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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