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Octagon Construction Using Compass Only


Date: 02/22/2002 at 15:52:08
From: Janci
Subject: Octagon using compass only

Can anyone help me to construct the vertices of a regular octagon 
using only a compass? The only thing you know about the octagon is 
the circumradius.


Date: 02/23/2002 at 18:31:45
From: Doctor Jubal
Subject: Re: Octagon using compass only

Hi Janci,

This type of construction is called a Mascheroni construction, after 
the Italian mathematician who showed that any construction that is 
possible with compass and straightedge is also possible with a movable 
compass alone (aside from things like straight lines).

Let me explain the notation I'll use below. I'll begin with a circle 
centered at point O, with point A on the circle.  This will eventually 
be the circumcircle of the octagon. For convenience, I'll let distance 
OA be unity.

In the first part of the construction, we'll construct a point B on 
circle OA such that BOA is a right angle. In the second part of the 
construction, we'll find the midpoint M of arc AB along circle O, so 
angle MOA is 45 degrees, and the remainder of the points of the 
octagon may be found simply by stepping off the distance AM along 
circle O.

Begin by drawing circle AO. Mark one of its intersections with circle 
OA as point C. Since the two circles have the same radius, ACO is an 
equilateral triangle. Continue by drawing circle CA, calling its point 
of intersection with circle AO point D, and then draw circle DA and 
call its point of intersection with circle AO point E. Just like 
triangle ACO, triangles ACD and ADE are equilateral, and all these 
equilateral triangles are congruent. Therefore, distance OD is the 
same as distance CE.  Both of these distances are twice the height of 
an equilateral triangle with unit edge, or sqrt(3).

Now draw circles OD and EC. Call either of their points of 
intersection point F. Because distance OD is equal to distance CE, 
point F is equidistant from O and E. But O and E both lie on circle 
AO, so point A is also equidistant from O and E. Therefore, line FA is 
the perpendicular bisector of OE, and we can apply the Pythagorean 
theorem to triangle OAF. OF is the hypoteneuse with length sqrt(3).  
OA has length 1, so AF has length sqrt(2). But sqrt(2) is the length 
of the hypoteneuse of an isoceles right triangle with unit legs.  
Therefore, draw circle AF, call either of its intersections with 
circle OA point B, and angle AOB will be right.

Halfway there. Now, we must find the midpoint of arc AB along circle 
O. Draw circle BO. Also draw a circle with radius AB centered at point 
O, and label its intersection with circle AO point G, such that ABOG 
is a parallelogram. Also label its intersection with circle BO point 
H, such that ABHO is a parallelogram congruent to ABOG. Points G, O, 
and H should be collinear.

Now draw circles HA and GB, labeling either of their intersections 
point J. Then draw circles centered and H and G with radius OJ. Call 
their point of intersection M. Now since points J and O are both 
equidistant from G and H, OJ is the perpendicular bisector of GH. But 
AB is parallel to GH, and point O is equidistant from A and B, so OJ 
is also the perpendicular bisector of AB. Point M, which is also 
equidistant from G and H, therefore lies on OJ, and so is equidistant 
from A and B, as any arc midpoint must be.

Moreover, for any parallelogram, the sum of the squares of the 
diagonals is twice the sum of the squares of the two side lengths.  
Applied to ABHO

  AH^2 + OB^2 = 2(AB^2 + BH^2)

Now because this is a parallelogram, BH = OA, and OA = OB, so

  AH^2 = 2 AB^2 + OA^2

and applying the Pythagorean theorem to triangle HOJ

  HJ^2 = OH^2 + OJ^2

and by construction, AH = HJ, and OH = AB, so

  OH^2 + OJ^2 = 2 AB^2 + OA^2

  OJ^2 = AB^2 + OA^2

Now, applying the Pythagorean theorem to triangle HOM

  HM^2 = OH^2 + OM^2

By construction, HM = OJ, and OH = AB, so

  OA^2 = OM^2

Therefore, M lies on circle OA in addition to being equidistant from 
A and B, and is the midpoint of their arc. A, M and B are three 
vertices of an octagon with circumcircle OA, and the rest may be found 
by stepping off distance AM along circle OA.

That was drawn-out, but Mascheroni constructions often are, leading to 
a corollary to Mascheroni's result: the straightedge, while not 
necessary, is awfully useful. If you want any of this explained or 
justified in greater detail, or if you find any part of the 
construction overly ambiguous, feel free to write back.

Does this help?  Write back if you'd like to talk about this more, or 
if you have any other questions.

- Doctor Jubal, The Math Forum
  http://mathforum.org/dr.math/   


Date: 02/24/2002 at 07:32:47
From: Janci
Subject: Octagon using compass only

Thank you very much. Your solution seemed unnecessarily difficult to 
me so I simplified it a little, but it was a very good hint.

I think that you can stop when G is constructed. You do not need H and 
J, because all you need is the length of OJ (you need it to be the 
radius of the last constructed circle - centered at point G) But this 
length is equal to sqrt(3), which you have (e.g. EC or OD).

Proof of this construction can be made by multiple use of the 
Pythagorean theorem.

But as I mentioned in the beginning I really appreciate your help.


Date: 02/24/2002 at 08:51:43
From: Doctor Jubal
Subject: Re: Octagon using compass only

Hi Janci,

I'm glad you found it useful.

Regarding your streamlining of the construction, it is good. From the 
very beginning I approached the problem in two steps: i) construct a 
right angle; ii) bisect it. After the completion of i) my computer 
screen had become such a mess that I started afresh with the right 
angle and set about bisecting it, which of course denied me easy 
access to the lengths I'd constructed in i), and I did not see this.
Thank you for the simplification.

- Doctor Jubal, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Constructions
High School Geometry
High School Triangles and Other Polygons

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