__POWs__

__POWs__

**What is a POW?**

POW is an acronym for Problem Of the Week. Every other week or so we are given a POW that is related to the subject at hand. These are worth more points than regular assignments and take longer to complete. They enhance problem solving skills and collaboration with our peers.

## How have POWs helped you grow mathematically?

POWs have helped me to understand a lot of mathematical concepts better than I would have otherwise. They were also helpful in getting me to collaborate more since it showed me that collaboration would be much easier. They were also nice because I couldn't really write about math before this, and by the end of the year it wasn't very difficult for me anymore.

## POW #5

**POW #5 Write-up**

Will Klumpenhower

Geometry

4-1-15

For this POW, we were told several weeks in advance what we were going to be doing. We were asked in a starter to find the size of a perfect 12 oz. soda can. This meant that we needed to find a model with the least amount of surface area. This problem was deceptively easy, as it turned out there were a multitude of possibilities.

First, I started to assess the problem. I knew that I would have to find the surface area of a cylinder. That would require two things: The area of the bases and the lateral area. I knew the volume was 355 cm3. I also knew that the equation for volume in this particular scenario was V=r2×h. Since I already knew the volume, I could plug in 355, and then solve for h, getting h=355/r2. I now had two variables to work with, the radius and the height. My objective now was to find numbers for these two variables that fit the volume equation. First I plugged in numbers for r, and then used what I got on one side of the equation to determine h. I put these in a table and checked to see if they worked with the surface area equation. I put these in Excel until I figured out the solution. I rounded to the hundredths place in order to not have huge numbers to deal with.

In the “A” column r stands for radius, in the “B” column h stands for height, and in the “C” column SA stands for surface area. As you can see, the lowest amount of aluminum is approximately 277.57 cm2. This was accomplished with a radius of 3.8 cm and a height of 7.83 cm. Of course there is a little assuming going on here. An actual soda can probably doesn't have these exact dimensions because it has to be somewhat sturdy and can’t just have aluminum as thin as a piece of paper all the way around. Also, a soda can isn’t an exact cylinder anyway, so it’s wrong in that respect too.

I struggled with finding an easy way to do this for a while, but when I figured it out, the rest was pretty easy. I wish I could have found an easy way to do it in a couple minutes, but I had to settle for punching buttons into my calculator over and over. Go with what works, I guess.

Will Klumpenhower

Geometry

4-1-15

For this POW, we were told several weeks in advance what we were going to be doing. We were asked in a starter to find the size of a perfect 12 oz. soda can. This meant that we needed to find a model with the least amount of surface area. This problem was deceptively easy, as it turned out there were a multitude of possibilities.

First, I started to assess the problem. I knew that I would have to find the surface area of a cylinder. That would require two things: The area of the bases and the lateral area. I knew the volume was 355 cm3. I also knew that the equation for volume in this particular scenario was V=r2×h. Since I already knew the volume, I could plug in 355, and then solve for h, getting h=355/r2. I now had two variables to work with, the radius and the height. My objective now was to find numbers for these two variables that fit the volume equation. First I plugged in numbers for r, and then used what I got on one side of the equation to determine h. I put these in a table and checked to see if they worked with the surface area equation. I put these in Excel until I figured out the solution. I rounded to the hundredths place in order to not have huge numbers to deal with.

In the “A” column r stands for radius, in the “B” column h stands for height, and in the “C” column SA stands for surface area. As you can see, the lowest amount of aluminum is approximately 277.57 cm2. This was accomplished with a radius of 3.8 cm and a height of 7.83 cm. Of course there is a little assuming going on here. An actual soda can probably doesn't have these exact dimensions because it has to be somewhat sturdy and can’t just have aluminum as thin as a piece of paper all the way around. Also, a soda can isn’t an exact cylinder anyway, so it’s wrong in that respect too.

