Dividing a Line Segment into Seven Equal Parts

Date: 07/04/2003 at 01:36:51
From: Erik
Subject: Dividing a line segment into seven equal parts

I need to divide a segment with the length of X into seven equal parts 
using only a compass and straightedge. 

I also have to construct a line segment the length squareroot of X. 

Date: 07/04/2003 at 12:12:45
From: Doctor Jaffee
Subject: Re: Dividing a line segment into seven equal parts

Hi Erik,

Suppose you have a segment AB whose length is X, and you want to 
divide it into seven equal parts. Draw AB along the bottom of a sheet 
of paper. Now draw the ray AC, where C is close to the top right 
corner of the paper. Locate point A1 on the ray about 1 inch from A.  
It doesn't have to be exact. Then mark off points A2,A3,A4,A5,A6, and 
A7 on ray AC so that the distance between each pair of points is 
exactly the same as the distance from A to A1.

Connect A7 to B. Then construct a line that goes through A6 and is 
parallel to BA7. Where this line intersects AB, call the point P. The 
distance from P to B should be exactly 1/7 of the segment AB.

You might also see Paul Kunkel's Geometry Construction Reference:

   Divide a line segment into n congruent line segments.
   http://whistleralley.com/construction/c8.htm 

Now for the second problem. If you have two legs of a right triangle 
whose lengths are sqrt(x - 1) and 1, the length of the hypotenuse will 
be sqrt(x).

I assume that you can construct a segment whose length is 1. Then you 
can construct any segment whose length is a whole number. So, suppose 
you want to construct a segment whose length is sqrt(11). According to 
my instructions, you would need a right triangle whose legs measure 1 
and sqrt(10). You can construct the leg of length 1, but not the leg 
of length sqrt(10). That is, not unless you construct another right 
triangle whose legs have lengths 1 and sqrt(9). This you can do since 
sqrt(9) = 3.

Give it a try and if you want to check your solutions with me or if 
you have difficulties or other questions write back to me and I'll try 
to help you some more.

Good luck,

- Doctors Jaffee and Sarah, The Math Forum
  http://mathforum.org/dr.math/