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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(5): Solution | Comment 8 of 10 |
(In reply to re(4): Solution by Brian Smith)

`Thanks for the explanation.`
`I noted the same about my expression.`
` 4*(r^2 - a^2)^2 + (2r^2 - b^2)^2 - 4*r^4 = 0`
` 4r^4 - 8a^2*r^2 + 4a^4 - 4b^2*r^2 + b^4 = 0`
` 4r^4 - 4(2a^2 + b^2)*r^2 + (2a^2 + b^2)^2 - 4a^2*b*2 = 0`
` [2r^2 - (2a^2 + b^2)]^2 - 4a^2*b*2 = 0`
` [(2r^2 - 2a^2 - b^2) - 2ab][(2r^2 - 2a^2 - b^2) + 2ab] = 0   2[r^2 - (a + b/2)^2 - (b/2)^2)][(2r^2 - 2a^2 - b^2) + 2ab] = 0`
`   |     Your expression        |`
` `

Edited on April 25, 2007, 2:47 pm
 Posted by Bractals on 2007-04-25 13:36:30

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