I know many such questions have already been asked but this one seemed to be a very messy and a lengthy question for me to solve. The question is as follows

Consider seven different points $P_1,P_2,P_3,P_4,Q_1,Q_2,Q_3$ in plane such that $P_1,P_2,P_3,P_4$ are on straight line $'l'$; and $Q_1,Q_2,Q_3$ are non collinear points and none of them lies on straight line $'l'$( $Q_1,Q_2,Q_3$ are on same side of line $'l'$)

From each of the three points $Q_1,Q_2,Q_3$ perpendicular lines are drawn to the straight lines formed by joining any two of the given points (excluding the point from which the perpendicular line is drawn). Find the maximum possible number of points of intersection of perpendicular lines ( excluding the points $Q_1,Q_2,Q_3$ ).

After a lot of messy work and a turmoil of 3 hours I am getting the answer as $283$. I pretty much think that the answer might be correct. But I wanted to know thoughts of members on this site over how to approach this problem without expending so much of time because such question was asked in our examination papers a few years back and I cannot afford so much time on a single question in exam. So please share your thoughts over this question

  • $\begingroup$ @Markus Scheuer Any Thoughts over this question? $\endgroup$ – Rohan Shinde Jan 26 '18 at 9:39
  • $\begingroup$ @robjohn Any thoughts over this question? $\endgroup$ – Rohan Shinde Jan 26 '18 at 9:57


Calculating no of perpendicular lines for each $Q_1,Q_2$ and $Q_3$ point

CAT $1$ - line made by joining $P$ points

CAT $2$- lines made by joining $1$ point from $P$ and $1$ point from $Q$

CAT $3$- lines made by joining points from $Q.$

For $Q_1$ point

CAT $1$ - since all the $P$ points lie in same line $'l'$ there is only $1$ single line and there will be only $1$ perpendicular drawn from $Q_1$.

CAT 2 - Total no. of perpendicular drawn from $Q_1$ is given by $^4C_1 * ^2C_1 =8$ ways. (selecting $1$ point from $P_1, P_2 ,P_3 , P_4$ and another point from $Q_2 ,Q_3$)

CAT $3$ – only $1$ perpendicular will be drawn to the line $Q_2Q_3 $

Total no. of perpendiculars drawn from $Q_1=1+8+1=10$

Total no of perpendiculars lines$ = 10*3 = 30$

No of intersection points from $30$ perpendicular lines = $^{30}C_2=435.$ But this is not true

Case $1$- some of these lines are parallel to each other therefore they will not intersect

(a) Lines from $Q_1 ,Q_2 ,Q_3$ to line $‘l'$.Total points to be cut$= ^3C_2=3$

(b) Lets connect a line between $Q1$ and $P_1$ the perpendicular are drawn from $Q_2$ and $Q_3$ to it will be parallel. such lines can be given by $^3C_1*^4C_1=12$

Case $2$: From each of the $3$ points ($Q_1,Q_2,Q_3$), $10 $ perpendicular lines are drawn through it. So it is the point of intersection of $10$ perpendicular lines. We have counted this point $^{10}C_2 =45$ times as a point of intersection of perpendicular lines. So the number of points of intersection should be cut by $3×45=135$(as it is mentioned that $Q_1,Q_2,Q_3$ have to be excluded)

Case $3$: The orthocenter of triangle formed by $Q_1Q_2Q_3 $has been counted $3$ times instead of $1$

Total number of intersection $= 435 - 3 - 12 - 135 - 2 = 283$

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  • $\begingroup$ Why does your answer turn out to be 286 $\endgroup$ – Rohan Shinde Jan 26 '18 at 13:08
  • $\begingroup$ @Manthanein i get 286 instead of 283. $\endgroup$ – NewGuy Jan 26 '18 at 13:09
  • $\begingroup$ i have calculated it i cant find the remaining 3 cases $\endgroup$ – NewGuy Jan 26 '18 at 13:10
  • $\begingroup$ I got it. Case 2. It should have been 45*3 instead of 44*3 $\endgroup$ – Rohan Shinde Jan 26 '18 at 13:10
  • $\begingroup$ yes it is mentioned in the question that you have to exclude Q1,Q2,Q3 $\endgroup$ – NewGuy Jan 26 '18 at 13:12


first we will calculate how many lines are made using the above points it comes from three categories:-

CAT 1- lines made by joining P point.

CAT 2- lines made by joining 1 point from P and 1 point from Q

CAT 3- lines made by joining points from Q.

CAT 1 - since all the P points lie in same line 'l' there is only 1 single line and the perpendiculars drawn from Q1,Q2,Q3 intersect in exactly 3 places.

CAT 2 - total no. of lines is given by 3C1 * 4C1 =12 ways. Perpendiculars are drawn from the other 2 points giving the total no. of intersection points= 12 * 2= 24

CAT 3 - total no. of lines is given by 3C2 = 3 ways. Perpendicular is drawn from the remaining point. Hence no of intersection = 3

total no of points = 3 + 24 + 3 = 30 points

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  • $\begingroup$ Well, I asked my teacher just few minutes ago my teacher said that 283 was the right answer $\endgroup$ – Rohan Shinde Jan 26 '18 at 8:56
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    $\begingroup$ If you dont mind can you post the solution or any hint related to it. $\endgroup$ – NewGuy Jan 26 '18 at 8:59
  • $\begingroup$ Well I made so many of cases even I don't remember where which concept I used. Moreover I don't maintain proper solution to my question so the answer I got, got messed up in the rough work $\endgroup$ – Rohan Shinde Jan 26 '18 at 9:01
  • $\begingroup$ math.stackexchange.com/questions/2298024/… use this method in your scenario i have calculated the answer comes out to be 283 $\endgroup$ – NewGuy Jan 26 '18 at 12:05
  • $\begingroup$ You can use that concept to answer this question here so that I understand what method you used for 4 collinear points. $\endgroup$ – Rohan Shinde Jan 26 '18 at 12:18

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