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A Common Vertex (Posted on 2004-07-12) Difficulty: 2 of 5
Three regular polygons, all with unit sides, share a common vertex and are all coplanar. Each polygon has a different number of sides, and each polygon shares a side with the other two; there are no gaps or overlaps. Find the number of sides for each polygon. There are multiple answers.

See The Solution Submitted by Brian Smith    
Rating: 3.0000 (1 votes)

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Solution "I got six" refined | Comment 17 of 18 |

From my previous post we have

k = 2*i*j/(i*j-2*(i+j))    with   2 < i < j < k

Therefore,

i*j-2*(i+j) > 0   =>   2*i/(i-2) < j

j < k   =>   j < 4*i/(i-2)

Hence,

max(i,2*i/(i-2)) < j < 4*i/(i-2)

This gives eight possible solutions:

i       j       k       solution

3      7      42      Y

3      8      24      Y

3      9      18      Y

3      10    15      Y

3      11    66/5   N

4      5      20      Y

4      6      12      Y

4      7      28/3   N


  Posted by Jerry on 2004-07-18 09:22:21
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