Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1964 IMO Problems/Problem 5"

(→‎Solution)
 
(4 intermediate revisions by 2 users not shown)
Line 6: Line 6:
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
Suppose, those five points are <math>(A, B, C, D, E)</math>. Now, we want to create some special structure. Let, we take the line <math>BC</math> and draw a perpendicular from <math>A</math> on <math>BC</math>, andd call it <math>P_1</math>. We can do this set up in <math>\binom{5}{1}\binom{4}{2}=30</math> ways. There will <math>30</math> such <math>P_i</math> s.
−
 
+
   
−
  There can me at most <math>10</math> intersections of these perpendiculars.  
+
Now, we will find how many other perpendiculars intersect the line. We can do this in total <math>20</math> ways. Why? See, can draw perpendiculars from <math>B</math> and <math>C</math> to other lines( we haven't counted the perpendicular from <math>B</math> to <math>AC</math> and perpendicular from <math>C</math> on <math>AB</math> , as they intersect <math>P_1</math> at the same point) in <math>5</math> ways for each. So, total <math>10</math> ways.
−
+
 
−
    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.  
+
Now, <math>5</math> perpendiculars from each <math>D</math> and <math>E</math> on the other lines except on <math>BC</math>( because in this case teh perpendiculars from <math>D</math> and <math>E</math> will be parallel to <math>P_1</math> , and so shall not intersect). So,total <math>10</math> cases.
−
  So, all the intersections of two perpendiculars are actually the orthocentre of the triangles formed by choosing all combinations of three points.  
+
From, these two cases we get <math>P_1</math> will be intersected at at most <math> 5*6*(5+5+5+5)=600 </math> 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.
+
 
 +
But, as we have passed this algorithm over all the five points, we have counted each intersection points twice. So, there are total <math>\frac{600}{2}=300</math> ways.
 +
 
 +
Now, as we had excluded  the orthocentres, we have to add now. There are total <math>\binom{5}{3}=10</math> orthocentres. Also we should add those vertices as these are also point of intersection of silimar perpendiculars, there are <math>5</math> such.
 +
 
 +
So, total ways <math>300+10+5=315</math>.
 +
 
 +
== See Also ==
 +
{{IMO box|year=1964|num-b=4|num-a=6}}

Latest revision as of 11:49, 29 January 2021

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 $(A, B, C, D, E)$. Now, we want to create some special structure. Let, we take the line $BC$ and draw a perpendicular from $A$ on $BC$, andd call it $P_1$. We can do this set up in $\binom{5}{1}\binom{4}{2}=30$ ways. There will $30$ such $P_i$ s.

Now, we will find how many other perpendiculars intersect the line. We can do this in total $20$ ways. Why? See, can draw perpendiculars from $B$ and $C$ to other lines( we haven't counted the perpendicular from $B$ to $AC$ and perpendicular from $C$ on $AB$ , as they intersect $P_1$ at the same point) in $5$ ways for each. So, total $10$ ways.

Now, $5$ perpendiculars from each $D$ and $E$ on the other lines except on $BC$( because in this case teh perpendiculars from $D$ and $E$ will be parallel to $P_1$ , and so shall not intersect). So,total $10$ cases. From, these two cases we get $P_1$ will be intersected at at most $5*6*(5+5+5+5)=600$ points.

But, as we have passed this algorithm over all the five points, we have counted each intersection points twice. So, there are total $\frac{600}{2}=300$ ways.

Now, as we had excluded the orthocentres, we have to add now. There are total $\binom{5}{3}=10$ orthocentres. Also we should add those vertices as these are also point of intersection of silimar perpendiculars, there are $5$ such.

So, total ways $300+10+5=315$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1964_IMO_Problems/Problem_5&oldid=143760"