Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 exptools 1/6/84; site ihnet.UUCP Path: utzoo!watmath!clyde!burl!mgnetp!ihnp4!ihnet!eklhad From: eklhad@ihnet.UUCP (K. A. Dahlke) Newsgroups: net.math Subject: A solution to the pentagon puzzle Message-ID: <145@ihnet.UUCP> Date: Thu, 2-Aug-84 23:06:46 EDT Article-I.D.: ihnet.145 Posted: Thu Aug 2 23:06:46 1984 Date-Received: Sat, 4-Aug-84 01:39:47 EDT Organization: AT&T Bell Labs, Naperville, IL Lines: 45 Here is a solution to the "divide the pentagon" problem presented earlier. Throughout this solution we will use S=sin(PI/10), and C=cos(PI/10). These values come up a lot. We will calculate S and C later. First we must find the area of the pentagon, so we know what to take half of. By drawing 5 lines, each from a vertex perpendicular to the opposite side, we divide the pentagon into 10 congruent right-triangles. Each triangle has an internal angle of pi/5, which is twice PI/10. By using the double-angle formulas, the area of each triangle is: 5*(5*(CC-SS)/2CS)/2. The area of the pentagon is: 62.5*(CC-SS)/CS. Now we divide the pentagon in half with a line segment. Place the pentagon with its base on the x axis (0,0 10,0). Draw the dividing line parallel to the base (at y=Y). The area of the lower portion is a simple integral: integral(y=0,Y)(10+2*y*S/C). In closed form: 10Y+YY*S/C. To divide the pentagon in half: (10Y+YY*S/C) = (62.5(CC-SS)/CS)/2. A more pleasant form: 2SSYY+20CSY-62.5(CC-SS) = 0. Quadriatic formula: Y = (-20CS +- sqrt(400CCSS+500CCSS-500SSSS))/4SS. simplifying: Y = 2.5*(-2C +- sqrt(14CC-5))/S. The length of the line is: 10+2*Y*S/C = 5*sqrt(14-5/CC). The time has come to compute S and C. There are many ways to compute these values, and they are all unpleasant. sin(4*(PI/10)) = C 2*(2CS)*(CC-SS) = C 4SCC-4SSS = 1 8SSS-4S+1 = 0 4SS+2S-1 = 0 S=(sqrt(5)-1)/4 Using CC=1-SS and substituting, the length of the line is: 5*sqrt(4+2*sqrt(5)). approximately 14.553466902253548081226618397 -- Karl Dahlke ihnp4!ihnet!eklhad