc

a

)(∑

cyc

1

a

)/(∑

cyc

a

)2

=

(∑

cyc

1

a

)/(∑

cyc

a

)

=∑

cyc

1

a=ab+ bc+ ca

abc,

where

L =a

a3 + b2c+ c2b+

b

b3 + c2a+ a2c+

c

c3 + a2b+ b2a

Now, applying Jensen’s inequality to the convex function f : (0,+∞) → Rdefined by f(t) = t3, we get 9(a3 + b3 + c3) ≥ (a+ b+ c)3. Taking intoaccount the well-known identity

a3 + b3 + c3 = (a+ b+ c)[(a2 + b2 + c2) − (ab+ bc+ ca)] + 3abc

= (a+ b+ c)[(a + b+ c)2 − 3(ab+ bc+ ca)] + 3abc

and again, on account of the constrain, we have

a3 + b3 + c3 = 1 − 3(ab + bc+ ca)] + 3abc;

and 9(a3 + b3 + c3) ≥ (a+ b+ c)3 becomes 27(abc − (ab+ bc+ ca)) + 8 ≥ 0from which follows ab+ bc+ ca ≤ 8

27 + abc. Finally, we have

a

a3 + b2c+ c2b+

b

b3 + c2a+ a2c+

c

c3 + a2b+ b2a≤ ab+ bc+ ca

abc≤ 1 +

8

27abc

Equality holds when a = b = c = 1/3 and we are done.


Nicolae Papacu, Slobozia, Romania


MC–6. Let m and n be distinct integer numbers. Find all functionsf : Z → R such that f(mx+ ny) = mf(x) + nf(y) for all x, y ∈ Z.

(IMAC – 2009)

Solution. We distinguish two cases. First, we suppose that m+ n 6= 1.Then, for x = y = 0 we obtain f(0) = (m+ n)f(0) from which followsf(0) = 0. For y = 0 we get f(mx)mf(x) for all x ∈ Z. For x = 0 we getf(ny) = nf(y) for all y ∈ Z. Now we havef(mnx+mny) = mf(nx) + nf(my) = mn(f(x) + f(y)), for all x, y ∈ Z andf(x+ y) = f(x) + f(y) or f(kx) = kf(x) for all k, x ∈ Z from which followsthat f(x) = cx, x ∈ Z and c ∈ R is solution of the claimed functionalequation.

Sorin Radulescu, Bucharest, Romania


MC–7. Let a, b, c be three positive real numbers such thata+ b+ c+

√abc = 2. Prove that

a

(b− 2)(c − 2)+

b

(c− 2)(a− 2)+

c

(a− 2)(b − 2)≥ 3

4

(Ibero Longlist – 2009)

Solution. Putting A =

√bc

a,B =

√ca

b, and A =

√ab

c, the constrain

becomes ABC +AB +BC + CA = 2 and the inequality claimed can bewritten as

1

(A+ 1)2 + 1+

1

(B + 1)2 + 1+

1

(C + 1)2 + 1≥ 3

4

Since,

ABC+AB+BC+CA = 2 ⇔ (A+1)(B+1)(C+1) = (A+1)+(B+1)+(C+1),


then exist three numbers α, β, γ ∈ [0, π] such that A+ 1 = tanα,B + 1 = tan β, C + 1 = tan γ, and α+ β + γ = π. Since A > 0, thentanα > 1 and α ∈ (π/4, π/2). Likewise, β ∈ (π/4, π/2) and γ ∈ (π/4, π/2).Moreover, we have

1

(A+ 1)2 + 1+

1

(B + 1)2 + 1+

1

(C + 1)2 + 1

=1

1 + tan2 α+

1

1 + tan2 β+

1

1 + tan2 γ

= cos2 α+ cos2 β + cos2 γ

Now we consider the function f : (π/4, π/2) → R defined by f(x) = cos2 x.Since, f ′(x) = − sin 2x y f ′′(x) = −2 cos 2x > 0 for all x ∈ (π/4, π/2), then fis convex in (π/4, π/2). Applying Jensen’s inequality, yields

cos2 α+ cos2 β + cos2 γ = f(α) + f(β) + f(γ) ≥ 3f

(α+ β + γ

3

)=

3

4

Equality holds when α = β = γ = π/3. That is, when A = B = C =√

3 − 1or a = b = c = 4 − 2

√3.

