There are eight balls arranged in such a way that one is in the centre while others form a circle around the first. How many lines have to draw in order to join each ball with both adjacent balls and also with the central balls? No line can intersect the other line also.

Answer 1

To solve this problem, we need to analyze how to connect the eight balls arranged in a circle with one central ball.

1. Arrangement:

   • We have one central ball (let's call it \(C\)).
   • There are seven other balls arranged in a circle around \(C\). Let's label these balls as \(B_1, B_2, B_3, B_4, B_5, B_6, B_7\).

2. Connections:

   • Each ball \(B_i\) (for \(i\) from 1 to 7) needs to be connected to \(C\).
   • Additionally, each ball \(B_i\) is adjacent to two other balls (specifically, \(B_{i-1}\) and \(B_{i+1}\), where \(B_1\) is adjacent to \(B_7\) and vice versa due to the circular arrangement).

3. Calculating Lines:

   • Lines from the central ball \(C\): There are 7 lines connecting \(C\) to each of the 7 surrounding balls: \(C\) to \(B_1\), \(C\) to \(B_2\), ..., \(C\) to \(B_7\).
   • Lines between adjacent balls: Each surrounding ball connects to its two adjacent balls. Specifically:
     • \(B_1\) connects to \(B_2\) and \(B_7\)
     • \(B_2\) connects to \(B_1\) and \(B_3\)
     • \(B_3\) connects to \(B_2\) and \(B_4\)
     • ...
     • \(B_7\) connects to \(B_6\) and \(B_1\)

   There are a total of 7 connections among the adjacent balls (one connection each for the pairs: \(B_1-B_2\), \(B_2-B_3\), ..., \(B_7-B_1\)).

4. Total Lines:

   • Lines from \(C\) to each \(B_i\): \(7\)
   • Lines between adjacent balls: \(7\)

Thus, the total number of lines required is:

\[ 7 + 7 = 14 \]

Therefore, the total number of lines that need to be drawn is \(\boxed{14}\).