The hexagon has five $90^o$ angles and one $270^o$ angle (see picture). Use a straight-line ruler to divide it into two equal-sized polygons.
Problem
Source: 2005 Oral Moscow Geometry Olympiad grades 8-9 p1
Tags: geometry, areas, rectangle
franzliszt
16.10.2020 23:50
Equal area? Or must shapes be congruent as well?
parmenides51
16.10.2020 23:55
just equal areas
franzliszt
17.10.2020 00:07
Are you allowed to make more than one cut?
parmenides51
17.10.2020 00:12
I don't think so, otherwise you could draw two bisectors (midlines) of each rectangle and it would be trivial
Bexultan
12.05.2023 08:43
Fact: Any line passing through the symmetry centre of a polygon, cuts it into two polygons with equal areas Let the pentagon be $ABCDE$. Construct the point $G$: the intersection of lines $BC$ and $EF$. Then you get two rectangles $ABGF$ and $CDGE$. Construct their centres with a ruler. Then the line passing through the centers bisects the area of the pentagon in half.
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