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Difference between revisions of "1964 IMO Problems/Problem 5"
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== Solution ==
 
== Solution ==
 
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There can me at most <math>10</math> intersections of these perpendiculars.  
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  We know the three perpendiculars of a triangle intersect at a point. Now, whenever we draw a perpendicular from a point(<math>A</math>) to the line joining two other points(<math>B</math> and <math>C</math>), we are forming a triangle <math>ABC</math>. And, obviously, that perpendicular intersects other two perpendiculars at exactly one point. So, all the intersections of two perpendiculars are actually the orthocentre of some triangle. So, there may be at most <math>{5}\choose{3}=10</math> trianglesand so, at most <math>10</math> intersections. As we are considering maximal case we avoid the coincidence of more than <math>3</math> perpendiculars at a point.
+
  There can me at most <math>10</math> 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(<math>A</math>) to the line joining two other points(<math>B</math> and <math>C</math>), we are forming a triangle <math>ABC</math>. 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 <math>\binom{5}{3}=10</math> trianglesand so, at most <math>10</math> intersections. As we are considering maximal case we avoid the coincidence of more than <math>3</math> perpendiculars at a point.

Revision as of 03:10, 24 February 2020

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

This problem needs a solution. If you have a solution for it, please help us out by adding it. 

  There can me at most $10$ 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($A$) to the line joining two other points($B$ and $C$), we are forming a triangle $ABC$. 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 $\binom{5}{3}=10$ trianglesand so, at most $10$ intersections. As we are considering maximal case we avoid the coincidence of more than $3$ perpendiculars at a point.
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