Diego Izquierdo, Madrid, Spain


MC–8. Let a, b be positive real numbers and let f : [a, b] → R be acontinuous function. Prove that there exits c ∈ (a, b) such that

1

2f(c) =

(c

a2 − c2+

c

b2 − c2

)∫ c

af(t) dt

(Spanish training for IMC – 2005)

Solution. Consider the function F : [a, b] → R defined by

F (x) = (x2 − a2)(x2 − b2)

∫ x

af(t) dt. The function F (x) is continuous in

[a, b], derivable in (a, b) and F (a) = F (b) = 0. Therefore, according to Rolle’stheorem, we have that there exists c ∈ (a, b) such that F ′(c) = 0. That is,

2c(c2 − b2)

∫ c

af(t) dt+ 2c(c2 − a2)

∫ c

af(t) dt+ (c2 − a2)(c2 − b2)f(c) = 0


or equivalently

2c[(c2 − a2) + (c2 − b2)

] ∫ c

af(t) dt+ (c2 − a2)(c2 − b2)f(c) = 0.

Dividing both sides by (c2 − a2)(c2 − b2), we get(

2c

c2 − a2+

2c

c2 − b2

)∫ c

af(t) dt+ f(c) = 0

and we are done.


Also solved by Mohammad W. Alomari, Jerash University, Jordan, ArnauMessegue, Barcelona Tech, Barcelona, Spain.

MC–9. Let A(z) =∑n

k=0 akzk be a polynomial of degree n with complex

coefficients having all its zeros in the disk C = {z ∈ C : |z| ≤√

6}. Show that

∣∣∣A(3z)∣∣∣ ≥

(3 +

√6

2 +√

6

)n ∣∣∣A(2z)∣∣∣

for any complex number z with |z| = 1.

(IMC Longlist – 2009)

Solution. Let z1, z2, . . . , zn be the zeros (not necessarily distinct) of A(z).Then zk = rke

iθk , 1 ≤ k ≤ n, 0 ≤ θ < 2π and rk ≤√

6. So, we can write

A(z) = C

n∏

k=1

(z − zk) = C

n∏

k=1

(z − rkeiθk)

For any z = eiθ, 0 ≤ θ < 2π, we have∣∣∣∣A(2z)

A(3z)

∣∣∣∣ =∣∣∣∣A(2eiθ)

A(3eiθ)

∣∣∣∣ =n∏

k=1

∣∣∣∣2eiθ − rke

iθk

3eiθ − rkeiθk

∣∣∣∣ =n∏

k=1

∣∣∣∣∣2ei(θ−θk) − rk3ei(θ−θk) − rk

∣∣∣∣∣

On the other hand, for 1 ≤ k ≤ n, we have∣∣∣∣∣2ei(θ−θk) − rk3ei(θ−θk) − rk

∣∣∣∣∣

2

=

(2ei(θ−θk) − rk3ei(θ−θk) − rk

)(2ei(θ−θk) − rk3ei(θ−θk) − rk

)

=4 + r2k − 4rk cos(θ − θk)

9 + r2k − 6rk cos(θ − θk)

≤(

2 + rk3 + rk

)2


Indeed, the last inequality

4 + r2k − 4rk cos(θ − θk)

9 + r2k − 6rk cos(θ − θk)≤ 4 + r2k + 4rk

9 + r2k + 6rk

is equivalent to

12rk(1 + cos(θ − θk)) ≥ 2r3k(1 + cos(θ − θk)),

or 6 ≥ r2k which is true because rk ≤√

6, 1 ≤ k ≤ n. Now, immediatelyfollows that

∣∣∣∣A(2z)

A(3z)

∣∣∣∣ ≤n∏

k=1

(2 + rk3 + rk

)≤

n∏

k=1

(2 +

√6

3 +√

6

)

=

(2 +

√6

3 +√

6

)n

Equality holds for the polynomial A(z) = (z +√

6)n and we are done.


Also solved by Jose Gibergans-Baguena, Barcelona Tech, Barcelona, Spain.

