1964 IMO Problems/Problem 5

Problem

Suppose five points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or coincident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maximum number of intersections that these perpendiculars can have.

Solution

   Suppose, those five points are . Now, we want to create some special structure. Let, we take the line  and draw a perpendicular from  on , andd call it . We can do this set up in  ways. There will  such  s.
    
   Now, we will find how many other perpendiculars intersect the line. We can do this in total  ways. Why? See, can draw perpendiculars from  and  to other lines( we haven't counted the perpendicular from  to  and perpendicular from  on  , as they intersect  at the same point) in  ways for each. So, total  ways. 
 Now,  perpendiculars from each  and  on the other lines except on ( because in this case teh perpendiculars from  and  will be parallel to  , and so shall not intersect). So,total  cases.
  From, these two cases we get  will be intersected at at most  points. 
  But, as we have passed this algorithm over all the five points, we have counted each intersection points twice. So, there are total  ways. 
 Now, as we had excluded  the orthocentres, we have to add now. There are total  orthocentres. Also we should add those vertices as these are also point of intersection of silimar perpendiculars, there are  such.

So, total ways .