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I have the following Mathematica result under the assumption that {ax, bx, px, ay, by, py} \[Element] Reals, and after refinement by Simplify:

ConditionalExpression[-((
  ay/Sqrt[(ax - px)^2 + (ay - py)^2] - by/
   Sqrt[(bx - px)^2 + (by - 
      py)^2] + (-(1/Sqrt[(ax - px)^2 + (ay - py)^2]) + 1/
      Sqrt[(bx - px)^2 + (by - py)^2]) py)/(
  ay (bx - px) + by px - bx py + ax (-by + py))), 
 Im[((ax - bx) (ax - px) + (ay - by) (ay - py) - 
     I Abs[ay bx - ax by - ay px + by px + ax py - bx py])/((ax - 
       bx)^2 + (ay - by)^2)] < 0 || 
  Re[((ax - bx) (ax - px) + (ay - by) (ay - py) - 
     I Abs[ay bx - ax by - ay px + by px + ax py - bx py])/((ax - 
       bx)^2 + (ay - by)^2)] < 0 || 
  Im[((ax - bx) (ax - px) + (ay - by) (ay - py) - 
     I Abs[ay bx - ax by - ay px + by px + ax py - bx py])/((ax - 
       bx)^2 + (ay - by)^2)] > 0 || 
  Re[((ax - bx) (ax - px) + (ay - by) (ay - py) - 
     I Abs[ay bx - ax by - ay px + by px + ax py - bx py])/((ax - 
       bx)^2 + (ay - by)^2)] > 1]

Pictorially, it's this:

enter image description here

If you consider the first part of the OR, can you confirm that the Im[...] < 0 is nothing more than the stuff in the absolute value brackets (Abs[ay bx - ax by - ay px + by px + ax py - bx py]) being greater than zero?

Similar simplifications can be made for the rest of the conditions. How do I get Mathematica to execute these simplifications on its own?

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1 Answer 1

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With the expression in the question designated z, the conditions in the ConditionalExpression are given by z[[2]], which can be simplified by

FullSimplify[ComplexExpand[z[[2]]], {ax, bx, px, ay, by, py} ∈ Reals]
(* ((ax - bx)^2 + (ay - by)^2) Abs[ay (bx - px) + by px - bx py + ax (-by + py)] > 0
|| ((ax - bx)^2 + (ay - by)^2) (ax^2 + ay^2 + bx px - ax (bx + px) + by py - ay (by + py)) < 0
|| ((ax - bx)^2 + (ay - by)^2) ((ax - bx) (bx - px) + (ay - by) (by - py)) > 0 *)

Is this the sort of thing you had in mind?

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  • $\begingroup$ Yes! I don't understand. How did this end up working when you put it this way? $\endgroup$
    – bzm3r
    Commented Sep 24, 2015 at 6:16
  • $\begingroup$ My theory on why this works, having read up on ComplexExpand: while I used Assumptions to specify that all variables are real in an Integrate command which created z, I did not use it when applying FullSimplify. $\endgroup$
    – bzm3r
    Commented Sep 24, 2015 at 6:39
  • $\begingroup$ Also, consider the result we get now: the first and third disjunct(((ax - bx)^2 + (ay - by)^2) Sqrt[(ay bx - ax by - ay px + by px + ax py - bx py)^2] > 0 || ((ax - bx)^2 + (ay - by)^2) Sqrt[(ay bx - ax by - ay px + by px + ax py - bx py)^2] < 0) is always true! This is because squaring a number always return a positive result, adding two positive numbers results in a positive number, and taking the square root returns a positive and negative root.So, the conditional is not even necessary because these two branches will always be true? $\endgroup$
    – bzm3r
    Commented Sep 24, 2015 at 6:49
  • $\begingroup$ @user89 Actually, squaring a real number does not always return a positive result. It could return zero. However, Mathematica could have dropped the second condition in your last comment, because that quantity is never negative. I have modified my answer to make this happen. $\endgroup$
    – bbgodfrey
    Commented Sep 24, 2015 at 12:12
  • $\begingroup$ You're right, and thank you very much for showing me this! $\endgroup$
    – bzm3r
    Commented Sep 25, 2015 at 1:04

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