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The 4th Romanian Master of Mathematics Competition Solutions

Day 1: Friday, February 25, 2011, Bucharest

Problem 1. Prove that there exist two functions

f, g: RR,

such that f g is strictly decreasing, while g f is strictlyincreasing.

(POLAND) ANDRZEJ KOMISARSKI & MARCIN KUCZMA

Solution. Let

A=

kZ

22k+1,22k

22k,22k+1

;

B=

kZ

22k,22k1

22k1,22k

.

Thus A= 2B, B= 2A, A= A, B=B, AB= , and finallyAB {0} =R. Let us take

f(x) =

x for x A;x for x B;0 for x= 0.

Take g(x) = 2f(x). Thus f(g(x)) = f(2f(x)) = 2x andg(f(x)) = 2f(f(x)) = 2x.

Problem 2. Determine all positive integers n for whichthere existsa polynomial f(x) with real coefficients, with thefollowing properties:

(1) for eachinteger k, the number f(k)isanintegerifandonly ifk is not divisible byn;

(2) the degree offis less than n.

(H UNGARY) GZ AKS

Solution. We will show that such polynomial exists if andonly ifn= 1 or nis a power of a prime.

We will use two known facts stated in Lemmata 1 and 2.

LEMMA 1. If pa is a power of a prime and k is an integer,

then (k1)(k2)...(k pa

+1)(pa1)! is divisible byp if and only

ifk is not divisible bypa.

Proof. First suppose that pa | kand consider

(k1)(k2) (k pa+1)(pa1)! =

k1pa1

k2pa2

k pa+11

.

In every fraction on the right-hand side, p has the samemaximal exponent in the numerator as in the denominator.

Therefore, the product (which is an integer) is not divisiblebyp.

Now suppose that pak. We have

(k1)(k2) (k pa+1)(pa1)! =

pa

k k(k1) (k p

a+1)(pa)!

.

The lastfraction is an integer. In the fraction pa

k,thedenom-

inator k is not divisible bypa.

LEMMA 2. If g(x) is a polynomial with degree less than nthen

n=0

(1)

n

g(x+n) = 0.

Proof. Apply induction on n. For n= 1 then g(x) is a con-stant and

1

0

g(x+1)

1

1

g(x) = g(x+1) g(x) = 0.

Now assume that n> 1 and the Lemma holds for n1. Leth(x) = g(x+1) g(x); the degree ofhis less than the degreeofg, so the induction hypothesis applies for g and n1:

n1=0

(1)

n1

h(x+ n1) = 0

n1=0

(1)

n1

g(x+n) g(x+n1)= 0

n1

0

g(x+n)+

n1=1

(1)

n11

+

n1

g(x+n) (1)n1

n1n1

g(x) = 0

n=0

(1)

n

g(x+n) = 0.

LEMMA 3. Ifn has at least two distinct prime divisors thenthe greatest common divisor ofn

1

,n

2

, . . . ,

nn1

is 1.

Proof. Suppose to the contrary that p is a common primedivisor of

n1

, . . . ,

nn1

. In particular, p| n1= n. Let abe the

exponent ofp in the prime factorization ofn. Since nhas atleast two prime divisors, we have 1 < pa < n. Hence, npa1and

npa

are listed among

n1

, . . . ,

nn1

and thus p | npa and

p | npa1

. But then p divides

npa

npa1

= n1pa1

, which

contradicts Lemma 1.

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Next we construct the polynomial f(x) when n= 1 or nisa power of a prime.

For n= 1, f(x) = 12 is such a polynomial.Ifn = pawhere p is a prime and a is a positive integer

then let

f(x) = 1p

x1

pa

1

= 1

p (x1)(x2) (x p

a+1)(pa

1)!

.

The degree of this polynomial is pa1 = n1.The number (k1)(k2)(kp

a+1)(pa1)! is an integer for any inte-

ger k, and, by Lemma 1, it is divisible byp if and only ifk isnot divisible bypa = n.

Finally we prove that ifn has at least two prime divisorsthen no polynomial f(x) satisfies (1,2). Suppose that somepolynomial f(x) satisfies (1,2), andapply Lemma 2 for g= fand x= kwhere 1 k n1. We get that

n

kf(0) = 0n,=k

(1)kn

f(k+).Since f(k),.. . ,f(1) and f(1),...,f(n k) are all integers,

we conclude thatn

k

f(0) is an integer for every 1 k n1.

