Conference rusure::math

     Is that with or without touching?
     
     Dan

    re -1
    
    Either way is satisfactory, but I would only consider the situation
    where they meet because these are the limit cases....
    
    Eugene

     It's about time you answered .1. :-)
     
     Dan


	(x + y + z) < (a + b + c)

	/ravi

>	(x + y + z) < (a + b + c)

Try x = y = z = 1, a = b = 1.1, c=2. Won't fit, either way.

    Why can't you determine r(A), the radius of the circle circumscribing
    triangle A, and see if r(B) < r(A)?
    
    Intuitively it seems that if the circumscribing circles fit inside
    one another then the base triangles fit inside one another.
    
      John

re .6; the counterexample in .5 is still relevant. Now if the triangles were 
equilateral...

This problem, as you may have noticed by now, is nontrivial. It is found, 
by the way, in "100 Problems in Elementary Mathematics" by Steinhaus, in
the 'Unsolved Problems' chapter. 

RE: .6 (John)
    
    <<relative radii of circumscribing circles>>
    
    No go John, try a triangle with sides (1, 1, 2-e) for a small e, and one
    of with sides of (1, 1, 1).  The second triangle has a moderate area
    and a compact circumscribing circle.  By shrinking e, we can give the
    first triangle as small an e with as large a circumscribing radius as
    we desire.  Obviously the circumscribing circle of the first is larger
    than the second, but the second couldn't possibly fit inside it since
    it has a greater area.
    
    				Topher

    re .7
    
    Actually the problem is in the section that contains some unsolved
    problems.  In other words, it might have been solved.  I have looked
    at the problem closely.  Though I have not found the solution, I
    have a feeling that this is one of the problems that the solution
    has been found.
    
    Eugene

    The phrase "fits within" is ambiguous.  Does it allow reflections
    or not?  I am assuming that it does allow them:
    
    Triangle B with sides of length (x,y,z) fits within triangle A 
    with sides of length (a,b,c) IFF:
    
    	The longest side of A (called A1) is greater than or equal to
    	the longest side of B (called B1).
    
    	AND: (
    		Given the triangle C formed by using the second largest
    		angle of A (called a2), the side A1 and the smallest angle
    		of B (using ASA), the second longest side of B (called B2) 
    		is less than or equal to the length of the side of C
    		opposite the angle a2.
    
    		OR:
    
    		Given the triangle D formed by using the smallest angle of 
    		A (called a3), the side A1 and the second largest angle
    		of B (using ASA); the smallest side of B (called B3) 
    		is less than or equal to the length of the side of D
    		opposite the angle a3.
	    )
    
    
    --  Barry

    .10 is not quite right.
    
    The following text:
    
>    	AND: (
>    		Given the triangle C formed by using the second largest
>    		angle of A (called a2), the side A1 and the smallest angle
>    		of B (using ASA), the second longest side of B (called B2) 
>    		is less than or equal to the length of the side of C
>    		opposite the angle a2.
>    
>    		OR:
>    
>    		Given the triangle D formed by using the smallest angle of 
>    		A (called a3), the side A1 and the second largest angle
>    		of B (using ASA); the smallest side of B (called B3) 
>    		is less than or equal to the length of the side of D
>    		opposite the angle a3.
>	    )

    Should be amended as follows:
    
    	AND: (
    		The smallest angle of B is less than or equal to the
    		smallest angle of A and ...
    		Given the triangle C formed by using the second largest
    		angle of A (called a2), the side A1 and the smallest angle
    		of B (using ASA), the second longest side of B (called B2) 
    		is less than or equal to the length of the side of C
    		opposite the angle a2.
    
    		OR:
    
    		The second largest angle of B is less than or equal
    		to the second largest angle of A and ...
    		Given the triangle D formed by using the smallest angle of 
    		A (called a3), the side A1 and the second largest angle
    		of B (using ASA); the smallest side of B (called B3) 
    		is less than or equal to the length of the side of D
    		opposite the angle a3.
    
    		OR:
    
    		The second largest angle of A is less than the second
    		largest angle of B and the smallest angle of A is less
    		than the smallest angle of B and the height of B over
    		it longest side is less than or equal to the height
    		of A over its longest side.
	    )

I got a little bit lost in the logic there, but I think this is a counter-
example:
	A: 78, 80, 21 degrees; longest side = 1
	B: 60, 60, 60 degress; all sides = .99
in that it seems to satisfy the conditions but won't nest.

    Re .12: "I'm confused"
    
    I am sympathetic.  I am working on more easily understood explanations,
    the concept I am using is pretty simple, but the formal conditions
    are a mess.  I could do better if I could find some of the Side
    to Angle formula for triangles.  You know, like given SSS what are
    the angles, and given ASA what are the remaining sides and angle?
    
    Anyone who has these please feel free to post.
    
    Anyway on to your counter example:
    
>	A: 78, 80, 21 degrees; longest side = 1
>	B: 60, 60, 60 degress; all sides = .99
>in that it seems to satisfy the conditions but won't nest.

    I do not believe that this meets all of the criterion.  Clearly
    it meets the first part of the "AND" since B's longest side is shorter
    than A's longest side.
    
    Since the second largest angle of B is less than the second largest
    angle of A, we construct a triangle consisting of the second largest
    angle of B (60), the longest side of A (1) and the smallest angle
    of A (21).  By my eyeballing of this triangle, it appears that the
    smallest side of B is NOT less than or equal to the length of the
    side opposite the 21 degree angle in our new triangle.
    
    My reasoning for this:
    
    	The new trangle (D) has angles (99, 60, 21) with the longest
    	side (1) being opposite the 99 degree angle.  Since 21 is the
    	smallest angle, the side opposite it must be somewhat smaller than
    	the side opposite the 99 degree angle.  Therefore it must be
    	significantly smaller than 1 which means it must certainly be smaller
    	than .99 also.
    
    
    --  Barry

RE: .13 (Barry)
    
    SSS formula:
    
    		    2    2    2
    		   b  + c  - a
    	A = arccos ------------
    		        2bc
    
    					Topher