Five Points on a plane.. 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...
 A: 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 
$$435-70-30-20=315$$
A: 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 $5×6×(5+5+5+5)=600$ ways. 
   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 $300+10+5=315$.
