A Near Diophantine Octagon Problem (Posted on 2007-04-22)

	Submitted by K Sengupta
Rating: 3.6667 (3 votes)
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We know that r is the radius of the circumcircle. Let H and W be the angles subtended by the sides of lengths a*sqrt(2) and b at the centre of the circle. Then, 4H + 4W = 2*pi and hence: W = pi/2 - H. Using cosine rule, we obtain: 2*a^2 = 2*r^2 - 2*r^2*cosH = 2*r^2(1-cos H); and: b^2 = 2*r^2(1-sin H) So, (a/b)^2 = (1/2)*(1-cos H)/(1-sin H) = t^2/(1-t)2; where t = tan H/2 Or, t = a/(a+ b) so, 1 - cosH = (2*t^2)/(1+ t^2) = 2a2/(2a2 + b2+ 2ab) Or, r =√((2a2 + b2+ 2ab)/2)...(#) Hence, r is a positive integer iff the expression on the RHS of (#) is a positive integer. Hence, from (#); b^2 + P^2 = Q^2; where P=2a+b and Q = 2r. Since, P = 2a+b it follows that P and b must possess the same parity. But we know that for any primitive pythagorean triplet (x,y,z); the parities of x and y must be different. Accordingly, it follows that (b, P, Q) cannot correspond to a primitive pythagorean triplet and both P and b must be even. For Part (i), a< b so that P = 2a+b<3b We know that the smallest primitive pythagorean triplet is (3,4,5). This generateds the triplet (6,8,10), which is in conformity with conditions of the problem. Now,(b,P,Q)=(6,8,10), yields 2a+b =8 and b =6, giving a=1. Consequently, minimum value of a is 1 whenever a< b. For Part (ii), we observe that: P = 2a+b> 3b. Since any pythagorean triplet must possess the form (k*(m^2 - n^2), 2kmn, k*(m^2-n^2), we obtain: 2kmn > 3(m^2 - n^2) 3m^2 - 2mn- n^2 < 0, so that: m< n(1+ sqrt(10))/3 = 1.3874n For n=2, we obtain m< 2.77 so that m=2, which is a contradiction since we cannot have b=0 For n=3, m< 4.1622, so that m=3, which is a contradiction For n=4, m<5.5496, so the minimum value of m is 4. Since, both b and P must be even, it follows that the minimum value of k is 2, giving: P= 2a+b = 2*2*3*4 = 48 and b = 2*(4^2 - 3^2) = 14 This yields a = 17. Consequently, the minimum value of b with a > b is 14.

	Subject	Author	Date
	re(7): Solution	Bractals	2007-04-26 13:43:02
	re(6): Solution	Brian Smith	2007-04-26 00:16:43
	re(5): Solution	Bractals	2007-04-25 13:36:30
	re(4): Solution	Brian Smith	2007-04-25 12:12:54
	re(3): Solution	Bractals	2007-04-24 13:23:57
	re(2): Solution	Brian Smith	2007-04-24 01:00:13
	re: Solution	Bractals	2007-04-23 13:47:11
	Solution	Brian Smith	2007-04-23 11:55:30
	non-analytical "solution"	Charlie	2007-04-22 15:40:57
	Analytic/Synthetic Solution	Bractals	2007-04-22 14:36:52