MathDB

2

Part of 2012 Iran Team Selection Test

Problems(6)

PE is bisector of BPC

Source: Iran TST 2012 -first day- problem 2

4/23/2012
Consider ω\omega is circumcircle of an acute triangle ABCABC. DD is midpoint of arc BACBAC and II is incenter of triangle ABCABC. Let DIDI intersect BCBC in EE and ω\omega for second time in FF. Let PP be a point on line AFAF such that PEPE is parallel to AIAI. Prove that PEPE is bisector of angle BPCBPC.
Proposed by Mr.Etesami
geometrycircumcircleincentertrigonometryangle bisectorpower of a point
Functional equation-We need pco!!!

Source: Iran TST 2012-First exam-2nd day-P5

4/24/2012
The function f:R≥0⟶R≥0f:\mathbb R^{\ge 0} \longrightarrow \mathbb R^{\ge 0} satisfies the following properties for all a,b∈R≥0a,b\in \mathbb R^{\ge 0}:
a) f(a)=0⇔a=0f(a)=0 \Leftrightarrow a=0
b) f(ab)=f(a)f(b)f(ab)=f(a)f(b)
c) f(a+b)≤2max⁡{f(a),f(b)}f(a+b)\le 2 \max \{f(a),f(b)\}.
Prove that for all a,b∈R≥0a,b\in \mathbb R^{\ge 0} we have f(a+b)≤f(a)+f(b)f(a+b)\le f(a)+f(b).
Proposed by Masoud Shafaei
functioninductionceiling functionbinomial coefficientsalgebra proposedalgebra
Polynomial functional equation!

Source: Iran TST 2012-Second exam-1st day-P2

5/12/2012
Let g(x)g(x) be a polynomial of degree at least 22 with all of its coefficients positive. Find all functions f:R+⟶R+f:\mathbb R^+ \longrightarrow \mathbb R^+ such that f(f(x)+g(x)+2y)=f(x)+g(x)+2f(y)   \forall x,y\in \mathbb R^+.
Proposed by Mohammad Jafari
algebrapolynomialfunctionalgebra proposed
2000 real numbers and roots of polynomials

Source: Iran TST 2012-Third exam-1st day-P2

5/15/2012
Do there exist 20002000 real numbers (not necessarily distinct) such that all of them are not zero and if we put any group containing 10001000 of them as the roots of a monic polynomial of degree 10001000, the coefficients of the resulting polynomial (except the coefficient of x1000x^{1000}) be a permutation of the 10001000 remaining numbers?
Proposed by Morteza Saghafian
algebrapolynomialpigeonhole principlealgebra proposed
Many circles and a perpendicular!

Source: Iran TST 2012-Second exam-2nd day-P5

5/13/2012
Points AA and BB are on a circle ω\omega with center OO such that π3<∠AOB<2π3\tfrac{\pi}{3}< \angle AOB <\tfrac{2\pi}{3}. Let CC be the circumcenter of the triangle AOBAOB. Let ll be a line passing through CC such that the angle between ll and the segment OCOC is π3\tfrac{\pi}{3}. ll cuts tangents in AA and BB to ω\omega in MM and NN respectively. Suppose circumcircles of triangles CAMCAM and CBNCBN, cut ω\omega again in QQ and RR respectively and theirselves in PP (other than CC). Prove that OP⊥QROP\perp QR.
Proposed by Mehdi E'tesami Fard, Ali Khezeli
geometrycircumcircletrapezoidIran
Non-intersecting broken lines

Source: Iran TST 2012-Third exam-2nd day-P5

5/16/2012
Let nn be a natural number. Suppose AA and BB are two sets, each containing nn points in the plane, such that no three points of a set are collinear. Let T(A)T(A) be the number of broken lines, each containing n−1n-1 segments, and such that it doesn't intersect itself and its vertices are points of AA. Define T(B)T(B) similarly. If the points of BB are vertices of a convex nn-gon (are in convex position), but the points of AA are not, prove that T(B)<T(A)T(B)<T(A).
Proposed by Ali Khezeli
rotationinequalitiescombinatorial geometrycombinatorics proposedcombinatorics