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A Near Diophantine Octagon Problem (Posted on 2007-04-22) Difficulty: 3 of 5
The cyclic octagon ABCDEFGH has the sides a√2, a√2, a√2, a√2, b, b, b and b respectively in that order. Each of a, b and r are positive integers, where r is the radius of the circumcircle.

Analytically determine:

(i) The minimum value of a with a < b

(ii) The minimum value of b with a > b

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Analytic/Synthetic Solution | Comment 1 of 10

Let alpha and beta be the central angles
subtended by sides a*sqrt(2) and b
respectively. From the law of cosines we have
                r^2 - a^2
  cos(alpha) = -----------
                   r^2

               2r^2 - b^2
  cos(beta) = ------------
                   2r^2
Since the sum of the central angles of an
octagon is 360 degrees, we have
  4*alpha + 4*beta = 360
         or
  alpha + beta = 90
Therefore,
  cos(alpha)^2 + cos(beta)^2 = 1
               or
    r^2 - a^2         2r^2 - b^2
  [-----------]^2 + [------------]^2 = 1
       r^2                2r^2
Thus, the integers a, b, and r must satisfy
the following constraints:
      0 < a < r 
 
      0 < b < r*sqrt(2)
      4*[r^2 - a^2]^2 + [2r^2 - b^2]^2 - 4*r^4 = 0
From these, I do not know how to find the integers
analytically. Using Perl I was able to find the
following answers:
For a < b,    (a,b,r) = (1,6,5)
For b < a,    (a,b,r) = (17,14,25)
 

  Posted by Bractals on 2007-04-22 14:36:52
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