I struggled with finding an easy way to do this for a while, but when I figured it out, the rest was pretty easy. I wish I could have found an easy way to do it in a couple minutes, but I had to settle for punching buttons into my calculator over and over. Go with what works, I guess.

## POW #4

**POW #4**

Will Klumpenhower

Isaac Jordan

Geometry

3-4-15

Almost Rubix Cube

For this week’s POW, we had to find the answers to the following questions.We had to find how many cubes have one face painted on a 5 by 5 by 5 cube with other 1 by 1 by 1 cubes inside. We also had to find how many smaller cubes have two sides painted, how many have three sides painted and how many have 4, 5, or 6 sides painted. Finally the last question asks how many cubes are not painted.

To start off we knew that it was going to be an easy POW because it just uses some basic math to solve the problem. The first step is to do the simple math to fuge out the solution, like in the first question, we were asked how many smaller cubes have one face painted. It was simple you figure out how many cubes have on side painted once you figure that out it was pretty simple you just multiply that by six because there are six more sides. The answer for number one is 45. For number two we had to find out how many cubes have two faces painted. It was really was easy because there are twelve sides on a square and three on a side. If you multiply three by six it is 36. And it asked how many of the smaller triangles don't have a face painted. the answer to that is 27 because there are smaller cubes inside of the outside and they are not painted. The last question asked how many cubes have 4, 5, or 6 sides painted.The answer to that is zero, there are no squares that have more than three sides painted.

Evaluation:

Isaac:

I think that this POW was a really easy POW because it just used common sense to solve it. I dont think that this POW really grow at all because I could have solved this in 5th grade just because all you use is basic math to solve it. This POW was the easiest one I have done out of them I solved really quick because it just used basic math. I think I deserve a 20-23 because we met all the criteria and got the right answers.

Will:

This POW was pretty easy. It only required some very basic math skills. I think the most complex concept you could relate it to is very basic pre-algebra. It honestly didn’t take very much effort to complete. However, it is good to have a project now and then that you actually have to focus your time on that isn’t just writing down answers on a worksheet. It is both more interesting and slightly more challenging than just a regular assignment. I feel like this POW was sort of a simpler assignment that makes me feel like school isn’t as hard as I think, which can be comforting. I think I deserve a 23- 24 out of 25, because I got it done on time with a correct answer, for what it’s worth.

Will Klumpenhower

Isaac Jordan

Geometry

3-4-15

Almost Rubix Cube

For this week’s POW, we had to find the answers to the following questions.We had to find how many cubes have one face painted on a 5 by 5 by 5 cube with other 1 by 1 by 1 cubes inside. We also had to find how many smaller cubes have two sides painted, how many have three sides painted and how many have 4, 5, or 6 sides painted. Finally the last question asks how many cubes are not painted.

To start off we knew that it was going to be an easy POW because it just uses some basic math to solve the problem. The first step is to do the simple math to fuge out the solution, like in the first question, we were asked how many smaller cubes have one face painted. It was simple you figure out how many cubes have on side painted once you figure that out it was pretty simple you just multiply that by six because there are six more sides. The answer for number one is 45. For number two we had to find out how many cubes have two faces painted. It was really was easy because there are twelve sides on a square and three on a side. If you multiply three by six it is 36. And it asked how many of the smaller triangles don't have a face painted. the answer to that is 27 because there are smaller cubes inside of the outside and they are not painted. The last question asked how many cubes have 4, 5, or 6 sides painted.The answer to that is zero, there are no squares that have more than three sides painted.

Evaluation:

Isaac:

I think that this POW was a really easy POW because it just used common sense to solve it. I dont think that this POW really grow at all because I could have solved this in 5th grade just because all you use is basic math to solve it. This POW was the easiest one I have done out of them I solved really quick because it just used basic math. I think I deserve a 20-23 because we met all the criteria and got the right answers.

