MHB Segments in a Circle: How Many Parts Can Form & What Will It Look Like?

  • Thread starter Thread starter Monoxdifly
  • Start date Start date
  • Tags Tags
    Circle Form parts
AI Thread Summary
A circle divided into 6 segments can form a maximum of 22 parts, as determined by the formula (c(c+1)/2) + 1, where c represents the number of cuts. The sequence of maximum parts follows a pattern: 2, 4, 7, 11, 16, and 22 for 1 to 6 segments. To achieve the maximum number of pieces, all lines must intersect without crossing at the same point. The discussion confirms that the initial assumption of 22 parts is accurate. Understanding these principles is essential for visualizing the configuration of segments in a circle.
Monoxdifly
MHB
Messages
288
Reaction score
0
If a circle has 6 segments, how many maximum parts which can be formed? I know that 1 segment makes 2 parts, 2 segments make 4 parts, and 3 segments makes 7 parts. Judging by the pattern, is the answer 22? What will the exact picture of the circle be? Thank you very much.
 
Mathematics news on Phys.org
Monoxdifly said:
If a circle has 6 segments, how many maximum parts which can be formed? I know that 1 segment makes 2 parts, 2 segments make 4 parts, and 3 segments makes 7 parts. Judging by the pattern, is the answer 22? What will the exact picture of the circle be? Thank you very much.

Looks like this sequence ...

$ 2, 4, 7, 11, 16, 22, ... , \dfrac{n^2+n+2}{2}, ...$
 
Last edited by a moderator:
The maximum number of pizza pieces formula is:

$$\frac{c(c+1)}{2} +1$$

where c is the number of cuts

You'll also notice that;

the maximum number of pieces formula -1 for every number = the triangle numbers

So that's why we add a +1 at the end of the formula

Substituting c for 6 gives us:

$$\frac{42}{2}+1= 21+1=22$$

So yes, you were correct

P.S. Just so you know, you need to cross all the lines with a line to make the largest number of pieces possible, however don't cross a line intersection
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top