Conference rusure::math

On the sheet of paper draw a long straight line segment, starting at X.
Mark one point A on the sphere and put A on X. Roll the sphere along the 
line until the point A contacts the line again, and call the point on the 
line Y. Bisect XY at Z. Roll the sphere back along the line until it contacts 
Z and mark B on the sphere. Using the compasses, measure the distance AB
and mark it on the paper. Bisect AB and you have the radius of the sphere.

No rolling allowed.

[Marking points on the sphere or paper is legal, as long as
you don't do any sliding or rolling to make points coincide -
that's not part of Euclidean constructions.  Don't ask me why:
ask Euclid.]

I tried .1, but found the constructed radius to be a bit short.
Apparently, the sphere must be rolled along a great circle.

So I tried again, but still found it too short.  Here's what I figure:
	|XY| = circumference of the sphere = 2*pi*R
	|AB| = R*sqrt(2)
	Constructed distance = |AB|/2 = R/sqrt(2)

If you roll right (with no slippage, along a great circle, etc.)
|AB| should equal 2 R.

However, valid rolling is not one of the fundamental operations
that Euclid allows you to perform with straightedge and compasses.

OK, what are the valid operations Euclid permits on the sphere? Is there any
operation involving the straight-edge on the sphere?

Lets see, you have to find the centre of the circle, draw a perpendicular from
there to the plane, a line thr the centre parallel to the plane, and a
perpendicular from one of the two points where this intersects to the plane.

The two points hitting the plane now are the ends of the requested linesegment.

Now, that just leaves two problems - finding the centre of the sphere, and
drawing a perpendicular from any point to a plane.

(Constructing the line parallel to the plane = constructing a right angle =
trivial).

Lets do the second operation first.  To construct a perpendicular from any
point not in a plane to the plane itself, set your compass at length longer
than the height of such a perpendicular and scribe three points on the plane
starting from the point.  These three points are on a circle whose centre is
the base of the perpendicular.  To find the centre of this circle intersect the
lines formed by constructing a rightangle in the middle of AB and in the middle
of BC.  (Again - trivial).

Now, back to the first problem - finding the centre of the sphere.  Pick any
three points on the sphere.  Find the centre of the circle that goes thru these
three points, and draw a perpendicular to the plane these three points lie
on, starting at the centre of this circle.  Repeat for another three points.
Both these perpendiculars go thru the centre of the sphere => they intersect
at the centre of the sphere (or everywhere in which case select three different
points and try again).

/Bevin

Yes, but isn't this a "solid sphere"?  That would make finding its center
rather difficult, right?

Select any point P on the sphere and, setting the compass at radius R, draw
a circle around P; on the circle select any point A and point B a distance
R from A. Construct Q on the sphere outside the circle so that AQ=BQ=R. On 
the paper, draw the triangle <R,R,R> and construct the altitude h of the
triangle. (Note that the center of <R,R,R> is at distance h/3 from each
side.) 

(Now imagine the plane through P, Q, and the center of the sphere. It 
bisects the two equilateral triangles PAB and QAB, forming the altitudes
h of each triangle. In this plane imagine the triangle with sides <PQ,h,h>.)

On the paper construct the triangle <PQ,h,h>. At distance h/3 from the apex
along each h-side construct perpindiculars, which meet at a point C. (The
two perpindiculars are the altitudes of the triangular pyramids with base
<R,R,R> and apex at the center of the sphere.) The distance from C to the
end points of the base of the triangle <PQ,h,h> is the radius of the
sphere. 

Ingenious!