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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)

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 Solution | Comment 3 of 10 |

Since the octagon is cyclic, the sides can be arranged in any order without affecting the values of a, b, or r.  Rearranging the sides in alternating order b, a√2, b, a√2, b, a√2, b, a√2 creates a symmetric equiangular octagon, equivalent to a square with sides 2a+b with the corners truncated by length a.

Draw two lines from the center, one to a vertex and the other as a perpendictular bisector of a b length side.  The triangle created has legs of b/2 and (2a+b)/2 and a hypotenuse of r.  Integer solutions can be created very easily by taking any integer Pythagorean triple and mapping the smallest member to b/2, the second to (2a+b)/2 and the largest to r.

PT(3,4,5) -> b=6, a=1, r=5  (smallest b>a)
PT(7,24,25) -> b=14, a=17, r=25 (smallest a>b)

 Posted by Brian Smith on 2007-04-23 11:55:30

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