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Difference between revisions of "1964 IMO Problems/Problem 5"
 
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== Solution ==
 
== Solution ==
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{{solution}}
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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.
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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.
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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.
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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.
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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.  
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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.
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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.
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So, total ways <math>300+10+5=315</math>.
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== See Also ==
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{{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
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