We can observe that every point can be connected with all the others. It would mean $n(n-1)$ connections for $n$ points. But in this way all the connections would overlap twice because we can go from point A to point B and from point B to point A (two directions but only one line). Therefore we must divide the previous result by $2$ which gives

$\frac{n(n-1)}{2}$ number of connecting lines.

For $\mathbf{n = 37}$ we have $\mathbf {666}$ connecting lines.

Read further if you are interested in a formal proof.

I am going to show by mathematical induction that the formula is true for all natural numbers.

Let $N(n)$ denote the number of segments between $n$ points on the circle.

Firstly , we must show that the formula holds true for n = 1.

$N(1) = \frac{1\cdot 0}{2} = 0$ , which means no connecting line for one point what is true ( base case is fullfiled).

Secondly , take arbitrary natural number $k$ and assume: $N(k) = \frac{k(k-1)}{2}$ . Assuming this, we must prove that the formula is also true for $k+1$ . Thus, we must show that $N(k+1) = \frac{(k+1)k}{2}$ . Let’s see how the number of connections changes if we add one point. We can easily notice that we are just connecting the new $(k+1)th$ point with all the others. It means we draw additional $k$ segments so we end up with $N(k) + k$ connecting lines. We must now check if $N(k) + k = N(k+1)$ . Indeed

$N(k) + k = \frac{k(k-1)}{2} + k = \frac{k(k-1)+2k}{2} = \frac{k(k+1)}{2} = N(k+1)$

So the induction step is also fullfiled. We have now fulfilled both conditions of the principle of mathematical induction.
Therefore, by the principle of mathematical induction the formula is therefore true for every natural number.

Since every two points connect a segment, thus the number of segments is: $\large\binom{37}{2} = 666$

See the sequence, if points = 4 then conecting lines = 3+2+1=6, and if points = 5 then conecting lines =4+3+2+1 =10. Conecting points = 37 then conecting lines =36+35+..............+2+1=666

A line is made by choosing 2 distinct points. Each of the 37 points on the circle can be connected with 36 other points, excluding itself. Thus the total number of line segment should be $37 \times 36$ . But no, we have double counted each line. Line $AB$ is the same as line $BA$ . Thus the answer will be divided by two $\implies \frac{37 \times 36}{2} = \boxed{666}$

Explicitly, the number of lines for $n$ points is equal to $\frac{n \times (n-1)}{2}$ which is actually $\binom{n}{2}$

Number of diagonals in a polygon of $n$ sides $=\frac{n(n-3)}{2}$

Number of sides= $n$

Number of total lines created

$=n + (\frac{n(n-3)}{2})$

$= 37 + \frac{37×34}{2} = \boxed{666}$

There are $37$ connected non-overlapping points.

We can consider it as a $37-$ sided polygon.

Number of lines $=$ Number of sides $+$ Number of diagonals

Number of lines $=37 + \dfrac{37(34)}{2}$ (Thanks Google!)

Number of lines $=37 + 629 =\boxed{666}$

If there are $n$ points, pick an arbitrary point $A_1$ and connect it with all other points which would form $n-1$ segments(because 1 segment for each point except $A_1$ itself) then pick an adjacent point $A_2$ and connect it with all other points which would also be $n-1$ segments but there is already a segment between $A_2$ and $A_1$ so to avoid overlap we subtract one so $n-2$ segments. keep doing this till the last point and sum all the segments and you'll have $(n-1)+(n-2)+(n-3)+\cdots+3+2+1$ putting $n=37$ gives $666$