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Constructing a Segment of a Given Length

Date: 09/09/2005 at 16:31:17
From: Ashley
Subject: Geometry Constructions

How do you construct the "8th root of 3" using a compass and a 

I was thinking that we may have to construct the "square root of 3" 
by using the Pythagorean Theorem so we have a triangle that has a 
hypotenuse of 2 and one leg of length 1.  This would make our other 
leg length "square root of 3".  Then I thought maybe you could 
somehow repeat this process two more times to make it the "8th root 
of 3".  But I didn't know where to go after the "square root of 3" 
part.  Please help!

Date: 09/09/2005 at 20:16:37
From: Doctor Greenie
Subject: Re: Geometry Constructions

Hi, Ashley--

Cool problem!

Using the Pythagorean Theorem is a good way to construct segments of 
length sqrt(n) where n is not a perfect square.  But I don't see how 
you are ever going to get the 8th root of a number that way.

Note that the 8th root of 3 is sqrt(sqrt(sqrt(3))).  That is,

  8th root of 3 = 3^(1/8) = {[3^(1/2)]^(1/2)}^(1/2)

So we can construct the 8th root of 3 if we can find a way to 
construct the square root of 3, then construct the square root of that 
number, and then construct the square root a third time.

Here is the thought I have about how we can do that:

  The length of the altitude to the hypotenuse of a right triangle
  is the geometric mean of the lengths of the two segments of the

So to construct the square root of 3 using this fact, we can do the 

  (1) draw segment AC, with B on AC, so that AB = 1 and BC = 3
  (2) construct a perpendicular to AC at B
  (3) find the point D on this perpendicular so that ADC is a right
      triangle with hypotenuse AC; i.e., with right angle at D

Then BD is the altitude to hypotenuse AC of triangle ADC; its length 
is the geometric mean of the lengths of AB and BC, which is

  sqrt(1*3) = sqrt(3)

Once we have the segment BD with length sqrt(3), we can repeat the 
construction with the two portions of the segment in step (1) having 
lengths 1 and sqrt(3); this construction will give us a segment of 
length sqrt(sqrt(3)) = 4th root of 3.

And then repeating the construction a third time will give us a 
segment of length 8th root of 3.

The required constructions are straightforward except for step (3); 
I have left that for you to think about a bit....

I hope this helps.  Thanks again for sending a question which provided 
me with some good mental exercise.  Please write back if you have any 
further questions about any of this.

- Doctor Greenie, The Math Forum 
Associated Topics:
College Constructions
High School Constructions

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