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 .
See Also
1964 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |