Question 766795
four points P,Q,R,S are such that PQ=10,QR=30,RS=15 and PS=m. If m is an integer and no three of these points lie on a straight line, what is the number of possible values of m?
<pre> 
{{{drawing(400,13600/59,-27,32,-17,17,circle(-15,0,.3),circle(15,0,.3),
locate(-15,0,Q),locate(15,0,R),line(-11,9.1651514,8,13.266499),
line(-15,0,15,0), circle(-15,0,10), circle(15,0,15),locate(-10,0,10),
locate(-11,12,P), locate(8,16,S),red(triangle(-25,0,-15,0,-20,0)),
red(triangle(30,0,15,0,20,0)), locate(-26.6,1,A), red(locate(-21,0,10)),
green(line(-15,0,-11,9.1651514),line(15,0,8,13.266499)),
locate(7.1,0,15), locate(9,7,15), red(locate(22.5,0,15)),
locate(-2.8,13.5,m), locate(30.5,1,D),
locate(-16,6,10),locate(-5,2.3,B),locate(-2.8,0,5), locate(-1,2.3,C)

 )}}}

Notice in the drawing that QR = QB+BC+CR = 10+5+15 = 30 units, as given.

Let's first of all assume that it would not matter if three of the
points P,Q,R,S could lie on a straight line.  Then  

P could be any point on the circle on the left.
S could be any point on the circle on the right.

Therefore PS is longest when P is at A and S is at D.
In that case the length of PS would be 

AD = AQ+QB+BC+CR+RD = 10+10+5+15+15 = 55 units

and PS is shortest when P is at B and S is at C.
In that case the length of PS would be 

BC = 5 units

Since m is an integer it could be any of these values

5,6,7,...,53,54,55

But when we require that no three of P,Q,R,S can lie on a
straight line, we must rule out 5 and 55, so the possible
values are 

6,7,8,...,52,53,54

There are 54 integers from 1 through 54 inclusive, and 
we cannot use 1,2,3,4 or 5, so there are 54-5 = 49 
possible integer values for m, the length of PS.

Answer: 49

Edwin</pre>