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Find My Angle, Said the Polygon!

J.M. (Bellingham) sent me this problem (see original source below). Though it involves straight-forward concepts in the geometry of a polygon, the problem pushes you to rethink what you know and do not know about regular polygons.


You are given the ends of three diagonals in a regular polygon:

And two questions to ponder:
  • What is the size of angle k?
  • How many sides does the polygon have?
  • Great problem to explore, J.M....Thanks!

     

    Source: Paul Stephenson, The Magic MathWorks Traveling Circus (and ATM, 2008)


    Hint: First, it may seem that you need more information...but you don't.

    The apparently-missing information is information you need to construct regarding the angles formed by diagonals in a regular polygon. Draw pictures...do some arithmetic...try special cases...pentagon, hexagon, octagon, etc....what do you notice?

     


    Solution Commentary: When submitting the problem, J.M. wrote: "This problem was a rich experience for me because it caused me to review and to notice some things about diagonals of regular polygons that I had not really attended to before. It was also fun to see that once the problem was truly understood that the solution was so straightforward."


    Rather than provide a solution, let's list key ideas for a n-sided regular polygon with diagonals from one vertex to all of the other non-adjacent vertices:
    • The measure of the interior angle at each vertext is....
    • The the measure of the exterior angle at each vertext is....
    • When the diagonal is connected to its first non-adjacent vertex, an isosceles triangle is formed with base angles equal to what measure...
    • In every case, the diagonals form a triangle where two of the angles sum to equal which exterior angle?
    • And perhaps the biggest piece of info needed to crack the problem, what are the sizes of the angles formed at the originating vertext...and are they related mathematically?
    A surprise is perhaps hidden above in one of these "hints"....Happy problem solving!



    J.S. (Whatcom)writes: "I have this problem that seems to make me take too much time to write when I think I have a solution to one of your problems. The problem I am thinking of ... involved regular polygons and angles therein. We were given the angles made by two diagonals from consecutive vertices with their respective sides of the regular polygon. A third angle was shown and marked as “k”. The two angles given were 30 degrees between one side and one of its diagonals and 25 degrees between a diagonal drawn from the next vertex and its side.

    Your comment about rethinking what one knows about regular polygons was helpful. It brought me to some conclusions that were perhaps known to me, but might have been filed in some remote area of what is left of my brain under “Sometimes Useful Things Concerning Regular Polygons”. On the other hand, I could be completely wrong.

    Among these thoughts were the following:

    1. The number of diagonals produced from any vertex is the same as that produced from any other vertex – this is of course true in any polygon.
    2. The smallest angle produced by a diagonal and its adjacent side in a regular polygon is congruent to any angle produced by two consecutive diagonals from that vertex.
    Since the drawing showed a 30 degree angle from one vertex and a 25 degree angle from another, then it seems there should be a 30 and a 25 from every vertex. If what I believe is true in no.2 above, then the smallest angle produced from any vertex would have to be either 5 degrees or 1 degree, provided we limit ourselves to whole number angles. If it is an angle of 5 degrees, that indicates a 36 sided regular polygon, and if it is 1 degree, it would be a 180 sided monster. I believe I can make a case for the smallest angle being 5X2-n or 2-n, but it is late.

    As for the value of k, unless I have missed something I can only offer a general solution and not a particular, at least from the drawing. I would think in one case it has to be a multiple of 5, and in another any positive whole number. Of course, if we go into the fractional values, we can develop any number of general solutions. I hope some of this makes sense.