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Two Polygons (Posted on 2008-10-29) Difficulty: 3 of 5
There are two regular polygons: one with X sides and one with Y sides with Y>X. Also, coincidentally, the latter has internal angles that are each (Y-X) degrees greater than the former. Also, X+Y is a perfect square.

What are X and Y?

See The Solution Submitted by Charlie    
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Solution solution | Comment 2 of 5 |
Such that Y and X are the number of sides of the two regular polygons, and the internal angle of a regular polygon is equal to (1 - 2/p)*180, such that p is the number of sides, we are given:
(1 - 2/Y)*180 = (1 - 2/X)*180 + (Y - X)

Simplifying...
(Y - X) = (1 - 2/Y)*180 - (1 - 2/X)*180
(Y - X) = 180*(Y - 2)/Y - 180*(X - 2)/X
(Y - X) = 180*((Y - 2)/Y - (X - 2)/X)
(Y - X) = 180*(XY - 2X)/XY - (XY - 2Y)/XY)
(Y - X) = 180/XY*((XY - 2X) - (XY - 2Y))
(Y - X) = 180/XY*(-2X + 2Y)
(Y - X) = 180/XY*2*(Y - X)
1 = 360/XY
XY = 360

The prime factors of 360 are 23, 32 and 5. And, given that X+Y is a perfect square and Y > X:

 X    Y 
  3 + 120 = 123
  4 +  90 =  94
  5 +  72 =  77
  6 +  60 =  66
  8 +  45 =  53
  9 +  40 =  49 <--
 10 +  36 =  46
 12 +  30 =  42
 15 +  24 =  39
 18 +  20 =  38

The only combination of factors that has X+Y as a perfect square, such that Y>X, is X = 9 and Y = 40.

 

Edited on October 29, 2008, 9:21 pm
  Posted by Dej Mar on 2008-10-29 11:11:25

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