MC–10. Compute

limn→∞

ln

[1

2n

n∏

k=1

(2 +

k

n2

)]

(Spanish Training for IMC – 2009)

Solution 1. It is easy to see that

n∏

k=1

(2 +

k

n2

)=

(2 +

1

n2

)(2 +

2

n2

)· · ·(2 +

n

n2

)

=(2n2 + 1)(2n2 + 2) · · · (2n2 + n)

n2n=

Γ(2n2 + n+ 1)

n2nΓ(2n2 + 1),

where Γ denotes the gamma function. From the asymptotic expansion:

ln Γ(x) =(x− 1

2

)lnx− x+ ln

√2π +

1

12x+O(x−3) (x→ ∞)


(see [1] p. 257, 6.1.41]), we find that

ln

[1

2n

n∏

k=1

(2 +

k

n2

)]

= ln

[Γ(2n2 + n+ 1)

2nn2nΓ(2n2 + 1)

]

= ln Γ(2n2 + n+ 1) − ln Γ(2n2 + 1) − n ln 2 − 2n lnn

∼ 1

4+

5

24n+O

(1

n2

)(n → ∞).

Hence,

limn→∞

ln

[1

2n

n∏

k=1

(2 +

k

n2

)]

=1

4.

[1] M. Abramowitz and I. A. Stegun (Editors), Handbook of MathematicalFunctions with Formulas, Graphs, and Mathematical Tables, AppliedMathematics Series 55, Ninth printing, National Bureau of Standards,Washington, D.C., 1972.

Chao-Ping Chen, School of Mathematics and Informatics, Henan PolytechnicUniversity, People’s Republic of China

Solution 2. We have that

ln

[1

2n

n∏

k=1

(2 +

k

n2

)]

=

n∑

k=1

ln

(1 +

k

2n2

),

and since

x− x2

2< ln(1 + x) < x ∀x > 0,

thenn∑

k=1

k

2n2− 1

2

n∑

k=1

k2

4n4< ln

[1

2n

n∏

k=1

(2 +

k

n2

)]<

n∑

k=1

k

2n2.

Butn∑

k=1

k

2n2=

1

2n2

n∑

k=1

k =n(n+ 1)

4n2−→ 1

4

andn∑

k=1

k2

4n4=

1

4n4

n∑

k=1

k2 =n(n+ 1)(2n + 1)

24n4−→ 0,


so

ln

[1

2n

n∏

k=1

(2 +

k

n2

)]−→ 1

4

Xavier Ros, Barcelona Tech, Barcelona, Spain

Solution 3. We begin with a lemma.

Lemma. Let f : I ⊆ R → R and let a ∈ I. If f is derivable in a then thesequence {xn}n≥1 defined by

xn =

n∑

k=1

f

(a+

k

n2

)− nf(a)

is convergent and limn→∞

xn =1

2f ′(a).

Proof. Since f is derivable in a, then f ′(a) = limx→a

f(x) − f(a)

x− a. This means,

as it is well known, that for all ǫ > 0 there exists δ = δ(ǫ) > 0 such that forall x ∈ (a− δ, a+ δ), (x 6= a) or 0 < |x− a| < δ is

∣∣∣∣f(x) − f(a)

x− a− f ′(a)

∣∣∣∣ < ǫ⇔ f ′(a) − ǫ <f(x) − f(a)

x− a< f ′(a) + ǫ

Let x = a+k

n2, (1 ≤ k ≤ n) then there exists n1 ∈ N such that for all n ≥ n1

we have 0 < |x− a| =k

n2≤ 1

n. Thus, for all k ∈ {1, 2, · · · , n}, we have

(f ′(a) − ǫ

) k

n2< f

(a+

k

n2

)− f(a) <

(f ′(a) + ǫ

) k

n2

Adding up the preceding inequalities, we get

(f ′(a) − ǫ

) n+ 1

2n<

n∑

k=1

f

(a+

k

n2

)− nf(a) <

(f ′(a) + ǫ

) n+ 1

2n

or (f ′(a) − ǫ

) n+ 1

2n< xn <

(f ′(a) + ǫ

) n+ 1

2n⇔

f ′(a)2n

− ǫ(n+ 1)

2n< xn − 1

2f ′(a) <

f ′(a)2n

+ǫ(n+ 1)

2n


Since limn→∞

f ′(a)2n

= 0, then immediately follows that

limn→∞

xn =1

2f ′(a)

�

Now, we set f : (−1,+∞) → R defined by f(x) = ln(1 + x) and a = 1 intothe previous lemma, and we get

xn =n∑

k=1

ln

(2 +

k

n2

)− n ln 2 = ln

[1

2n

n∏

k=1

(2 +

k

n2

)]

Since f ′(x) =1

1 + xthen f ′(1) =

1

2and

limn→∞

ln

[1

2n

n∏

k=1

(2 +

k

n2

)]

=1

2f ′(1) =

1

4


Also solved by Angel Plaza, Universidad de Las Palmas de Gran Canaria,Spain, Jose Gibergans-Baguena, Barcelona Tech, Barcelona, Spain.

