Determining all ( a , b ) on the unit circle such that 2 x + 3 y + 1 ≤ a ( x + 2 ) + b ( y + 3 ) for all ( x , y ) in the unit diskIn the middle of another problem, I came up with the following inequality which needed to be solve for ( a , b ) : 2 x + 3 y + 1 ≤ a ( x + 2 ) + b ( y + 3 )for all ( x , y ) ∈ R 2 with x 2 + y 2 ≤ 1..Here the solution for ( a , b ) must be a subset of the unit circle, and I believe that it is a singleton.

Question

Determining all   (  a  ,  b  ) on the unit circle such that   2  x  +  3  y  +  1  ≤  a  (  x  +  2  )  +  b  (  y  +  3  ) for all   (  x  ,  y  ) in the unit diskIn the middle of another problem, I came up with the following inequality which needed to be solve for   (  a  ,  b  ) :  2  x  +  3  y  +  1  ≤  a  (  x  +  2  )  +  b  (  y  +  3  )for all   (  x  ,  y  )  ∈        R    2   with       x    2    +      y    2    ≤  1..Here the solution for   (  a  ,  b  ) must be a subset of the unit circle, and I believe that it is a singleton.

Yair Boyle · Accepted Answer

Rewrite   2  x  +  3  y  +  1  ≤  a  (  x  +  2  )  +  b  (  y  +  3  ) as   (  2  −  a  )  x  +  (  3  −  b  )  y  ≤  2  a  +  3  b  −  1. Solve for the intersection of the line   (  2  −  a  )  x  +  (  3  −  b  )  y  =  2  a  +  3  b  −  1 with       x    2    +      y    2    =  1 to get that the   x-coordinates of the intersection are            −      2              a        2            −      3      a      b      +      5      a      +      6      b      −      2      ±              −        (        b        −        3                  )          2                (        3                  a          2                +        12        a        b        +        8                  b          2                −        12        )                            a        2            −      4      a      +              b        2            −      6      b      +      13        .If this has two real solutions, then the inequality cannot be satisfied for all   (  x  ,  y  ) in the unit circle, so we need to find the intersection of   −  (  b  −  3      )    2    (  3      a    2    +  12  a  b  +  8      b    2    −  12  )  ≤  0 and       a    2    +      b    2    ≤  1. Simultaneously solving these equations, we find that the real solutions for   (  a  ,  b  ) are   (      2          13        ,      3          13        ) and   (  −      2          13        ,  −      3          13        ). It&#039;s easy to see that the first solution is valid and the second is invalid - plug in   (  x  ,  y  )  =  (  0  ,  0  ), for instance.

Determining all ( a , b ) on the unit circle such that 2 x + 3 y + 1 &#

Answered question

Answer & Explanation

New Questions in Algebra I