cutting the pie p.o.w.
Problem Statement:
The main component we are supposed to incorporate is the use of In-Out tables to better understand the problem. We were able to incorporate this within the problem by using the number of chords in a circle (the in portion) to determine the largest amount of segments that could possible be in the circle. The In-Out table was used to try and find a formula that could be used for all the chords in a circle and how many segments that would be. As well as finding patterns to could help in finding the formula for the problem.
The main component we are supposed to incorporate is the use of In-Out tables to better understand the problem. We were able to incorporate this within the problem by using the number of chords in a circle (the in portion) to determine the largest amount of segments that could possible be in the circle. The In-Out table was used to try and find a formula that could be used for all the chords in a circle and how many segments that would be. As well as finding patterns to could help in finding the formula for the problem.
Process:
As a class we first looked at this problem in simpler terms, by making a single line across the circle, or a diameter, to see how many segments that would make. We then discussed how we knew that two segments was the highest amount of segments we could get with a single line, and from there we started to explore the problem on our own.
As a class we first looked at this problem in simpler terms, by making a single line across the circle, or a diameter, to see how many segments that would make. We then discussed how we knew that two segments was the highest amount of segments we could get with a single line, and from there we started to explore the problem on our own.
To continue this process, I began to increase the number of lines through the circle (or chords) by one, and with that I looked for the maximum number of pieces I could find using a certain amount of chords. For two chords, the maximum amount of pieces I found was four. For three chords, the maximum amount of pieces I found was seven. For four chords, the maximum amount of pieces was eleven. At this point I was able to see a constant increase occurring in the In-Out table, and was able to predict what the next maximum pieces would be.
Solution:
In the In-Out table (below) it shows how the number of cuts done to the circle relate to the maximum number of pieces. The number of cut 'zero' was added to help further explain and provide information as to how one could create an equation from the information. If one looks at how the maximum number of pieces increases with each cut done, one can see that the maximum number of pieces increases chronologically. For example, from one to two, the difference is a single unit, or one. From two to four, there is a two piece difference, from four to seven a three piece difference, and so on and so forth. With this information I was able to conclude that five cuts had sixteen pieces and six cuts had twenty-two pieces. With this information, I tried to create a formula that could be used for all of the different chords in the circle. I first tried to multiply the number of chords and add or subtract it by a certain amount of numbers. I tried to multiply the number of chords by two and subtract from there, but it was never consistent to what the maximum number of pieces. Then Mr. Corner reminded us that when we are looking for a formula to include +1 for the increase of the cuts to the circle. With this I was reminded about the quadratic equation, ax^(2)+bx+c, and I concluded that the +c was the +1 that was mentioned before. From there I tried to find the variables of a and b by using the quadratic equation and having it equal a certain number of maximum pieces, in this case 7, so the x variable would be 3. To find the variables, I began with ax^(2) + bx + 1 = 0, I then had a equal zero and x equal three so the equation looked like 0(3)^2 + b(3) + 1 = 0. From here I multiplied the factors out so it looked like 0 + 3b + 1 = 0, I subtracted one from both sides and divided by three so that b equaled negative one third. I then did the same process for the other variable and had gotten negative one third. I was able to conclude that a equaled b. With this information, I replaced both a and b by positive one third to make sure that the answer was positive and replaced x by three to test to see if this worked. The equation 1/3(3)^2 + 1/3(3) + 1 = 7 was incorrect because a third of nine would be three and a third of three would be one and one added to one added to three equals five and not seven. I then increased the factor value of a and b so that a and b was one half, in hopes that this would also increase the answer at the end, thus making the equation equal to 1/2(3)^2 + 1/2(3) + 1 = 7. This successfully worked, because half of nine is four and five tenths and half of three is one and five tenths and combine both to equal six, then add one to equal seven. I mentally went through the list to see if this applied for all the In-Out sections, and when it successfully worked with the rest of sections I looked for the maximum number of pieces for ten chords. The equation would have been .5(10)^2 + .5(10) + 1 = x, ten squared would be one hundred and half of a hundred would fifty. Then half of ten would be five and five added to fifty would be fifty-five, then one would be added so the maximum number of pieces would be fifty-six.
In the In-Out table (below) it shows how the number of cuts done to the circle relate to the maximum number of pieces. The number of cut 'zero' was added to help further explain and provide information as to how one could create an equation from the information. If one looks at how the maximum number of pieces increases with each cut done, one can see that the maximum number of pieces increases chronologically. For example, from one to two, the difference is a single unit, or one. From two to four, there is a two piece difference, from four to seven a three piece difference, and so on and so forth. With this information I was able to conclude that five cuts had sixteen pieces and six cuts had twenty-two pieces. With this information, I tried to create a formula that could be used for all of the different chords in the circle. I first tried to multiply the number of chords and add or subtract it by a certain amount of numbers. I tried to multiply the number of chords by two and subtract from there, but it was never consistent to what the maximum number of pieces. Then Mr. Corner reminded us that when we are looking for a formula to include +1 for the increase of the cuts to the circle. With this I was reminded about the quadratic equation, ax^(2)+bx+c, and I concluded that the +c was the +1 that was mentioned before. From there I tried to find the variables of a and b by using the quadratic equation and having it equal a certain number of maximum pieces, in this case 7, so the x variable would be 3. To find the variables, I began with ax^(2) + bx + 1 = 0, I then had a equal zero and x equal three so the equation looked like 0(3)^2 + b(3) + 1 = 0. From here I multiplied the factors out so it looked like 0 + 3b + 1 = 0, I subtracted one from both sides and divided by three so that b equaled negative one third. I then did the same process for the other variable and had gotten negative one third. I was able to conclude that a equaled b. With this information, I replaced both a and b by positive one third to make sure that the answer was positive and replaced x by three to test to see if this worked. The equation 1/3(3)^2 + 1/3(3) + 1 = 7 was incorrect because a third of nine would be three and a third of three would be one and one added to one added to three equals five and not seven. I then increased the factor value of a and b so that a and b was one half, in hopes that this would also increase the answer at the end, thus making the equation equal to 1/2(3)^2 + 1/2(3) + 1 = 7. This successfully worked, because half of nine is four and five tenths and half of three is one and five tenths and combine both to equal six, then add one to equal seven. I mentally went through the list to see if this applied for all the In-Out sections, and when it successfully worked with the rest of sections I looked for the maximum number of pieces for ten chords. The equation would have been .5(10)^2 + .5(10) + 1 = x, ten squared would be one hundred and half of a hundred would fifty. Then half of ten would be five and five added to fifty would be fifty-five, then one would be added so the maximum number of pieces would be fifty-six.
Evaluation:
In this problem I learned to recall what I had previously learned to try and solve problems that I am currently working on. As well as using an In-Out table to solve and find a formula that applies to all the factors in it. I used Seek Why and Prove to solve this problem because I was able to use the information that I learned and the hint that Mr. Corner gave me to find a formula that could find the maximum number of segments in a circle. From there I used the quadratic equation to find a more definite formula, and was able to prove that this formula was true using the information I knew and information I gathered. This is why I believe I deserve a 10/10.
In this problem I learned to recall what I had previously learned to try and solve problems that I am currently working on. As well as using an In-Out table to solve and find a formula that applies to all the factors in it. I used Seek Why and Prove to solve this problem because I was able to use the information that I learned and the hint that Mr. Corner gave me to find a formula that could find the maximum number of segments in a circle. From there I used the quadratic equation to find a more definite formula, and was able to prove that this formula was true using the information I knew and information I gathered. This is why I believe I deserve a 10/10.