Cut Wire (Posted on 2005-05-20)

	Submitted by Bractals
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If r is the apothem (inradius) of a regular n-gon, then area(n-gon) = n*r^2*tan(180/n) perimeter(n-gon) = 2*n*r*tan(180/n) ,where angles are in degrees. sum-of-areas = area(p-gon) + area(q-gon) perimeter(p-gon) + perimeter(q-gon) = constant A little calculus will show that the sum-of-areas is minimized when the apothems are equal. *** This is the crux of the problem - the rest is algebra and trig. *** Therefore, perimeter(p-gon) perimeter(q-gon) apothem = ------------------ = ------------------ 2*p*tan(180/p) 2*q*tan(180/q) If 2*perimeter(q-gon) = 3*perimeter(p-gon), then p*tan(180/p) perimeter(p-gon) 2 -------------- = ------------------ = --- q*tan(180/q) perimeter(q-gon) 3 If p = 2*q, then from (2*q)*tan(180/[2*q]) 2 ---------------------- = --- q*tan(180/q) 3 or 3*tan(90/q) = tan(2*[90/q]) = 2*tan(90/q)/[1 - tan(90/q)^2] or tan(90/q)^2 = 1/3 or 90/q = 30 Therefore, q = 3

	Subject	Author	Date
	Puzzle Thoughts	K Sengupta	2023-06-16 00:29:18
	re-interpretation	Larry	2005-05-21 19:10:30
	Agree	Old Original Oskar!	2005-05-21 01:03:58
	intuition	Larry	2005-05-20 22:31:26
	Jer & Charlie	Bractals	2005-05-20 21:01:59
	solution	Charlie	2005-05-20 19:15:03
	Calculus?	Jer	2005-05-20 19:07:25