Difference between revisions of "1964 IMO Problems/Problem 5"
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== 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. | ||
| + | |||
| + | 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. | ||
| + | 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. | ||
| + | |||
| + | 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 12:49, 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 
. 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 | ||
