# How many lines are determined by 8 points, no more than 3 of which are collinear? How many triangles are determined by the same points if no more than 4 are coplanar?

I count 14 lines

#### Explanation:

Collinear points

We have 8 points (and just to have them named, we'll have $A , B , C , D , E , F , G , H$)

A line is defined by 2 points, and so we can have lines $A B , A C , A D , \ldots , G H$. But we are being limited by being told that no more than 3 points are on the same line.

Let's consider the case where we have $A , B , C$ on the same line - that means that $A B , A C , B C$ are all the same line. No other point can be on this line and since 2 points are needed to define a line, any other point (say for instance, $D$) can't be on a line with any two of $\left(A , B , C\right)$.

Let's go one further. Let's take line $D E$. It can run through $A$, or $B$, or $C$, but not any more than one of them (otherwise it'll be collinear with all three points, $A , B , C$). So let's assume that we have line $A D E$. We know points $B , C$ are not collinear with this and just like before, any of the remaining points ($F , G , H$) can't be collinear with any two points $A , D , E$.

So what lines have we made so far that have 3 points collinear?

$A B C , A D E$

Following this pattern, what other lines with 3 collinear points can we make? As I'm working through this, I'm keeping in mind that the maximum number of times I'll be able to list each point is three:

$A F G , B D F , C D G , B E G , H C E$

I don't think I can make any more lines like this, so we have 7 with three collinear points.

And now let's make the remaining lines. These will be two point lines that haven't been connected by our three point lines:

$A H , B H , C F , D H , E F , F H , G H$

And so I count 14 lines in total.