Regular polygons meeting at a pointN regular polygons with E edges each meet at a point with no intervening space. Show that N = 2 E E − 2 E = 2 N N − 2 .I did this part considering that the internal angles must add up to 2 π, i.e. ( 1 − 2 E ) π × N = 2 π. I am unable to explain the following:Using the result given above, show that the only possibilities are E = 3 , 4 , 6

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Regular polygons meeting at a pointN regular polygons with E edges each meet at a point with no intervening space. Show that   N  =            2      E              E      −      2          E  =            2      N              N      −      2      .I did this part considering that the internal angles must add up to   2  π, i.e.       (    1    −                  2        E              )    π  ×  N  =  2  π. I am unable to explain the following:Using the result given above, show that the only possibilities are   E  =  3  ,  4  ,  6

Karli Moreno · Accepted Answer

Step 1If   E  &amp;gt;  6 then your second equation tells you something impossible about N.EDIT: Perhaps this was too subtle, so here are the details.Suppose   E  &amp;gt;  6. Then by the second equation in the question,  2  N      /    (  N  −  2  )  &amp;gt;  6,   2  N  &amp;gt;  6  N  −  12,   4  N  &amp;lt;  12,   N  &amp;lt;  3.Step 2But if N regular polygons meet at a point with no intervening space, then   N  ≥  3, so we have a contradiction to our assumption that E was greater than 6.So,   E  ≤  6.Also, there are no polygons with fewer than 3 edges, so   E  ≥  3.Well, that doesn&#039;t leave very many values of E, does it? And   E  =  5 leads to   N  =  10      /    3, which is absurd.So, the only possibilities are   E  =  3  ,  4  ,  6.

N regular polygons with E edges each meet at a point with no intervening space. Show that N=(2E)/(E-2), E=(2N)/(N-2). Explain the following: Using the result given above, show that the only possibilities are E=3,4,6.

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