Interesting Combinatorics Problem :: Help

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 $435-120 = \boxed{315}$.

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...

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 try to find the number of other perpendiculars which intersect $P_1$.

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. So, total $10+10=20$ 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 $300+10+5=315$.