MathDB

2

Part of 2010 Turkey MO (2nd round)

Problems(2)

Turkey NMO 2010 P2

Source:

12/15/2010
Let PP be an interior point of the triangle ABCABC which is not on the median belonging to BCBC and satisfying ∠CAP=∠BCP. BP∩CA={B′} , CP∩AB={C′}\angle CAP = \angle BCP. \: BP \cap CA = \{B'\} \: , \: CP \cap AB = \{C'\} and QQ is the second point of intersection of APAP and the circumcircle of ABC. B′QABC. \: B'Q intersects CC′CC' at RR and B′QB'Q intersects the line through PP parallel to ACAC at S.S. Let TT be the point of intersection of lines B′C′B'C' and QBQB and TT be on the other side of ABAB with respect to C.C. Prove that ∠BAT=∠BB′Q ⟺ ∣SQ∣=∣RB′∣\angle BAT = \angle BB'Q \: \Longleftrightarrow \: |SQ|=|RB'|
geometrycircumcircleratiogeometry proposed
Turkey NMO 2010 P5

Source:

12/15/2010
For integers aa and bb with 0≤a,b<2010180 \leq a,b < {2010}^{18} let SS be the set of all polynomials in the form of P(x)=ax2+bx.P(x)=ax^2+bx. For a polynomial PP in S,S, if for all integers n with 0≤n<2010180 \leq n <{2010}^{18} there exists a polynomial QQ in SS satisfying Q(P(n))≡n(mod201018),Q(P(n)) \equiv n \pmod {2010^{18}}, then we call PP as a good polynomial. Find the number of good polynomials.
algebrapolynomialmodular arithmeticnumber theory proposednumber theory