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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(4): Solution | Comment 7 of 10 |
(In reply to re(3): Solution by Bractals)

I started by looking at PT of the form (2k+1, 2k^2+2k, 2k^2+2k+1).  They have the property that the longer leg and the hypotenuse are consecutive, which makes the ratio between the legs large as k increases.  At k=3 the ratio was already greater than 3,which gave PT(7,24,25).

Proving this was the smallest with a ratio greater than 3 is as simple as listing all the smaller PT and showing their ratios are less than 3.  (3,4,5)-4/3=1.333; (6,8,10)-8/6=1.333; (5,12,13)-12/5=2.4; (8,15,17)-15/8=1.875.

BTW, I was also able to prove your expression of a,b, and r is satisfied by my right triangle by direct substitution.

 Posted by Brian Smith on 2007-04-25 12:12:54
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