Challenge problem - - - sum of the measures of the internal angles of polygons

lookagain

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Consider all of and only those polygons of four or more sides which have some combination
of internal angles of just 90 degrees and 270 degrees. Include squares, and rectangles that
are not squares, even though those don't have any internal angles of 270 degrees.

Edit Think polyominoes (the ones without the "holes" in them).

Let \(\displaystyle \ n\ \) = the number of sides of a polygon


Using the fact that the sum of the measures of the internal angles of any simple, closed
(convex or concave) polygon equals \(\displaystyle \ (n - 2)(180 \ degrees), \ \)**
and limiting yourself to the group of polygons described above, show that the number of
internal 90 degree angles is always four more than the number of internal 270 degree
angles for each polygon.

(My intent for the solution is along the lines of basic/intermediate algebra.)


** Source: http://en.wikipedia.org/wiki/Polygon
 
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