[Search for users] [Overall Top Noters] [List of all Conferences] [Download this site]

Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

119.0. "Dividing a Pentagon in half" by HARE::STAN () Tue Aug 07 1984 23:21

Newsgroups: net.math
Path: decwrl!amd!dual!zehntel!ihnp4!ihuxs!eagle
Subject: Puzzle (forwarded)
Posted: Thu Aug  2 07:02:22 1984

From: Michael Rios (at AT&T Bell Labs, I think)
Re: A personal headache.

When I was in high school, I was a very active member of the Math Team (much
to the other members' annoyance, most of the time).  We would always (to keep
in practice) generate problems to give to each other and set a time limit
(typically a week) to solve them.  The only problem ever to defeat me was one
that I made up myself:

	A regular pentagon of side 10 has a line drawn inside it, parallel
	to one side, which divides the pentagon into two sections of equal
	area.  What is the length of this line (no decimal approximations,
	please)?

Could someone please help me with this?

("Is this the right room for an arguement?" "I've told you once.")

                                              Michael Rios
                                              Chicago, Il.
                                              Earth

("No you haven't.")
-- 
				John T. Blumenstein
				ihuxr!eagle
T.RTitleUserPersonal
Name		DateLines
119.1HARE::STANTue Aug 07 1984 23:2154
From:	ROLL::USENET       "USENET Newsgroup Distributor"    4-AUG-1984 22:03  
To:	RANI::LEICHTERJ
Subj:	USENET net.math newsgroup articles

Newsgroups: net.math
Path: decwrl!decvax!mcnc!unc!ulysses!mhuxl!ihnp4!ihnet!eklhad
Subject: A solution to the pentagon puzzle
Posted: Thu Aug  2 20:06:46 1984

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