MathDB

2

Part of 2014 Iran MO (3rd Round)

Problems(5)

Continuous function

Source: Iran 3rd round 2014-Algebra exam-P2

8/31/2014
Find all continuous function f:R≥0→R≥0f:\mathbb{R}^{\geq 0}\rightarrow \mathbb{R}^{\geq 0} such that : f(xf(y))+f(f(y))=f(x)f(y)+2  ∀x,y∈R≥0f(xf(y))+f(f(y)) = f(x)f(y)+2 \: \: \forall x,y\in \mathbb{R}^{\geq 0}
Proposed by Mohammad Ahmadi
functioninductionalgebra proposedalgebra
exchange sequence of natural number

Source: Iranian 3rd round Number Theory exam P2

9/22/2014
We say two sequence of natural numbers A=(a1,...,ana_1,...,a_n) , B=(b1,...,bnb_1,...,b_n)are the exchange and we write A∼BA\sim B. if 503∣ai−bi503\vert a_i - b_i for all 1≤i≤n1\leq i\leq n. also for natural number rr : ArA^r = (a1r,a2r,...,anra_1^r,a_2^r,...,a_n^r). Prove that there are natural number k,mk,m such that : ii)250≤k250 \leq k
iiii)There are different permutations π1,...,πk\pi _1,...,\pi_k from {1,2,3,...,5021,2,3,...,502} such that for 1≤i≤k−11\leq i \leq k-1 we have πim∼πi+1\pi _i^m\sim \pi _{i+1}
(15 points)
modular arithmeticnumber theory proposednumber theory
Tennis tournament

Source: Iranian 3rd round Combinatorics exam P2 - 2014

9/25/2014
In a tennis tournament there are participants from nn different countries. Each team consists of a coach and a player whom should settle in a hotel. The rooms considered for the settlement of coaches are different from players' ones. Each player wants to be in a room whose roommates are all from countries which have a defense agreement with the player's country. Conversely, each coach wants to be in a room whose roommates are all from countries which don't have a defense agreement with the coach's country. Find the minimum number of the rooms such that we can always grant everyone's desire.
proposed by Seyed Reza Hosseini and Mohammad Amin Ghiasi
combinatorics proposedcombinatorics
Pit challenge!

Source: Iran 3rd round 2014 - final exam problem 2

9/30/2014
Consider a flat field on which there exist a valley in the form of an infinite strip with arbitrary width ω\omega. There exist a polyhedron of diameter dd(Diameter in a polyhedron is the maximum distance from the points on the polyhedron) is in one side and a pit of diameter dd on the other side of the valley. We want to roll the polyhedron and put it into the pit such that the polyhedron and the field always meet each other in one point at least while rolling (If the polyhedron and the field meet each other in one point at least then the polyhedron would not fall into the valley). For crossing over the bridge, we have built a rectangular bridge with a width of d10\frac{d}{10} over the bridge. Prove that we can always put the polyhedron into the pit considering the mentioned conditions.
(You will earn a good score if you prove the decision for ω=0\omega = 0).
combinatorics unsolvedcombinatorics
Cyclic quadrilateral

Source: Iranian 3rd round Geometry exam P2 - 2014

9/27/2014
△ABC\triangle{ABC} is isosceles(AB=AC)(AB=AC). Points PP and QQ exist inside the triangle such that QQ lies inside PAC^\widehat{PAC} and PAQ^=BAC^2\widehat{PAQ} = \frac{\widehat{BAC}}{2}. We also have BP=PQ=CQBP=PQ=CQ.Let XX and YY be the intersection points of (AP,BQ)(AP,BQ) and (AQ,CP)(AQ,CP) respectively. Prove that quadrilateral PQYXPQYX is cyclic. (20 Points)
geometrygeometric transformationreflectioncircumcircleangle bisectorgeometry proposed