2010 QEDMO 7th
Part of QEDMO
Subcontests
(12)m skyscrapers in square grid nxn
Let a city be in the form of a square grid which has n×n cells, each of which contain a skyscraper . At first the m skyscrapers burn, but the fire spreads: everyone skyscraper that has at least two burning neighboring houses (by neighboring houses we mean only houses that border the house along a street, not just at a corner) immediately gets fire. Prove that when in the end the whole city burns down, of must have been m≥n.[hide=original wording in German]
Eine Stadt habe die Form eines quadratischen Gitters, welches n × n Zellen habe, von denen jede ein Hochhaus enthalte. Anfangs brennen m der Hochh¨auser, doch der Brand pflanzt sich fort: Jedes Hochhaus, das mindestens zwei brennende Nachbarh¨auser hat (unter Nachbarh¨ausern verstehen wir dabei nur H¨auser, die entlang einer Straße an das Haus angrenzen, nicht nur an einer Ecke), f¨angt sofort Feuer. Man beweise: Wenn am Ende die gesamte Stadt abgebrannt ist,muss m ≥ n gewesen sein. a_\pi(1)/ b_\pi(1)} <a_\pi(2)/ b_\pi(2)} < ...< a_\pi(n)/ b_\pi(n)}
Let (a1,a2,...,an) and (b1,b2,...,bn) be two sequences of positive real numbers. Let π be a permutation of the set {1,2,...,n}, for which the sum aπ(1)(bπ(1)+bπ(2)+...+bπ(n))+aπ(2)(bπ(3)+bπ(3)+...+bπ(n))+...+aπ(n)bπ(n) is minimal.
Proce for this permutation π, that bπ(1)aπ(1)≤bπ(2)aπ(2)≤...≤bπ(n)aπ(n)Application: In an idealized role-playing game you fight against n opponents at the same time. In order to minimize the damage you suffer yourself, you should first take care of your opponent for the ratio of the time it takes to defeat him (if you only focus on him), and the damage it does per second is minimal; next, one should fight the opponent with the second smallest such ratio, and so on. equal periodic sequences
Let m and n be two natural numbers and let d=gcd(m,n) their greatest common divisor.
Let a1,a2,... and b1,b2,... be two sequences of integers which are periodic with periods m and n respectively (this means that ai+m=ai and bi+n=bi for all natural numbers i≥1, note that there could be smaller periods).
Prove that if the two sequences on the first m+n−d terms match (i.e. ai=bi for all i∈{1,2,...,m+n−d}), then they are the same (so ai=bi for all natural i≥1). sum with permutations of 1,2.,...n
Let a1,a2,...,an be positive real numbers. Furthermore, let Sn denote the set of all permutations of set {1,2,...,n}. Prove thatπ∈Sn∑aπ(1)(aπ(1)+aπ(2))...(aπ(1)+aπ(2)+...+aπ(n))1=a1a2...an1 concurrent wanted, 6 circles, equal into 3 pairs, tangencies given
Let ABC be a triangle. Let x1 and x2 be two congruent circles, which touch each other and the segment BC, and which both lie within triangle ABC, and for which it also holds that x1 touches the segment CA, and that x2 is the segment AB. Let X be the contact point of these two circles x1 and x2. Let y1 and y2 two congruent circles that touch each other and the segment CA, and both within of triangle ABC, and for which it also holds that y1 touches the segment AB, and that y2 the segment BC. Let Y be the contact point of these two circles y1 and y2. Let z1 and z2 be two congruent circles that touch each other and the segment AB, and both within triangle ABC, and for which it also holds that z1 touches the segment BC, and that z2 the segment CA. Let Z be the contact point of these two circles z1 and z2. Prove that the straight lines AX,BY and CZ intersect at a point. words of max 3n letters in alphabet of n letters
An alphabet has n letters. A word is called differentiated if it has the following property fulfilled: No letter occurs more than once between two identical letters. For example with the alphabet {a,b,c,d} the word abbdacbdd is not, the word bbacbadcdd is differentiated.
(a) Each differentiated word has a maximum of 3n letters.
(b) How many differentiated words with exactly 3n letters are ther c (a,b) c(b,a) = 1 = c(a,-a)
Let c:Q−{0}→Q−{0} a function with the following properties (for all x,y,a,b∈Q−{0} and x=1):
a) c(x,1−x)=1
b) c(ab,y)=c(a,y)c(b,y)
c) c(y,ab)=c(y,a)c(y,b)
Show that then c(a,b)c(b,a)=1=c(a,−a) also holds.