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
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There can me at mostintersections of these perpendiculars. We know the three perpendiculars of a triangle intersect at a point. Now, whenever we draw a perpendicular from a point(
) to the line joining two other points(
and
), we are forming a triangle
. And, obviously, that perpendicular intersects other two perpendiculars at exactly one point. So, all the intersections of two perpendiculars are actually the orthocentre of the triangles formed by choosing all combinations of three points. So, there may be at most
trianglesand so, at most
perpendiculars at a point.
intersections of these perpendiculars.
We know the three perpendiculars of a triangle intersect at a point. Now, whenever we draw a perpendicular from a point(
) to the line joining two other points(
and 
), we are forming a triangle 
. And, obviously, that perpendicular intersects other two perpendiculars at exactly one point.
So, all the intersections of two perpendiculars are actually the orthocentre of the triangles formed by choosing all combinations of three points.
So, there may be at most 
trianglesand so, at most 
perpendiculars at a point.