MC–11. Let a, b be positive integers. Prove that

ϕ(ab)√ϕ2(a2) + ϕ2(b2)

≤√

2

2,

where ϕ(n) is the Euler’s totient function.


Solution. Let a =

p∏

i=1

pαii

q∏

j=1

qβj

j and b =

p∏

i=1

pαii

r∏

k=1

rγkk being αi, βj and γk

nonnegative integers. Then,

ϕ(a) = a

p∏

i=1

(1 − 1

pi

) q∏

j=1

(1 − 1

qj

), ϕ(b) = b

p∏

i=1

(1 − 1

pi

) r∏

k=1

(1 − 1

rk

)

On the other hand, ϕ(ab) = ab

p∏

i=1

(1 − 1

pi

) q∏

j=1

(1 − 1

qj

) r∏

k=1

(1 − 1

rk

)and

ϕ(a2) = a2p∏

i=1

(1 − 1

pi

) q∏

j=1

(1 − 1

qj

), ϕ(b2) = b2

p∏

i=1

(1 − 1

pi

) r∏

k=1

(1 − 1

rk

)


Since

q∏

j=1

(1 − 1

qj

) r∏

k=1

(1 − 1

rk

)≤ 1,then

q∏

j=1

(1 − 1

qj

) r∏

k=1

(1 − 1

rk

)≤

√√√√q∏

j=1

(1 − 1

qj

) r∏

k=1

(1 − 1

rk

)

Taking into account GM-QM inequality, we have

ϕ(ab) ≤√ϕ(a2)ϕ(b2) ≤

√ϕ2(a2) + ϕ2(b2)

2

from which the statement follows.



MC–12. If xi > 0 and αi ∈ R, then

Vn(x, α) =

∣∣∣∣∣∣∣∣∣

xα11 xα1

2 . . . xα1n

xα21 xα2

2 . . . xα2n

......

. . ....

xαn1 xαn

2 . . . xαnn

∣∣∣∣∣∣∣∣∣

is called a generalized Vandermonde determinant of order n. Let0 < x1 < x2 < . . . < xn, and α1 < α2 < . . . < αn be real numbers. Prove thatVn(x, α) > 0.


Solution. We will argue by mathematical induction. The case when n = 1trivially holds. For n = 2, using Lagramnge’s mean value theorem, we have

V2(x, α) =

∣∣∣∣xα1

1 xα12

xα21 xα2

2

∣∣∣∣ = xα11 xα2

2 − xα21 xα1

2

= xα11 xα1

2 (xα2−α12 xα2−α1

1 ) = xα11 xα1

2 (x2 − x1)dtα2−α1

dt

∣∣∣t=ξ

= xα11 xα1

2 (x2 − x1)(α2 − α1)ξα2−α1−1 > 0,


where x1 < ξ < x2. Suppose that the inequality holds for n− 1 with n ≥ 2.Namely,

Vn−1(y, α) =

∣∣∣∣∣∣∣∣∣

yα12 yα1

3 . . . yα1n

yα22 yα2

3 . . . yα2n

......

. . ....

yαn2 yαn

3 . . . yαnn

∣∣∣∣∣∣∣∣∣

> 0,

where 0 < y2 < y3 < . . . < yn and α2 < α3 < . . . < αn. Let

f(xn) =

∣∣∣∣∣∣∣∣∣

1 1 . . . 1

xα2−α11 xα2−α1

2 . . . xα2−α1n

......

. . ....

xαn−α11 xαn−α1

2 . . . xαn−α1n

∣∣∣∣∣∣∣∣∣

Applying Lagrange’s mean value theorem again, we get

f(xn) = f(xn) − f(xn−1) = (xn − xn−1)f′(ξn)

= (xn − xn−1

∣∣∣∣∣∣∣∣∣∣∣

1 1 . . . 0

xα2−α1

1 xα2−α1

2 . . . dxα2−α1

dt

∣∣∣t=ξ

......

. . ....

xαn−α1

1 xαn−α1

2 . . . dxαn−α1

dt

∣∣∣t=ξ

∣∣∣∣∣∣∣∣∣∣∣

= (xn − xn−1)

∣∣∣∣∣∣∣∣∣

1 1 . . . 1 0xα2−α1

1 xα2−α1

2 . . . xα2−α1

n−1 (α2 − α1)ξα2−α1−1n

......