Will:

This POW was pretty easy. It only required some very basic math skills. I think the most complex concept you could relate it to is very basic pre-algebra. It honestly didn’t take very much effort to complete. However, it is good to have a project now and then that you actually have to focus your time on that isn’t just writing down answers on a worksheet. It is both more interesting and slightly more challenging than just a regular assignment. I feel like this POW was sort of a simpler assignment that makes me feel like school isn’t as hard as I think, which can be comforting. I think I deserve a 23- 24 out of 25, because I got it done on time with a correct answer, for what it’s worth.

## POW #2

POW Write-up #2

Will Klumpenhower

Geometry

For this week’s POW we had an interesting problem. We had been learning a lot about triangle ratios, so Caitlyn applied that in our POW. We had 4 rods of different distances: 6, 4, 3, and 2 inches. We also had an unlimited supply of other rods, all whole numbers and from 1-20 inches. We had to use two of these extras along with the four originals to create as many pairs of similar triangles as we could. Similar triangles are triangles that have corresponding sides that can be multiplied or divided in order to have the side lengths of the other triangle.

When I started this problem, I thought that I could solve it by finding triangles that worked and seeing if they could have a similar pair. This has harder than I previously imagined, so I decided to just use plain old math. I eventually found 2 pairs. The first one was a triangle that had side lengths of 2, 3, and 4 inches, and the similar triangle I found for it had 4, 6, and 8 inches for the side lengths. To find this one, I used 3 of the sides I already knew that made a triangle, and then doubled all the lengths to get a 2:1 ratio and a similar triangle. I then realized that this principle would work with another ratio, 3:1. My next pair had side lengths of 2, 3, and 4 again, but the others were 6, 9, and 12. in this triangle, I just multiplied all the original side lengths by 3 to get the new ones. After this, I couldn’t find any more. The reason for this was because it seemed to me that it was impossible for any of the triangle ratios I came up with to work. We had learned earlier that to have a triangle, the sum of the 2 shorter side lengths had to be greater than the longest side length. Well, I already used 2, 3, and 4, and I couldn’t find any other ratios with that, I had to find another one. 6, 4, and 2 didn’t work, because 4+2=6. 3, 4, and 6 worked as a triangle, but I needed to have 2 in the other triangle somewhere. I still had 2 other rods, but remember, they have to be whole numbers. So if I have 3, 4, and 6, my ratio to the other triangle could be either 2:1 or 3:1. If I have 2:1, then my other side lengths would have to be 1.5, 2, and 3. That doesn’t work because I need all whole numbers. If I have 3:1, the my other side lengths would be 1, 1.3, and 2. That doesn’t work either. This was very interesting to me as I know some of my classmates got more, and it makes me wonder how they did it.

An interesting addition to the problem would be to have one more extra rod that could be anything you wanted. I think that many more opportunities would open up and while you would get more answers, I’m not sure if it would be any easier. You certainly would be able to make many more triangles, but it would be difficult to find all of the possibilities.

This POW was not as interesting as some of the previous ones, but it was still challenging and helpful. I think that POWs with more interesting basises are easier to do and do well. It did leave me with some questions as to how many triangles were actually possible, as I don’t think I got all of them. It would be interesting to have a debriefing in class to explain how it was possible to come up with more answers. I thought I did a good job on this, despite my falling short of the maximum amount of pairs. I think I deserve a 23 out of 25 for this POW*, to account for not getting all the pairs. I still think I did a good job with the process of solving it, despite the aforementioned issue with the maximum number of pairs.

Will Klumpenhower

Geometry

For this week’s POW we had an interesting problem. We had been learning a lot about triangle ratios, so Caitlyn applied that in our POW. We had 4 rods of different distances: 6, 4, 3, and 2 inches. We also had an unlimited supply of other rods, all whole numbers and from 1-20 inches. We had to use two of these extras along with the four originals to create as many pairs of similar triangles as we could. Similar triangles are triangles that have corresponding sides that can be multiplied or divided in order to have the side lengths of the other triangle.

