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# Interesting Combinatorics Problem :: Help

There are 5 Points in a plane from each points Perpendicular's are drawn to the Lines Joining other Points. what is maximum number of point of Intersection of these perpendiculars ?

Explain it to Me Please ! Thanks

Answer is $$315$$

Note by Karan Shekhawat
3 years, 2 months ago

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Just for the sake of convenience, let us name the points as $$A,B,C,D$$ and $$E$$. Consider any point as a reference point and WLOG take it to be $$A$$. A total of six lines can be made among the remaining points. So there would be six perpendiculars drawn from $$A$$. Thus there would be a total of of thirty lines. Thus the maximum number of points of intersection are $$\displaystyle \binom{30}{2} = 435$$.

As stated before there would be six lines through each of the points and we have counted the intersection points of these lines separately. Therefore we have counted$$5$$ points as $$\displaystyle 5 \times \binom{6}{2} = 75$$. Also all the perpendiculars from to a line would be parallel and would not have a point of intersection. A total of $$\displaystyle \binom{5}{2} =10$$ lines can be drawn between any two given points and there would a set of three parallel lines on each of these. Therefore we have counted an extra of $$30$$ points. Consider any of the $$\displaystyle \binom{5}{3} =10$$ triangles formed by these points. Just by a simple analysis, it is not a difficult task to realize that the orthocentres of these triangles would lie on these lines. We have counted each of these points thrice i.e. the intersection of three lines thereby counting an additional of $$20$$ points. Thus we have counted a total of $$75 -5 +30 +20 =120$$ points extra. Therefore, the total number of maximum possible points of intersections are $$435-120 = \boxed{315}$$.

- 3 years, 2 months ago

Thanks ! Many-Many Time To You "The great Sudeep " :)

- 3 years, 2 months ago

Assuming no trhee points are collinear and the pentagon is not regular, the number of Perpendiculars is $$5 \cdot {4 \choose 2} = 30$$ . Hence, the maximum number of intersection point is at most $$\frac{30\cdot 29}{2} = 435$$

But we have to notice that:

• if the perpendiculars are the altitudes of a triangle, then they are concurrent. Therefore the total number must be reduced by $2 \cdot {5 \choose 3} = 20$

• if the perpendiculars are drawn from three different points, and perpendicular to the two remaining, then they don't intersect each other (because they are parallel to each other). Therefore the total number must be reduced by $3 \cdot {5 \choose 3} = 30$

• if the perpendiculars are drawn all from the same point, they do not intersect each other in any other point (otherwise two sides would be parallel). Therefore the total number must be reduced by $\left( {6 \choose 2} -1 \right) \cdot 5 = 70$

These considerations reduce the total number by, respectively, $$20$$, $$30$$ and $$70$$ points. Hence the maximum number of points is $$N \leq 435-20-30-70 = 315$$. Now we should show that it's possible to achive it... Maybe with a diagram...

- 3 years, 2 months ago

Thanks a lot ! But sorry I did not understand The Third Point (70) . Can You Please explain it more in word's Thanks

- 3 years, 2 months ago

If we pick one of the five point, let's say $$A$$, the perpendiculars we draw from $$A$$ are $${6 \choose 2}$$ and since two lines cannot intersect in 2 different points only, alle these perpendiculars intersect in $$A$$ only. They could intersect in infinitely many points, but we can assume this is not happening. Therefore, we have to subtract the points of intersection of 6 lines with themselves, but one (that is $$A$$ ). Repeating the process for other 4 point we obtain the result above.

- 3 years, 2 months ago

Comment deleted Jan 18, 2015

- 3 years, 2 months ago

My weakest topic is combinatorics hence I fear I can't help.

- 3 years, 2 months ago