Construct an angle of 3 degrees with straightedge and compass.

Here's a fairly simple construction for a 6-degree angle, which you can
easily bisect to get a 3-degree one.

              A
                     T
       F              B
           X      Y    Z


       E              C

              D

ABCDEF is a regular hexagon, X and Y the midpoints of AE and AC, and Z
the point where XY meets the circumcircle.

T lies on the both the circles
     1) center A, through Y
and  2) center Y, through Z.

Then angle BAT is 6 degrees.

-- John Conway <conway@math.Princeton.EDU>

I have just noticed a slight modification to my construction, which
yields a 3-degree angle very nicely, viz;

                A
                          T
                         U
   F                         B

           X          Y      Z




   E                         C



                D

ABCDEF is a regular hexagon, X and Y the midpoints of AE and AC, and Z
the point where XY meets the circumcircle.

The circle center A through X and Y meets the circle center Y through
Z in T and meets AB in U.  Then angle TXU is 3 degrees.

-- John Conway <conway@math.Princeton.EDU>  (geometry.puzzles, October
   1995)

[ Editor: Conway's later posting with a proof, geometry.puzzles,
11 October 1995. ]

   A few people have asked for a proof of my construction for a 3
degree angle.  I shall give a slightly modified version of it below,
with proof.

          A-------P-------B
           \             /
            \    U      /
             \  T      /
         Z----Y-------Q
               \     /
                \   /
                 \ /
                  C

   ABC is an equilateral triangle, with midpoints P,Q,Y, while Z is
where QY produced hits the circumcircle.  The circle center A through
Y and Q meets AQ in T, and meets the circle center Y through Z in U.

   then angle TYU equals 3 degrees.

  [Note:  for the benefit of those who want to compare with the
original construction,  A,C,Y,T,U  are really the same points as
before, and the distance  YZ  is also the same as it was before.]

    Proof.  The point  Y  divides  QZ  in golden section; that is to
say,  YZ/QZ  =  (root5 - 1)/2 =  s, say.  I shall prove this later.
Then, since  AY:AU:YU  ::  1:1:s,  the angle  YAU  is  36 degrees.

But angle  YAT = angle  YAQ  =  30 degrees, so  angle TAU = 6 degrees.

But this is the angle subtended at the center (A)  of the circle
YTUP  by two points  (T,U)  on that circle.  The  angle  TYU  they
subtend at the point  Y  on the circumference will therefore be
exactly half of it, as desired.

     I find it easier to consider the golden section assertion with
the triangle in another position:

            C

                   Y

                          A

                   Q

            B
                   Z

  Taking  A = (1,0),  C = (-1/2, (root3)/2), B = (-1/2, -(root3)/2)
 (think of the cube roots of unity!)

we see that  Y = (1/4, (root3)/4)  , Q = (1/4, -(root3)/4), and so  Z
must be  (1/4, -something),  from which (since it lies on the unit
circle), we find  "something" = (root15)/4.

This is enough for the desired assertion.

     I would like to ask two questions:

     1)  can anyone find a simpler construction than this for a 3
         degree angle?

     2)  I learned of the assertion that the mid-points of two edges
of an equilateral triangle divide the chord of the circumcircle that
they determine in golden section, only within the last 10 years.  It
enables us to give some very simple constructions (as here, for
example) for various "pentagonal" things.  Can anyone provide a
reference?

-- John Conway <conway@math.Princeton.EDU>