Olympiad problems

2002 BAMO, 2

Tags: Tiling , combinatorics , hecagon , octagon , rhombus

In the illustration, a regular hexagon and a regular octagon have been tiled with rhombuses. In each case, the sides of the rhombuses are the same length as the sides of the regular polygon. (a) Tile a regular decagon ($10$-gon) into rhombuses in this manner. (b) Tile a regular dodecagon ($12$-gon) into rhombuses in this manner. (c) How many rhombuses are in a tiling by rhombuses of a $2002$-gon? Justify your answer. [img]https://cdn.artofproblemsolving.com/attachments/8/a/8413e4e2712609eba07786e34ba2ce4aa72888.png[/img]

2018 Czech-Polish-Slovak Junior Match, 2

Tags: hecagon , parallel , diagonal , geometry , perimeter

A convex hexagon $ABCDEF$ is given whose sides $AB$ and $DE$ are parallel. Each of the diagonals $AD, BE, CF$ divides this hexagon into two quadrilaterals of equal perimeters. Show that these three diagonals intersect at one point.

1983 Brazil National Olympiad, 2

Tags: geometry , Equilateral Triangle , 3D geometry , pyramid , hecagon

An equilateral triangle $ABC$ has side a. A square is constructed on the outside of each side of the triangle. A right regular pyramid with sloping side $a$ is placed on each square. These pyramids are rotated about the sides of the triangle so that the apex of each pyramid comes to a common point above the triangle. Show that when this has been done, the other vertices of the bases of the pyramids (apart from the vertices of the triangle) form a regular hexagon.