Question

What is the maximum number of non-overlapping sectors that can be created when a circle is crossed by 3 straight lines?
A. 3
B. 4
C. 6
D. 7

Answer 1

Hint: The maximum number of parts into which a circle (or any convex plane figure) can be divided with n straight lines is $c(n) = 1 + \dfrac{{n(n + 1)}}{2}$ . We shall use this formula to gain the exact number sectors that can be created.

Complete step-by-step answer:
Consider first the number $l(n)$ of parts into which a straight line segment can be divided with n points. Obviously it is $l(n) = n + 1$ . Then consider that a straight line either divides a convex plane figure into two parts (if it intersects it) or leaves it undivided (if it doesn’t intersect it). If the figure gets divided, the two parts are still convex because they are the intersection of a convex figure and a semi-plane, which is also a convex figure; the intersection of two convex figures is convex. so we can repeat the same reasoning for another straight line intersecting those parts , and so on. Start with a convex figure (for instance a circle, but any convex figure will do). Divide it with n-1 straight lines, always taking care to get the maximum number of parts, which we call $c(n - 1)$ . Then cross the figure with one more straight line. On this n-th straight line we shall see the following:
i. the section of the original figure, which will be a straight line segment
ii. the traces of the previous n-1 straight lines, which will be n-1 points
iii. the sections of the intersected parts, which will be the parts into which the traces of the previous lines divide the section of the original figure
If we position the ${n^{th}}$ is the best possible way, we will intersect $l(n - 1) = n$ parts, because $l(n - 1) = n$ is the maximum number of parts into which a straight line segment can be divided with n-1 points. So we intersect n parts and leave $c(n - 1) - n$ parts undivided, obtaining a total of $c(n) = 2n + c(n - 1) - n = c(n - 1) + n$ parts.
Starting with $c(0) = 1$ (with zero lines the original figure remains in one piece) and using the recursive equation $c(n) = c(n - 1) + n$ We obtain;
$c(n) = 1 + \sum\limits_{i = 1}^n i = 1 + \dfrac{{n(n + 1)}}{2}$ .
If we put $n = 3$ in the formula $c(n) = 1 + \dfrac{{n(n + 1)}}{2}$ we will get the number of sectors that can be formed.

\therefore c(n) = 1 + \dfrac{{n(n + 1)}}{2}\\\ \Rightarrow c(3) = 1 + \dfrac{{3(3 + 1)}}{2}\\\ \Rightarrow c(3) = 1 + \dfrac{{3 \times 4}}{2}\\\ \Rightarrow c(3) = 1 + 3 \times 2\\\ \Rightarrow c(3) = 1 + 6\\\ \Rightarrow c(3) = 7 \end{array}$$ We can also do this question without using any formula and still get the same result as; The first line can separate the circle into at most 2 sectors, because the line is straight and only has 1 sector to divide. The second line can separate the circle into at most 4 sectors because it is a straight line and only has 2 sectors to divide. When drawing the third line, there are 4 sectors which could be divided, but because the line is straight it can only reach 3 of the sectors, thus dividing 3 of the original 4 sectors into 6 sectors and leaving one of the original 4 sectors untouched, which totals 7 sectors. Note: One cannot remember each and every formula everytime so it's better to work on the reasoning part and we can also do these types of questions just by taking each possibility under consideration, but it's better to use the formula if you have more number of lines like for example if n becomes 10.