. . ....

...xαn−α1

1 xαn−α1

2 . . . xαn−α1

n−1 (αn − α1)ξαn−α1−1n

∣∣∣∣∣∣∣∣∣

,

where xn−1 < ξn < xn. Applying the same procedure, it is easy to obtain

f(xn) =

n∏

j=k−1

(xj − xj−1)

∣∣∣∣∣∣∣∣∣

1 1 . . . 0

xα2−α1

1 (α2 − α1)ξα2−α1−12 . . . (α2 − α1)ξ

α2−α1−1n

......

. . ....

xαn−α1

1 (αn − α1)ξα2−α1−1n . . . (αn − α1)ξ

αn−α1−1n

∣∣∣∣∣∣∣∣∣

=

n∏

j=2

(xj − xj−1)

n∏

i=2

(αi − α1)

∣∣∣∣∣∣∣∣∣

ξα2−α1−12 ξα2−α1−1

3 . . . ξα2−α1−1n

ξα3−α1−12 ξα3−α1−1

3 . . . ξα2−α1−1n

......

. . ....

ξαn−α1−12 ξαn−α1−1

3 . . . ξαn−α1−1n

∣∣∣∣∣∣∣∣∣

=

n∏

j=2

[(αj − α1)(xj − xj−1)ξ

−(α1+1)]Vn−1(ξ, α),


where 0 < x1 < x2 < . . . < xn−1 < ξn < xn and α2 < α3 < . . . < αn. Forgeneral n from the preceding, we have

Vn(x, α) = f(xn)n∏

j=1

xα1j

= Vn−1(ξ, α)

n∏

j=1

xα1j

n∏

j=2

[(αj − α1)(xj − xj−1)ξ

−(α1+1)]> 0,

where 0 < x1 < ξ1 < x2 < . . . < xn−1 < ξn < xn and α1 < α2 < . . . < αn.Therefore, by the PMI the statement is proved.



MC–13. Let X be a set of cardinal n. Determine

∑

(A,B)⊆X×X

card (A ∩B)


Solution. Given A ⊆ X and an integer i, let us calculate the number ofways to choose B ⊆ X such that

card(A ∩B) = i.

Let k be the number of elements of A. The set B must have i elements in Aand some other elements not in A. Then, to calculate the number of ways tochoose B, we have

(ki

)ways to choose i elements from A, and 2n−k ways to

choose the others that are not in A. Hence, we have

2n−k

(k

i

)

ways to choose B. Therefore, given A ⊆ X such that card(A) = k,

∑

B⊆X

card(A ∩B) =

k∑

i=0

i2n−k

(k

i

)= 2n−k

k∑

i=0

i

(k

i

).


Now, for each k, we can choose A in(nk

)different ways, and then,

∑

(A,B)⊆X×X

card(A ∩B) =∑

A⊆X

∑

B⊆X

card(A ∩B) =

=n∑

k=0

(n

k

) ∑

B⊆X

card(A ∩B) =n∑

k=0

(n

k

)2n−k

k∑

i=1

i

(k

i

).

Now, taking derivatives in both sides of the well known equality

(1 + x)k =

n∑

i=0

xi

(k

i

),

we get that

k(1 + x)k−1 =

n∑

i=1

ixi−1

(k

i

),

and setting x = 1,k∑

i=1

i

(k

i

)= k2k−1.

Hence,

n∑

k=0

(n

k

)2n−k

k∑

i=1

i

(k

i

)=

n∑

k=0

(n

k

)2n−kk2k−1 = 2n−1

n∑

k=0

k

(n

k

)= 2n−1n2n−1,

and then, ∑

(A,B)⊆X×X

card(A ∩B) = n4n−1.

Xavier Ros, Barcelona Tech, Barcelona, Spain


MC–14. Compute ∫ ∞

1

dt

2[t] + 3[t]2 + [t]3,

where [x] represents the integer part of x.

(Spanish Training Team for IMC – 2008)


Solution 1.∫ ∞

1

dt

2[t] + 3[t]2 + [t]3= lim

N→∞

∫ N

1

dt

2[t] + 3[t]2 + [t]3

= limN→∞

N−1∑

k=1

∫ k+1

k

dt

2[t] + 3[t]2 + [t]3

= limN→∞

N−1∑

k=1

1

2k + 3k2 + k3

= limN→∞

N−1∑

k=1

(1/2

k− 1

k + 1+

1/2

k + 2

)

= limN→∞

1

2− 1

4− 1/2

N+

1/2

N + 1=

1

4.

