MathDB

1

Part of 2014 Romania Team Selection Test

Problems(5)

Nice collinearity!

Source: Romania TST 2014 Day 1 Problem 1

1/21/2015
Let ABCABC be a triangle, let A′{A}', B′{B}', C′{C}' be the orthogonal projections of the vertices AA ,BB ,CC on the lines BCBC, CACA and ABAB, respectively, and let XX be a point on the line AA′A{A}'.Let γB\gamma_{B} be the circle through BB and XX, centred on the line BCBC, and let γC\gamma_{C} be the circle through CC and XX, centred on the line BCBC.The circle γB\gamma_{B} meets the lines ABAB and BB′B{B}' again at MM and M′{M}', respectively, and the circle γC\gamma_{C} meets the lines ACAC and CC′C{C}' again at NN and N′{N}', respectively.Show that the points MM, M′{M}', NN and N′{N}' are collinear.
geometrypower of a pointradical axis
Centroid homothety!

Source: Romania TST 2014 Day 2 Problem 1

1/21/2015
Let ABCABC be a triangle and let XX,YY,ZZ be interior points on the sides BCBC, CACA, ABAB, respectively. Show that the magnified image of the triangle XYZXYZ under a homothety of factor 44 from its centroid covers at least one of the vertices AA, BB, CC.
geometrygeometric transformationhomothetygeometry unsolved
Concurrent lines

Source: Romania ,4th TST 2014,Problem 1

12/5/2014
Let △ABC\triangle ABC be an acute triangle of circumcentre OO. Let the tangents to the circumcircle of △ABC\triangle ABC in points BB and CC meet at point PP. The circle of centre PP and radius PB=PCPB=PC meets the internal angle bisector of ∠BAC\angle BAC inside △ABC\triangle ABC at point SS, and OS∩BC=DOS \cap BC = D. The projections of SS on ACAC and ABAB respectively are EE and FF. Prove that ADAD, BEBE and CFCF are concurrent.
Author: Cosmin Pohoata
geometrycircumcircletrigonometryangle bisectorprojective geometrygeometry proposed
Angle bisectors and concurrency!

Source: Romania TST 2014 Day 3 Problem 1

1/21/2015
Let ABCABC be an isosceles triangle, AB=ACAB = AC, and let MM and NN be points on the sides BCBC and CACA, respectively, such that ∠BAM=∠CNM\angle BAM=\angle CNM. The lines ABAB and MNMN meet at PP. Show that the internal angle bisectors of the angles BAMBAM and BPMBPM meet at a point on the line BCBC.
geometrysimilar trianglesgeometry unsolved
Collinearity!

Source: Romania TST Day 5 Problem 1

1/21/2015
Let ABCABC a triangle and OO his circumcentre.The lines OAOA and BCBC intersect each other at MM ; the points NN and PP are defined in an analogous way.The tangent line in AA at the circumcircle of triangle ABCABC intersect NPNP in the point XX ; the points YY and ZZ are defined in an analogous way.Prove that the points XX , YY and ZZ are collinear.
geometrycircumcircleprojective geometrytrigonometry