3
Part of 2001 All-Russian Olympiad
Problems(5)
K inside parallelogram ABCD => NAK = NCK
Source: All-Russian MO 2001 Grade 9 #3
1/2/2012
A point is taken inside parallelogram so that the midpoint of is equidistant from and , and the midpoint of is equidistant form and . Let be the midpoint of . Prove that the angles and are equal.
geometryparallelogramcircumcirclegeometry unsolved
Russian geometry
Source:
7/27/2003
Let be a point on the longest side of a triangle . The perpendicular bisectors of and intersect and respectively in and . Prove that the circumcenter of lies on the circumcircle of triangle .
geometrycircumcircleparallelogramsimilar triangles
orthocenter
Source:
12/17/2011
Points inside an acute-angled triangle are selected on the altitudes from respectively so that the sum of the areas of triangles , and is equal to the area of triangle . Prove that the circumcircle of triangle passes through the orthocenter of triangle .
geometrycircumcirclefunctioncyclic quadrilateralgeometry unsolved
Convex polygons
Source: Russian 2001
4/5/2006
There are two families of convex polygons in the plane. Each family has a pair of disjoint polygons. Any polygon from one family intersects any polygon from the other family. Show that there is a line which intersects all the polygons.
combinatorics unsolvedcombinatorics
2001 towns in a country
Source: All-Russian MO 2001 Grade 11 #7
1/3/2012
The towns in a country are connected by some roads, at least one road from each town, so that no town is connected by a road to every other city. We call a set of towns dominant if every town not in is connected by a road to a town in . Suppose that each dominant set consists of at least towns. Prove that the country can be partitioned into republics in such a way that no two towns in the same republic are connected by a road.
graph theorycombinatorics unsolvedcombinatorics