Where we have used that [t] = k for t ∈ [k, k + 1), and the decomposition

into simple fractions of1

2k + 3k2 + k3.

Angel Plaza, Universidad de Las Palmas de Gran Canaria, Spain

Solution 2. Let f(t) =1

2[t] + 3[t]2 + [t]3. If n ≤ t < n+ 1, the [t] = n and

∫ n+1

nf(t) dt =

∫ n+1

n

dt

2n+ 3n2 + n3

=1

n(n+ 1)(n + 2)

∫ n+1

ndt =

1

n(n+ 1)(n + 2)

Therefore,

∫ n

1f(t) dt =

n−1∑

k=1

∫ k+1

kf(t) dt =

n−1∑

k=1

1

k(k + 1)(k + 2)=

1

4− 1

2n(n+ 1)

Taking limits when n→ ∞, we have∫ ∞

1

dt

2[t] + 3[t]2 + [t]3= lim

n→∞

∫ n

1f(t) dt

= limn→∞

(1

4− 1

2n(n+ 1)

)=

1

4

and we are done.



Also solved by Arnau Messegue, Jose Gibergans-Baguena, Barcelona Tech,Barcelona, Spain.

MC–15. Find all triplets (x, y, z) of real numbers such that

x2 +√y2 + 12 =

√y2 + 60,

y2 +√z2 + 12 =

√z2 + 60,

z2 +√x2 + 12 =

√x2 + 60.

(Spanish First Stage Contest – 2009)

Solution. Putting x2 = a, y2 = b and z2 = c, yields

a+√b+ 12 =

√b+ 60,

b+√c+ 12 =

√c+ 60,

c+√a+ 12 =

√a+ 60.

⇔a =

√b+ 60 −

√b+ 12,

b =√c+ 60 −

√c+ 12,

c =√a+ 60 −

√a+ 12.

Now we consider the function f : [0,+∞) → R defined by

f(t) =√t+ 60 −

√t+ 12 =

48√t+ 60 +

√t+ 12

. Since for 0 ≤ u < v is

48√u+ 60 +

√u+ 12

>48√

v + 60 +√v + 12

,

then f is decreasing and the same occurs with f(f(f(t))). On the otherhand, from f(b) = a, f(c) = b and f(a) = c we get f(f(f(a))) = a which ispossible if and only if f(a) = a. Next, we find the fixed points of f. That is,we have to solve the equation

√t+ 60 −

√t+ 12 = t⇔ 48 − t2 = 2t

√t+ 12

Since 48 − t2 ≥ 0 then t ∈ [0, 4√

3] from which follows

t4 − 4t3 − 144t2 + 2304 = (t− 4)(t3 − 144t− 576) = 0

Since for all t ∈ [0, 4√

3] is t3 < 144t+ 576, then t = 4 is the only fixed pointof f and the solutions are (a, b, c) = (4, 4, 4). Taking into account thatx2 = a, y2 = b, z2 = c, we get that the solutions of the system are

(2, 2, 2), (−2, 2, 2), (2,−2, 2), (2, 2,−2),(−2,−2, 2), (−2, 2,−2), (2,−2,−2), (−2,−2,−2).




MC–16. Let p and q be two prime numbers and let r be a whole number.Find all possible values of p, q, r for which

1

p+ 1+

1

q + 1− 1

(p + 1)(q + 1)=

1

r

(OMCC – 2011)

Solution. We have,

1

p+ 1+

1

q + 1− 1

(p+ 1)(q + 1)=

p+ q + 1

(p + 1)(q + 1)

The fractionp+ q + 1

(p+ 1)(q + 1)will be equal to

1

rwith r integer, when

p+q+1|(p+1)(q+1) ⇐⇒ p+q+1|(p+1)(q+1)− (p+q+1) ⇐⇒ p+q+1|pq

Since the only divisors of pq are 1, p, q, pq, then p+ q + 1|pq in the followingcases: (1) p+ q + 1 = 1 which is impossible; (2) p+ q + 1 = p which is alsoimpossible; (3) p+ q + 1 = q impossible; and (4) p+ q + 1 = pq.In the last case we have

pq − p− q − 1 = 0 ⇐⇒ pq − p− q + 1 = 2 ⇐⇒ (p− 1)(q − 1) = 2

The only solutions of the preceding equation are (2, 3) and (3, 2), andtherefore, the unique numbers that satisfy the statement are p = 2, q = 3and r = 2, and also p = 3, q = 2 y r = 2.