When I started this problem, I thought that I could solve it by finding triangles that worked and seeing if they could have a similar pair. This has harder than I previously imagined, so I decided to just use plain old math. I eventually found 2 pairs. The first one was a triangle that had side lengths of 2, 3, and 4 inches, and the similar triangle I found for it had 4, 6, and 8 inches for the side lengths. To find this one, I used 3 of the sides I already knew that made a triangle, and then doubled all the lengths to get a 2:1 ratio and a similar triangle. I then realized that this principle would work with another ratio, 3:1. My next pair had side lengths of 2, 3, and 4 again, but the others were 6, 9, and 12. in this triangle, I just multiplied all the original side lengths by 3 to get the new ones. After this, I couldn’t find any more. The reason for this was because it seemed to me that it was impossible for any of the triangle ratios I came up with to work. We had learned earlier that to have a triangle, the sum of the 2 shorter side lengths had to be greater than the longest side length. Well, I already used 2, 3, and 4, and I couldn’t find any other ratios with that, I had to find another one. 6, 4, and 2 didn’t work, because 4+2=6. 3, 4, and 6 worked as a triangle, but I needed to have 2 in the other triangle somewhere. I still had 2 other rods, but remember, they have to be whole numbers. So if I have 3, 4, and 6, my ratio to the other triangle could be either 2:1 or 3:1. If I have 2:1, then my other side lengths would have to be 1.5, 2, and 3. That doesn’t work because I need all whole numbers. If I have 3:1, the my other side lengths would be 1, 1.3, and 2. That doesn’t work either. This was very interesting to me as I know some of my classmates got more, and it makes me wonder how they did it.

An interesting addition to the problem would be to have one more extra rod that could be anything you wanted. I think that many more opportunities would open up and while you would get more answers, I’m not sure if it would be any easier. You certainly would be able to make many more triangles, but it would be difficult to find all of the possibilities.

This POW was not as interesting as some of the previous ones, but it was still challenging and helpful. I think that POWs with more interesting basises are easier to do and do well. It did leave me with some questions as to how many triangles were actually possible, as I don’t think I got all of them. It would be interesting to have a debriefing in class to explain how it was possible to come up with more answers. I thought I did a good job on this, despite my falling short of the maximum amount of pairs. I think I deserve a 23 out of 25 for this POW*, to account for not getting all the pairs. I still think I did a good job with the process of solving it, despite the aforementioned issue with the maximum number of pairs.

## POW #1

POW #1 Write-up

Will Klumpenhower

Geometry

1-7-15

Slices of Pie

You are given a pie that you have to cut with straight lines that go all the way through it,but they don’t have to have the same shape. How many pieces can you make with a given number of cuts?

When we received this POW, I started working on it immediately. I figured out within the class period that the maximum for 4 cuts was 16. I already knew that 1 cut was 2, 2 cuts was 4, and 3 cuts was 7. Then the next day, I figured out that 5 cuts made 16 at the most. I started to make a diagram to show these findings simply and had cranked out a pretty good diagram when Caitlyn showed us on the board what the answer was, which sort of surprised me. I was also really surprised that I hadn’t realized it before. The equation used to find the maximum pieces of pie was the number of cuts plus the former number of pieces. 1+1=2, 2+2=4, 3+4=7, etc. So I put my findings in a Geogebra document.

We also had to figure out what the maximum for 10 cuts would be, using our formula, f(n) =f(n-1)+n. Plug in 10 for n and you get 56. It’s pretty basic algebra.

An extension to the problem could be having all the cuts form a regular triangle, in sets of three. This would pose some interesting questions, but would probably result in a less complicated formula.

I thought the problem was more interesting than the last POW we had, and thus I think I did better on it and it kept me working on it more. It reminded me of the chess problem, my favorite so far. It was maybe a little too easy, but I didn’t mind. I think I should receive a 25 out of 25 for the problem. I think I did a good job and I feel I expressed my ideas well.

