Question 1144732
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            As the problem is formulated, it is not exactly clear what this sum,  AB + BC,  means.


            I will interpret it as the sum of the lengths of segments  AB  and  BC.


            My solution is different from the solution by the tutor @greenestamps, and gives another answer.



<pre>
It is a version of the classic Fermat's problem : 

    on the plane, a straight line L and two points A and C are given in the same  of the two half-planes.

    Find the shortest path from A to C under the imposed condition, that this path must touch the line L at some point B.


In our case the points are two given points A and C; the straight line L is the x-axis, 
and we are searching for the point B on this line (on this x-axis)


The solution (very well and widely known) is THIS :  


    - take the point C; reflect it symmetrically about x-axis to get its mirror image C'; 

    - then connect the point A with C' by a straight line (which is unique);

    - its intersection with x-axis will be the point B, you are looking for.


So, everything is very easy.


    - The mirror image of the pont C(8,3) is the point C'(8,-3);

    - the straight line connecting A(-4,9) with C'(8,-3) has the slope  m = {{{((-3)-9)/(8-(-4))}}} = {{{(-12)/12}}} = -1;

    - the equation of this straight line is  y = 9 - 1*(x-(-4) = 9 - (x+4) = -x + 5;

    - its intersection with the x-aqxis is  at x = 5  (x-intercept);

    - therefore, the point B has coordinates B = (5,0);

    - hence,  k = 5  is the  <U>ANSWER</U>  to the problem's question.
</pre>

Answered, solved, explained in all details and completed.


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As I presented here this solution for you, it is, usually, one full-scale session of a school Math circle.


So, it is as if you visited a first-class school Math circle today at a local University . . .