Five Points on a plane..

Question

There are 5 points on a plane. From each point, perpendiculars are drawn to the line joining the other points. What is the maximum number of points of intersection of these perpendiculars ? I cant think of the logic please help me out...

Answer 1

There are totally $\displaystyle \binom{5}{2}=10$ lines joining any two of the five points. From each of the five point, $\displaystyle \binom{4}{2}$ perpendicular lines can be drawn. So there are totally $5\times 6$ perpendicular lines drawn.

$30$ lines can intersect at at most $\displaystyle \binom{30}{2}=435$ points. However, this is not the answer, as:

(1) From each of the five points, $6$ perpendicular lines are drawn through it. So it is the point of intersection of 6 perpendicular lines. We have counted this point $\displaystyle \binom{6}{2}=15$ times as a point of intersection of perpendicular lines. So the number of points of intersection should be cut by $5\times 14=70$.

(2) some of these perpendicular lines are parallel to each other. Let $A$, $B$, $C$, $D$ and $E$ be the five points. The three perpendicular lines from $C$, $D$ and $E$ to $AB$ are parallel to each other and do not intersect. So the number of points of intersection should be further cut by $10\times 3=30$.

(3) The five points are the vertices of $\displaystyle \binom{5}{3}=10$ triangles. The altitudes of each triangle intersect at a single point, which is the orthocentre of the triangle. We have triple counted these intersection points. So the number of points of intersection should be further cut by $10\times 2=20$.

Therefore, the maximum number of points of intersection is

$$435-70-30-20=315$$