Will Klumpenhower

Geometry

1-7-15

Slices of Pie

You are given a pie that you have to cut with straight lines that go all the way through it,but they don’t have to have the same shape. How many pieces can you make with a given number of cuts?

When we received this POW, I started working on it immediately. I figured out within the class period that the maximum for 4 cuts was 16. I already knew that 1 cut was 2, 2 cuts was 4, and 3 cuts was 7. Then the next day, I figured out that 5 cuts made 16 at the most. I started to make a diagram to show these findings simply and had cranked out a pretty good diagram when Caitlyn showed us on the board what the answer was, which sort of surprised me. I was also really surprised that I hadn’t realized it before. The equation used to find the maximum pieces of pie was the number of cuts plus the former number of pieces. 1+1=2, 2+2=4, 3+4=7, etc. So I put my findings in a Geogebra document.

We also had to figure out what the maximum for 10 cuts would be, using our formula, f(n) =f(n-1)+n. Plug in 10 for n and you get 56. It’s pretty basic algebra.

An extension to the problem could be having all the cuts form a regular triangle, in sets of three. This would pose some interesting questions, but would probably result in a less complicated formula.

I thought the problem was more interesting than the last POW we had, and thus I think I did better on it and it kept me working on it more. It reminded me of the chess problem, my favorite so far. It was maybe a little too easy, but I didn’t mind. I think I should receive a 25 out of 25 for the problem. I think I did a good job and I feel I expressed my ideas well.

**Second Semester**

## POW #2

Will Klumpenhower

POW #2

Triangle Congruence

In class lately, we have been studying triangle congruence. We found that if you know a certain amount of a triangle’s parts, you could have enough information to make an identical triangle. If you have three sides of a triangle, then it is enough. This is called Side-Side-Side, or SSS, the latter being more commonly used. However, if you have three angles, this is not enough, because the angle could be floating out in space somewhere. Our POW was to find if it is possible to make two triangles if you know five of the parts.

After studying the problem, I realized that you would have to have three angles in it. I knew this because if you had three sides, it would automatically be a congruence shortcut. So I knew I needed three angles and two sides. Instead of trying to figure this out with actual numbers, which would be almost impossible, I used congruence marks.

I had struggled with this problem with a while and never would have solved it if Caitlyn hadn’t given us a bunch of help in class. With this immense guidance, I finally figured it out. If you keep the same angles, but changed up the position of the sides. Eventually, I got to a position where the triangles had five out of six parts in common, but were also different. It is really hard to show this without a diagram, so I included one on the page attached.

An interesting extension for this problem would just to expand it a little, by trying to figure out two different triangles with four, three, two, and one common parts. It wouldn’t be as hard, but it would take a bit more time.

In this, I learned that I need to take more time out of my homework schedule to do the problem. The hardest part of this was again the writing it all down. I am more used to just having a problem and then writing down a quick, concise answer. The POW requires more thinking and more writing. I think I deserve a 25 to 27 out of 30 on this assignment, because my diagrams could have been a bit better.

POW #2

Triangle Congruence

In class lately, we have been studying triangle congruence. We found that if you know a certain amount of a triangle’s parts, you could have enough information to make an identical triangle. If you have three sides of a triangle, then it is enough. This is called Side-Side-Side, or SSS, the latter being more commonly used. However, if you have three angles, this is not enough, because the angle could be floating out in space somewhere. Our POW was to find if it is possible to make two triangles if you know five of the parts.

After studying the problem, I realized that you would have to have three angles in it. I knew this because if you had three sides, it would automatically be a congruence shortcut. So I knew I needed three angles and two sides. Instead of trying to figure this out with actual numbers, which would be almost impossible, I used congruence marks.

