MathDB

3

Part of 2011 Romania Team Selection Test

Problems(4)

Line through incentres is perpendicular to bisectrix of AEB

Source: Romanian TST 2011

4/4/2012
Let ABCABC be a triangle such that AB<ACAB<AC. The perpendicular bisector of the side BCBC meets the side ACAC at the point DD, and the (interior) bisectrix of the angle ADBADB meets the circumcircle ABCABC at the point EE. Prove that the (interior) bisectrix of the angle AEBAEB and the line through the incentres of the triangles ADEADE and BDEBDE are perpendicular.
geometrycircumcircleincenterperpendicular bisectorgeometric transformationangle bisector
no cycles of even length

Source: 2011 Romania TST,problem 7

2/4/2012
Given a positive integer number nn, determine the maximum number of edges a simple graph on nn vertices may have such that it contain no cycles of even length.
combinatorics unsolvedcombinatorics
A combinatorial geometry problem

Source: BMO&amp;IMO TST

12/31/2011
Given a set LL of lines in general position in the plane (no two lines in LL are parallel, and no three lines are concurrent) and another line ℓ\ell, show that the total number of edges of all faces in the corresponding arrangement, intersected by ℓ\ell, is at most 6∣L∣6|L|.
Chazelle et al., Edelsbrunner et al.
geometrycombinatorics unsolvedcombinatoricsProbabilistic Method
Prove that JM, KN and LP are concurrent

Source: Romanian TST 2011

4/9/2012
The incircle of a triangle ABCABC touches the sides BC,CA,ABBC,CA,AB at points D,E,FD,E,F, respectively. Let XX be a point on the incircle, different from the points D,E,FD,E,F. The lines XDXD and EF,XEEF,XE and FD,XFFD,XF and DEDE meet at points J,K,LJ,K,L, respectively. Let further M,N,PM,N,P be points on the sides BC,CA,ABBC,CA,AB, respectively, such that the lines AM,BN,CPAM,BN,CP are concurrent. Prove that the lines JM,KNJM,KN and LPLP are concurrent.
Dinu Serbanescu
geometryincentersearchgeometry proposed