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Given 5 points in space such that no three of them are colinear and no four of them are coplanar. If we consider all the planes containing any 3 of these 5 points, and the intersections of all these planes taken two by two, how many lines will we obtain, at most?

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At first, my idea was just to say that there are $C(3,5) = 10$ (binomial coefficient) different planes. And then any combination of 2 planes must give a different intersection line: therefore $C(2,10)=45$ should be the answer?

However, I sense that this is naive and overly simple. Could I be missing something?

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  • $\begingroup$ You have to decrease the final number. Imagine plane A is made of P1,P2,P3 and plane B is made of P1,P2,P4. So, common line is P1,P2. Now another plane C is P1,P2,P5; the common line with planes A and B is still P1,P2 $\endgroup$
    – Ripi2
    Commented Dec 16, 2020 at 0:39
  • $\begingroup$ Indeed, that's why I sensed my approach was too simplistic. How shall we go about solving the problem, then? $\endgroup$
    – vin sands
    Commented Dec 16, 2020 at 0:42
  • $\begingroup$ Count the number N of planes sharing P1,P2. If N>2 then N-1 lines must be dismmised. Repeat with P1,P3, or P1,P4,or P3,P5, etc $\endgroup$
    – Ripi2
    Commented Dec 16, 2020 at 0:44
  • $\begingroup$ @vinsands One way to continue, is to consider how many times you have counted each line in the count of $C(2,10)=45$ lines. You have counted each line once for every pair of planes that contain in. But as the comment above illustrates, many lines are contained in three planes, and hence in three pairs of planes. Can you continue from here? $\endgroup$
    – Servaes
    Commented Feb 16, 2021 at 11:06

2 Answers 2

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OK, based on the past comments, here is how I see it:

There are 10 pairs of points, i.e. through each pair we can fit 3 planes (the planes going through that pair as well as each of the other 3 points separately). Indeed, then there is only 1 common intersection to these 3 planes. Therefore, the planes formed this way would have just 10 intersections that coincide with an existing pair of points (out of the theoretical 30 'intersections' we could form with the naive approach I first used).

There remains another 15 planes intersections which do not coincide with any existing pairs of points (these are the intersections of 123-145, 123-245, 123-345, 124-135, 124-235, 124-345, 125-134, 125-234, 125-345, 134-235, 134-245, 135-234, 135-245, 145-234, 145-235). Therefore, IF (but are we sure of this?) none of these intersection are also the same, we would have 10+15=25 total intersection for these 10 planes? Is that correct?

This is intriguing: is this a famous result which generalizes to N points?

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You are right that there are $\binom{5}{3}= \color{red}{10}$ planes BUT ...

The $3$ planes $(1,2,3),(1,2,4),(1,2,5)$ will intersect in the line $(1,2)$.

Can you fix things from here ?

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