### A Crooked Polygon

Here's a curious problem from a 2008 Moscow Olympiad:

Find a polygon and point O on its boundary such that every straight line through O cuts the polygon into two pieces of equal area.

The construction starts with placing point O at the origin and forming a polygon for which O lies on the boundary. There are three steps:

- Draw an isosceles triangle AOB.
- Add an isosceles trapezoid CDEF, with area equal to that of ΔAOB. Observe that, since Area(CDEF) = Area(DOE) - Area(AOF), AO² + EO² = DO², so that there is some freedom in choosing the trapezoid.
- Add isosceles trapezoids GHIJ and KLMN both equal in area to ΔAOB and shifted a little outwards from O. It's important to have segments BO and MN that do not intersect.

A proof is based on the following observation. A line through a vertex of a triangle divides the triangle into two smaller ones. The areas of the latter are in the ratio of their bases. This applies to triangles with a vertex at O (say, in particular, to ΔAOB, ΔGOJ, and ΔHOI.) But the bases of all triangles cut by the same straight line are divided in the same proportion, implying the required equality of the areas. (When the line coincides with one of the axes the claim holds by construction.)

[...] 来源：http://www.mathteacherctk.com/blog/2012/05/a-crooked-polygon/ Posted in Brain Storm Tags: 趣题, 几何, 惊奇数学事实Trackback: [...]

June 3rd, 2012 at 9:06 am[...] Alexander Bogomolny has sent in a solution to a puzzle from the 2008 Moscow Olympiad, featuring a crooked polygon. [...]

July 3rd, 2012 at 2:01 am