Guillem Alsina Oriol, Barcelona Tech, Barcelona, Spain


MC–17. Solve in R the following system of equations:

√x+

√y +

√z = 3

x√x+ y

√y + z

√z = 3

x2√x+ y2√y + z2√z = 3


(Spanish Training Team Second Stage – 2008)

Solution. Applying Cauchy’s inequality (~u · ~v)2 ≤ ||~u||2||~v||2 to the vectors~u = ( 4

√x, 4

√y, 4

√z) and ~v = (x 4

√x, y 4

√y, z 4

√z), we get

(x√x+ y

√y + z

√z)2 ≤ (

√x+

√y +

√z)(x2√x+ y2√y + z2√z)

or 32 ≤ 3 · 3. Equality holds when vectors ~u and ~v are collinear. That is,when

x 4√x

4√x

=y 4√y

4√y

=z 4√z

4√z

= k

or, x = y = z = k. Then, from√x+

√y +

√z = 3, we get 3

√k = 3. That is,

k = 1 and the only solution is (x, y, z) = (1, 1, 1).

Jose Gibergans-Baguena, Barcelona Tech, Barcelona, Spain

Also solved by Jose Luis Dıaz-Barrero, Barcelona Tech, Barcelona, Spain,Oscar Rivero Salgado, Barcelona Tech, Barcelona, Spain.

MC–18. Find the biggest positive integer that can not be written in theform 5a+ 503b where a are b nonnegative integer numbers.

(IMAC Longlist – 2007)

Solution. Let A = {5a+ 503b | a, b ∈ Z+}. We have to find the biggestpositive integer not belonging to the set A. To do it, we imagine the positiveintegers written in an array with five rows and infinite columns like thefollowing:

C1 C2 · · · C101 · · · C202 · · · C302 · · · C402 C403 · · ·F1 1 6 · · · 501 · · · 1006 · · · 1506 · · · 2006 2011 · · ·F2 2 7 · · · 502 · · · 1007 · · · 1507 · · · 2007 2012 · · ·F3 3 8 · · · 503 · · · 1008 · · · 1508 · · · 2008 2013 · · ·F4 4 9 · · · 504 · · · 1009 · · · 1509 · · · 2009 2014 · · ·F5 5 10 · · · 505 · · · 1010 · · · 1510 · · · 2010 2015 · · ·

The multiples of 503 least than 2515 that are the product of 5 by 503 are503, 1006, 1509 and 2012 respectively (marked in bold in the array). Toknown in which row and column lie theses numbers it is easy to observe that,for instance, 503 = 100 × 5 + 3 will be in row F3 and column Cx where x isthe solution of the equation 5x− 2 = 503. That is, in the column C101.


Likewise, we get the rows and columns where lie the other multiples of 503.That is, C202, C302 y C403.

Once the array is build up it is easy to realize that all the numbers in thelast row belong to A and all the numbers of the preceding rows bigger thanthe ones marked in bold belong to A too. Since the rightest number is 2012,then the number searched is 2007. Now, will be suffice to see that 2007 doesnot belong to A. Indeed, suppose that there exist two nonnegative integers aand b such that 5a+ 503b = 2007 (Notice that 0 ≤ b ≤ 3). Since2007 = 5 × 503 − (5 + 503), then 5 × 503 = 5(a + 1) + 503(b + 1) and5|503(b + 1). Since mcd(5, 503) = 1, then 5|b+ 1 which impossible becauseb+ 1 < 5. Contradiction and we are done.



