Conference rusure::math

    Drop perpendiculars from points A and B onto the plane giving
    points C and D.  Now if P is a point in the plane such that
    the triangles ACP and BDP are similar the lines AP and BP will
    make equal angles with the plane.

    The locus of points P will lie on a line in the plane at right
    angles to CD; this line will intersect CD at Q where ACQ and BDQ
    are similar triangles.

    - Jim

     Are A and B both on the same side of plane M?  If so, then
     the following may give the answer.  Let B' be the "mirror
     image" of B through M, i.e., drop a perpendicular from point
     B to plane M and extend it its own length to the other side
     of M and call that point B'.  So if the perpendicular from B
     to M intersects plane M at point Q, then BQB' is a straight
     line and BQ = QB'.  Now connect A and B' with a staright
     line.  Will the intersection of AB' with plane M be the
     desired point?
     
     Dan                        
                               /M
                      B       /
                       \     /
                        \   /
                         \ /
             A------------/------B'
                         / P?
                        /

	Another pair for you to ponder:

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	The focus F and the directrix d1d2 of a parabola are given, but not the
	curve itself.  Also given is a straight line m that would intersect the
	parabola if it were present.  Using straightedge and compass, determine
	a point P on the line m where the parabola would cross it.
	-----------------------------------------------------------------------
	If no vertex of a simple polyhedron is incident with exactly 3 edges,
	prove that at least 8 of the faces must be triangles.

						Thanks in advance,
							Steve Barkley.

	Euler says:	G = E + 2, where E is # edges
	                             and G is # faces & vertices
	
	I say:		G = T + X, where T is # triangles & vertices of deg 3.
				     and X are the rest.

	4*E >= 3*T + 4*X = 4*G - T = 4*E + 8 - T

	i.e.	T >= 8.

	Since there are no vertices of degree 3, there must be at least 8
	triangles.

    Note 1005.2 gets one point of the locus (are there more?).
    
    Note 1005.1 is wrong.  Consider M as the xy plane, and the points
    A = (-1,0,1), B = (2,0,2):
    
    		     z
    		     ^	 B
    		     |	/|
    		   A | / |		C = (-1,0,0)
    		   |\|/  |		D = (2,0,0)
    	        ---C-O---D-> x
    
    The triangles ACO and BDO are similar, so (OC)/(AC) = (OD)/(BD).
    Note 1005.1 implies that for points P = (0,y,0), ACP and BDP are
    similar, so that (PC)/(AC) = (PD)/(BD).  But
    
    	(OP)^2 + (OC)^2   (PC)^2   (PD)^2   (OP)^2 + (OD)^2
    	--------------- = ------ = ------ = ---------------
    	     (AC)^2       (AC)^2   (BD)^2        (BD)^2
    
    Now subtracting [(OC)/(AC)]^2 = [(OD)/(BD)]^2 gives:
    
    	(OP)^2   (OP)^2
    	------ = ------
    	(AC)^2   (BD)^2
    
    So that (AC) = (BD), and the perpendiculars from A and B to the plane
    must be the same length.  Hmmm...
    
    	-	-	-	-	-	-	-	-	-
    
    The points P satisfy: (PC)/(AC) = (PD)/(BD), or (PC) = R*(PD), where
    R = (AC)/(BD).  Thus, the original problem can be reduced to the following:
    
    Given two points C and D in the plane, and a positive real R, find
    the locus of points P in the plane such that (PC) = R*(PD).
    
    Without loss of generality, we can assume C = (R,0), and D = (1,0)
    (by rotating and scaling the plane).  Now, let P = (x,y) so that:
    
    	sqrt((x-R)^2 + y^2) = R sqrt((x-1)^2 + y^2)
    
    	...
    
    	 2    2 R x      2
        y  =  -----  -  x
    	      R + 1
    
    Which is some moderately strange curve.

     re .5
     
>>    	   2    2 R x      2
>>        y  =  -----  -  x
>>    	        R + 1
>>    
>>    Which is some moderately strange curve.
     
     Let a = R/(R + 1).  Then you have
     
          y^2 = 2ax - x^2
     
          x^2 - 2ax + y^2 = 0
     
          x^2 - 2ax + a^2 + y^2 = a^2
     
          (x - a)^2 + y^2 = a^2
     
     Don't you recognize this moderately strange symmetrical
     curve?  :-)                                 -----------
     
     Dan