Conference rusure::math

sin r= XB/CX
sin i= XA/DX

so XA/DX = 2 XB/CX  --- (1)

     2    2     2
   CX  = a  + XB    --- (2)
   
     2    2     2
   DX  = a  + AX    --- (3)

   XA + XB = 2a      -- (4)

four indept. equations, four unknowns CX,DX,XB,XA .  QED

    

        Or, let AX = x, then sin i = 2 sin r becomes
        
               x                     2a - x
        --------------- = 2  ----------------------
        sqrt(a^2 + x^2)      sqrt(a^2 + (2a - x)^2)
        
        Or, squaring both sides,
        
           x^2           (2a - x)^2
        --------- = 4 ----------------
        a^2 + x^2     a^2 + (2a - x)^2
        
        Multiply by both denominators to get a fourth degree
        polynomial for x in terms of a.
        
        Dan

   This would lead to the condition that sin(i)^2 = 4*sin(r).
   Expressing this in terms of AX = x, with AD = 1, AB = 2, we find

	x^2         (2-x)^2
       ----- = 4 * ---------
       1+x^2       1+(2-x)^2


    Or

	3 x^4 - 12 x^3 + 15 x^2 - 16 x + 16 = 0

    Which doesn't seen to have a simple closed form solution.

    A usable numerical solution would be about 1.538264244165589...

    - Jim

	3 x^4 - 12 x^3 + 15 x^2 - 16 x + 16 = 0

suppose it has integral root,let w be a solution 

i.e. 3 w^4  - 12 w^3 + 15 w^2 - 16 w = - 16

i.e  (3 w^3  - 12 w^2 + 15 w   - 16) w = - 16

LHS is a product of two integers.
then w is a divisor of  16

so possible solutions are 16, 8, 4, 2, 1

substitute back each divisor in the equation, if the equation has
an integral solution, it will be found among these.

non of these substitutions leads to a solution, so the equation has
a non integral solution.

assume the equation has a rational solution, w= p/q where p,q are relative
primes, then 

       p^4       p^3       p^2
    3  --- - 12  ---- + 15 --- - 16 p/q + 16 =0
       q^4       q^3       q^2

or  3 p^4 - 12 q p^3 + 15 q^2 p^2 - 16 q^3 p + 16 q^4 =0 -- (1)

or  3 p^4 = - ( - 12 p^3 + 15 q p^2 - 16 q^2 p + 16 q^3) q

ie  q divides 3 p^4, i.e q divides 3  since p,q are relative primes.

similarly work on p, from equation (1)

 16 q^3 = - ( 3 p^3 - 12 q p^2 + 15 q^2 p - 16 q^3) p 

so p divides 16 q^3, i.e. p divides 16 since p,q are relative primes.

so possible roots are

p/q such that p= 16,8,4,2,1  and q= 3,1

i.e.  16/3 or 8/3 or 4/3 or 2/3 or 1/3

back substitution, leads to no solution, (unless my substitution was wrong)
then the solution of this in not in the integral or rational field.

well , now i feel better.

/Nasser

    

        A look at the problem in .0 shows there can't be any
        "reasonable" integral solutions.  Neither x=0 nor x=a
        nor x=2a will give angles with sin i = 2 sin r.
        
        A quick check shows that the polynomial has no repeated
        roots (its derivative doesn't divide evenly into it). 
        Since the polynomial has degree four and real
        coefficients, there will be either 0, 2, or 4 real roots. 
        I doubt there will be no real roots, esp. since a
        previous reply gave one to many decimal places. :-)  Of
        the real roots, one would correspond to x between a and 2a
        and the others would be outside the rectangle (x < 0a or
        x > 2a).
        
        Dan

>       A quick check shows that the polynomial has no repeated
>       roots (its derivative doesn't divide evenly into it). 

    Actually the test for repeated roots is if the discriminant
    vanishes or not.

    You can find out if a pair if polynomials have a common factor
    by evaluating the resultant of the polynomials, a determinant
    formed from their coefficients.

    Suppose your polnomials are a(x) and b(x) of degrees m and n.
    Then there will be a common factor if you can form

	a(x)*c(x) = b(x)*d(x)

    where deg(c) < deg(b) and deg(d) < deg(a).

    If you can solve the resulting system of simultaneous equations
    for c and d's coeficients there's a common factor.

    To get the discriminant, take the resultant of p'(x) and n*p(x)-x*p'(x).

    The resultant matrix has cyclically shifted coefficient vectors in its
    columns like this example of a cubic and quadratic.

    | a3    b2       |
    | a2 a3 b1 b2    |
    | a1 a2 b0 b1 b2 |
    | a0 a1    b0 b1 |
    |    a0       b0 |

    The a's are repeated by the degree of b and the b's are repeated by the
    degree of a.  (a3 is the leading coefficient.)

    - Jim

        Oops!
        
        I didn't follow all of .-1, but the condition I meant to
        use isn't whether p'(x) divides p(x), it is whether
        gcd(p'(x),p(x)) is nontrivial.  If p(x) has a factor of
        (x - root)^n (n >= 2) then p'(x) has a factor of (x - root)^(n-1)
        and so (x - root) is a common factor, which the gcd algorithm
        will find.
        
        Dan