MathDB

6

Part of 2018 Iran Team Selection Test

Problems(3)

iran tst 2018 combinatorics

Source: Iranian TST 2018, second exam day 2, problem 6

4/17/2018
a1,a2,…,ana_1,a_2,\ldots,a_n is a sequence of positive integers that has at least 2n3+1\frac {2n}{3}+1 distinct numbers and each positive integer has occurred at most three times in it. Prove that there exists a permutation  b1,b2,…,bnb_1,b_2,\ldots,b_n of aia_i 's such that all the nn sums bi+bi+1b_i+b_{i+1} are distinct (1≤i≤n1\le i\le n , bn+1≡b1b_{n+1}\equiv b_1 )
Proposed by Mohsen Jamali
combinatoricsIranIranian TSTSequence
Iran combinatorics

Source: Iranian TST 2018, first exam day 2, problem 6

4/8/2018
A simple graph is called "divisibility", if it's possible to put distinct numbers on its vertices such that there is an edge between two vertices if and only if number of one of its vertices is divisible by another one.
A simple graph is called "permutationary", if it's possible to put numbers 1,2,...,n1,2,...,n on its vertices and there is a permutation π \pi such that there is an edge between vertices i,ji,j if and only if i>ji>j and π(i)<π(j)\pi(i)< \pi(j) (it's not directed!)
Prove that a simple graph is permutationary if and only if its complement and itself are divisibility.
Proposed by Morteza Saghafian .
combinatoricsgraph theory
Geometry from Iran TST

Source: Iranian TST 2018, third exam day 2, problem 6

4/19/2018
Consider quadrilateral ABCDABCD inscribed in circle ω\omega . P≡AC∩BDP\equiv AC\cap BD. EE, FF lie on sides ABAB, CDCD respectively such that APE^=DPF^\hat {APE}=\hat {DPF} . Circles ω1\omega_1, ω2\omega_2 are tangent to ω\omega at XX , YY respectively and also both tangent to the circumcircle of △PEF\triangle PEF at PP . Prove that: EXEY=FXFY\frac {EX}{EY}=\frac {FX}{FY}
Proposed by Ali Zamani
geometryIranIranian TST