Straight distance between two towns 
The map shows the distance of four towns as shown in the figure. Compare quantity A with quantity B. A: Straight distance between Austen and Seburg
B: Straight distance between Coaltown and Woodland
Is quantity B always greater than quantity A?  
What I think is that if Austen=A, Coaltown=C, Seburg=S and Woodland=W then |18-x|< AS< |18+s| and |19-x|< CW< |19+x|. So, AS and CW can vary. But, my instinct says CW>AS.
A: The answer is NOT always.

If C’ is the point on AW such that $CC’ \bot SW$, then CC’ is the shortest, and any other lines (like CW) will be longer that CC’ and hence longer than AS.


However, if we shift the whole  SW line left one unit such that $CW \bot SW$, then CW is the shortest, and any other lines (like CC’) will be longer that CW and hence AS = CC’ will be longer.
A: Mark a point $P$ 1 km to the west of $W$, so that it's 18 km from $S$.  (I'm assuming $AC$ is always parallel to $SW$.)  By the symmetry of the picture, $|AS|=|CP|$.  Now we can see that $|CW| >|CP|=|AS|$.
A: If $a$ is the length CS and $t$ the angle ACS = angle WSC, then the squared distances, using the law of cosines, of CW and AS are respectivelty
$$ (CW)^2=a^2-38a \cos t+361,\\ 
(AS)^2=a^2-36 a \cos t +324.$$
Hence the square of the (presumably) longer minus that of the shorter is
$$-2a \cos t+37.\tag{1}$$
Now if $37/(2a)<1$ we can find a first quadrant angle $t$ [in $(0,\pi/2)$] for which $(1)$ holds with equality, and then slightly larger and smaller $t$ make that inequality go one or the other way. Conclude the two distances are not necessarily comparable (without also knowing something more such as the length CS).
