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 A Near Diophantine Octagon Problem (Posted on 2007-04-22)
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.)
 re(6): Solution | Comment 9 of 10 |
(In reply to re(5): Solution by Bractals)

Nice job on the factoring, but now you got me curious about another set of possible solutions, those which satisfy the other factor you created: (2r^2 - 2a^2 - b^2) + 2ab = 0.

I rewrote this as 2r^2 = a^2 + (a-b)^2.  From the identity 2*5^2 = 7^2 + 1^2, I found sets (a,b,r)=(7,6,5) or (7,8,5).

Since r is the smallest value, I wonder if these represent an improper octagon, joining every other corner or every third corner?!

 Posted by Brian Smith on 2007-04-26 00:16:43

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