By dint of Lemma 3, the greatest common divisor ofn1

,n

2

, . . . ,

nn1

is 1. Hence, there will exist some integers

u1, u2, . . . ,un1 for which u1n

1

+ +un1 nn1= 1. Thenf(0) =

n1k=1

uk

n

k

f(0) =

n1k=1

uk

n

k

f(0)

is a sum of integers. This contradicts the fact that f(0) is notan integer. So such polynomial f(x) does not exist.

Alternative Solution. (I. Bogdanov) We claim the answeris n= p for some prime pand nonnegative.

LEMMA. For every integers a1, . . . , an there exists an integer-valued polynomial P(x) of degree < n such that P(k) = akfor all 1 k n.

Proof. Induction on n. Forthe base case n= 1onemaysetP(x) = a1. For the induction step, suppose that the polyno-mial P1(x) satisfies the desired property for all 1 k n1.Then set P(x) = P1(x)+ (anP1(n))

x1n1

; since

k1n1

= 0 for

1

k

n

1 and n1n1

= 1, the polynomial P(x) is a soughtone. Now, if for some n there exists some polynomial f(x)

satisfying the problem conditions, one may choose someinteger-valued polynomial P(x) (of degree < n1) coincid-ing with f(x) at points 1,..., n 1. The difference f1(x) =f(x) P(x) also satisfies the problem conditions, thereforewe may restrict ourselves to the polynomials vanishing atpoints 1,..., n 1 that are, the polynomials of the formf(x) = cn1

i=1 (x i) for some (surely rational) constant c.

Let c= p/q be its irreducible form, and q=dj=1 pjj be theprime decomposition of the denominator.

1. Assume that a desired polynomial f(x) exists. Sincef(0) is not an integer, we have q(1)n1(n1)! and hencepjj

(1)n1(n1)! for some j. Hencen1

i=1

(pjj

i)

(

1)n1(n

1)!

0 (mod pjj

),

therefore f(pii

) is not integer, too. By the condition (i), thismeans that n | pi

i, and hence n should be a power of a

prime.

2. Now let us construct a desired polynomial f(x) for anypower of a prime n= p. We claim that the polynomial

f(x) = 1p

x1n1

= n

px

x

n

fits. Actually, consider some integer x. From the first repre-sentation, the denominator of the irreducible form of f(x)

may be 1 or p only. Ifp

x, then the prime decompositionof the fraction n/(px) contains p with a nonnegative expo-nent; hence f(x) is integer. On the other hand, ifn= p | x,then the numbers x1, x2,..., x(n1) contain the sameexponents of primes as the numbers n1,n2,...,1respec-tively; hence the number

x1n1

=n1

i=1 (x i)n1i=1 (n i)

is not divisible byp. Thus f(x) is not an integer.

Problem 3. A triangle ABC is inscribed in a circle . Avariable line chosen parallel to BC meets segments AB,

AC at points D, E respectively, and meets at points K, L(where D lies between K and E). Circle 1 is tangent to thesegments K Dand BD and also tangent to , while circle 2is tangent to the segments LE and C Eand also tangent to.Determine the locus, as varies, of the meeting point of thecommon inner tangents to 1 and 2.

(RUSSIA) VASILYMOKIN & FEDOR IVLEV

Solution. Let P be the meeting point of the common in-ner tangents to 1 and 2. Also, let b be the angle bisec-tor ofB AC. Since K L BC, b is also the angle bisectorofK AL.

LetH

be the composition of the symmetryS

with respectto band the inversionI of centre Aand ratio AK AL(it isreadily seen that S and I commute, so since S2 = I2 = id,then also H2 = id, the identical transformation). The ele-ments of the configuration interchanged byH are summa-rized in Table I.

Let O1 and O2 be the centres of circles 1 and 2. Sincethe circles 1 and 2 are determined by their construc-tion (in a unique way), they are interchanged byH, there-fore the rays AO1 and AO2 are symmetrical with respect

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to b. Denote by1 and 2 the radii of1 and 2. SinceO1AB = O2AC, we have 1/2 = AO1/AO2. On theother hand, from the definition ofP we have O1P/O2P =1/2 = AO1/AO2; this means that AP is the angle bisectorofO1AO2 and therefore ofB AC.

The limiting, degenerated, cases are when the parallelline passes through A when P coincides with A; respec-

tively when the parallel line is BC when P coincides withthe foot A BC of the angle bisector ofB AC (or anyother point on BC). By continuity, any point P on the opensegment A A is obtained for some position of the parallel,therefore the locus is the open segment A A of the angle bi-sector bofB AC.

point K point Lline K L circlerayAB rayACpoint B point Epoint C point D

segment B D segment ECarc B K segment E Larc C L segment DK

TABLE I: Elements interchanged byH.

A

B C

D EK L

O1

O2P

b