MathDB

5

Part of 2017 Iran Team Selection Test

Problems(3)

A difficult geometry from Iranian TST 2017

Source: Iranian TST 2017, first exam day 2, problem 5

4/6/2017
In triangle ABCABC, arbitrary points P,QP,Q lie on side BCBC such that BP=CQBP=CQ and PP lies between B,QB,Q.The circumcircle of triangle APQAPQ intersects sides AB,ACAB,AC at E,FE,F respectively.The point TT is the intersection of EP,FQEP,FQ.Two lines passing through the midpoint of BCBC and parallel to ABAB and ACAC, intersect EPEP and FQFQ at points X,YX,Y respectively. Prove that the circumcircle of triangle TXYTXY and triangle APQAPQ are tangent to each other.
Proposed by Iman Maghsoudi
geometryIranIranian TSTcircumcircle
2017 Iran TST2 day2 p5

Source: 2017 Iran TST second exam day2 p5

4/24/2017
k,nk,n are two arbitrary positive integers. Prove that there exists at least (k−1)(n−k+1)(k-1)(n-k+1) positive integers that can be produced by nn number of kk's and using only +,−,×,÷+,-,\times, \div operations and adding parentheses between them, but cannot be produced using n−1n-1 number of kk's.
Proposed by Aryan Tajmir
combinatoricsIranIranian TST
Polynomial from Iran TST 2017

Source: 2017 Iran TST third exam day2 p5

4/27/2017
Let {ci}i=0∞\left \{ c_i \right \}_{i=0}^{\infty} be a sequence of non-negative real numbers with c2017>0c_{2017}>0. A sequence of polynomials is defined as P−1(x)=0 , P0(x)=1 , Pn+1(x)=xPn(x)+cnPn−1(x).P_{-1}(x)=0 \ , \ P_0(x)=1 \ , \ P_{n+1}(x)=xP_n(x)+c_nP_{n-1}(x). Prove that there doesn't exist any integer n>2017n>2017 and some real number cc such that P2n(x)=Pn(x2+c).P_{2n}(x)=P_n(x^2+c).
Proposed by Navid Safaei
IranIranian TSTpolynomialalgebra