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2009 Italy TST

2009 Italian TST

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0% found this document useful (0 votes)
282 views2 pages

2009 Italy TST

2009 Italian TST

Uploaded by

yurtman
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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AoPS Community 2009 Italy TST

Italy TST 2009


www.artofproblemsolving.com/community/c5511
by littletush

Day 1

1 Let n, k be positive integers such that n ≥ k. n lamps are placed on a circle, which are all off.
In any step we can change the state of k consecutive lamps. In the following three cases, how
many states of lamps are there in all 2n possible states that can be obtained from the initial
state by a certain series of operations?
i)k is a prime number greater than 2;
ii) k is odd;
iii) k is even.

2 ABC is a triangle in the plane. Find the locus of point P for which P A, P B, P C form a triangle
whose area is equal to one third of the area of triangle ABC.

3 Find all pairs of integers (x, y) such that

y 3 = 8x6 + 2x3 y − y 2 .

Day 2

1 Let n be an even positive integer. An n-degree monic polynomial P (x) has n real roots (not
necessarily distinct). Suppose y is a positive real number such that for any real number t < y,
we have P (t) > 0. Prove that
1 1
P (0) n − P (y) n ≥ y.

2 Two circles O1 and O2 intersect at M, N . The common tangent line nearer to M of the two
circles touches O1 , O2 at A, B respectively. Let C, D be the symmetric points of A, B with
respect to M respectively. The circumcircle of triangle DCM intersects circles O1 and O2 at
points E, F respectively which are distinct from M . Prove that the circumradii of the triangles
M EF and N EF are equal.

3 Two persons, A and B, set up an incantation contest in which they spell incantations (i.e. a
finite sequence of letters) alternately. They must obey the following rules:
i) Any incantation can appear no more than once;
ii) Except for the first incantation, any incantation must be obtained by permuting the letters

© 2019 AoPS Incorporated 1


AoPS Community 2009 Italy TST

of the last one before it, or deleting one letter from the last incantation before it;
iii)The first person who cannot spell an incantation loses the contest. Answer the following
questions:
a) If A says ’ST AGEP REIM O’ first, then who will win?
b) Let M be the set of all possible incantations whose lengths (i.e. the numbers of letters in
them) are 2009 and containing only four letters A, B, C, D, each of them appearing at least
once. Find the first incantation (arranged in dictionary order) in M such that A has a winning
strategy by starting with it.

© 2019 AoPS Incorporated 2


Art of Problem Solving is an ACS WASC Accredited School.

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