MC–19. Let x1, x2, · · · , xn be real numbers. Prove that,

(1

n

n∑

k=1

cosh xk

)2

≥ 1 +

(1

n

n∑

k=1

sinhxk

)2

(IMAC Longlist – 2007)

Solution 1. The inequality is equivalent to

(n∑

k=1

coshxk

)2

−(

n∑

k=1

sinhxk

)2

≥ n2 (6)

The left-hand side of (6) may be written as follows:

(n∑

k=1

coshxk

)2

−(

n∑

k=1

sinhxk

)2

=

[(n∑

k=1

cosh xk

)

−(

n∑

k=1

sinhxk

)][(n∑

k=1

cosh xk

)

+

(n∑

k=1

sinhxk

)]

=

[n∑

k=1

(cosh xk − sinhxk)

][n∑

k=1

(coshxk + sinhxk)

]


=

[n∑

k=1

(exk + e−xk

2− exk − e−xk

2

)][ n∑

k=1

(exk + e−xk

2+exk − e−xk

2

)]

=

[n∑

k=1

e−xk

][n∑

k=1

exk

]

Mohammad W. Alomari, Jerash University, Jordan

Solution 2. First, we write the inequality claimed in the most convenientform (

n∑

k=1

coshxk

)2

−(

n∑

k=1

sinhxk

)2

≥ n2

and we argue by mathematical induction. The statement obviously holdswhen n = 1. For n = 2, we also have

(coshx1 + cosh x2)2 − (sinhx1 + sinhx2)

2

= 2 + 2(cosh x1 cosh x2 − sinhx1 sinhx2) = 2 + 2 cosh(x1 − x2) ≥ 4

Assume that the inequality holds for n− 1. We should prove(

n∑

k=1

cosh xk

)2

−(

n∑

k=1

sinhxk

)2

≥ n2.

Indeed, the LHS of the preceding inequality can be written as

(n−1∑

k=1

cosh xk + cosh xn

)2

−(

n−1∑

k=1

sinhxk + sinhxn

)2

=

(n−1∑

k=1

coshxk

)2

+ 2cosh xn

n−1∑

k=1

cosh xk + cosh2 xn

−(

n−1∑

k=1

sinhxk

)2

− 2 sinhxn

n−1∑

k=1

sinhxk − sinh2 xn

≥ (n− 1)2 + 1 + 2

n−1∑

k=1

(cosh xk coshxn − sinhxk sinhxn)

= (n− 1)2 + 1 + 2

n−1∑

k=1

cosh(xn − xk) ≥ n2,

and we are done.




MC–20. Let a, b, c be distinct nonzero complex numbers. Prove that

a2(1 + b2c2)

(a− b)(a− c)+

b2(1 + a2c2)

(b− a)(b− c)+

c2(1 + a2b2)

(c− a)(c− b)

is an integer and determine it.

(Spanish Training Team for IMC – 2002)

Solution 1. The proposed expression, say S, may be written asS = S1 + S2, where

S1 =a2

(a− b)(a− c)+

b2

(b− a)(b− c)+

c2

(c− a)(c− b)

S2 = a2b2c2(

1

(a− b)(a− c)+

1

(b− a)(b− c)+

1

(c− a)(c− b)

).

Note that, since (a− b) + (b− c) + (c− a) = 0, then S2 = 0. We shall showthat S1 = 1 from where we conclude that S = S1 = 1:

S1 =a2(b− c) − b2(a− c) + c2(a− b)

(a− b)(a− c)(b− c)

=a2(b− c) − b2(a− b) − b2(b− c) + c2(a− b)

(a− b)(a− c)(b− c)

=(a2 − b2)(b− c) + (a− b)(c2 − b2)

(a− b)(a− c)(b− c)

=(a− b)(b− c)(a + b− c− b)

(a− b)(a− c)(b− c)= 1.

Jose Gabriel Alonso (student, 4◦ E.S.O. , Colegio Garoe) and Angel Plaza,Universidad de Las Palmas de Gran Canaria, Spain

Solution 2. We claim that the rational expression in the statement is equalto 1. Indeed, it is well known from the theory of divided differences [1] that

f(z0, z1, · · · , zn) =

n∑

j=0

f(zj)∏

k=0k 6=j

1

zj − zk


Applying the preceding result to the function f(z) = z2 + a2b2c2, the LHS ofthe above identity equals to

f(a, b, c) =f(b, c) − f(a, b)

c− a=

1

c− a

[(b+ c) − (a+ b)

]= 1

and the RHS is

f(a)

(a− b)(a− c)+

f(b)

(b− a)(b− c)+

f(c)

(c− a)(c− b)

a2(1 + b2c2)

(a− b)(a− c)+

b2(1 + a2c2)

(b− a)(c− a)+

c2(1 + a2b2)

(c− a)(c− b),

and we are done.[1.] Isaacson, E., Keller, H. B., Analysis of Numerical Methods. Dover(1994).



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