I had struggled with this problem with a while and never would have solved it if Caitlyn hadn’t given us a bunch of help in class. With this immense guidance, I finally figured it out. If you keep the same angles, but changed up the position of the sides. Eventually, I got to a position where the triangles had five out of six parts in common, but were also different. It is really hard to show this without a diagram, so I included one on the page attached.

An interesting extension for this problem would just to expand it a little, by trying to figure out two different triangles with four, three, two, and one common parts. It wouldn’t be as hard, but it would take a bit more time.

In this, I learned that I need to take more time out of my homework schedule to do the problem. The hardest part of this was again the writing it all down. I am more used to just having a problem and then writing down a quick, concise answer. The POW requires more thinking and more writing. I think I deserve a 25 to 27 out of 30 on this assignment, because my diagrams could have been a bit better.

## POW #1

Will Klumpenhower

POW Write-up 1

The Four Knights

In this week’s POW, we were given a problem. Four knights in Chess, two black and two white, were on the corners of a 3 square by 3 square chessboard. The white ones were on the bottom side, and the black ones were on the top side. Is it possible, using only real chess moves, to have the black knights and the white knights switch places? If so, do it in the least number of moves.

I immediately assumed it was possible, because it just didn’t seem right for there not to be a way. In chess, knights are very versatile pieces and I thought that it was almost obvious that the problem could be solved. Having determined this, I set to work on the problem.

When I looked at the board, I saw that there were four vacant spaces that a knight could occupy. There was one vacant spot that no matter what, a knight could not ever be there, or move from it once it was: the center square. according to the rules of chess, A knight has to move in a pattern of two squares either up, down, and to the side, and then one square up down or to the side, depending on the original move. A knight can jump over other pieces. So since it is impossible to have a starting point that is two spaces away from the middle, it was impossible for a knight to be placed there. So I knew I had to use the edges for this puzzle.

Since it is difficult to display how I came to my answer, I will try to explain it as best I can. I solved the problem the day I got it. I moved all the knights around in a circle until they came to a point where they had switched places. I moved each piece four times to come to a total of 16 moves. I know that this is correct because I cross-referenced my answer with my classmates. Several of them were in their spots at only about six moves, but the way they were placed made it impossible to put their counterparts in their spots. If this is still confusing, see the paper attached.

An interesting extension to this problem would be to make the board 4 by 4 and see if it was still possible. I think it still would be, but it would be much more difficult, since there are no borders. I think the borders helped a bit in the original, since they made it less confusing.

I learned from this project that the problem was slightly more complicated than I thought, but still not too difficult. I learned also that while I figured out the problem quickly, it was much harder for me to describe how I solved it than I thought. This made it difficult for me to get it done on time. I think I deserve somewhere from 20 to 25 points out of 30 on this problem, since, while I understood and solved the problem well and easily, I was not the most punctual on getting it written down.

POW Write-up 1

The Four Knights

In this week’s POW, we were given a problem. Four knights in Chess, two black and two white, were on the corners of a 3 square by 3 square chessboard. The white ones were on the bottom side, and the black ones were on the top side. Is it possible, using only real chess moves, to have the black knights and the white knights switch places? If so, do it in the least number of moves.

I immediately assumed it was possible, because it just didn’t seem right for there not to be a way. In chess, knights are very versatile pieces and I thought that it was almost obvious that the problem could be solved. Having determined this, I set to work on the problem.

When I looked at the board, I saw that there were four vacant spaces that a knight could occupy. There was one vacant spot that no matter what, a knight could not ever be there, or move from it once it was: the center square. according to the rules of chess, A knight has to move in a pattern of two squares either up, down, and to the side, and then one square up down or to the side, depending on the original move. A knight can jump over other pieces. So since it is impossible to have a starting point that is two spaces away from the middle, it was impossible for a knight to be placed there. So I knew I had to use the edges for this puzzle.

Since it is difficult to display how I came to my answer, I will try to explain it as best I can. I solved the problem the day I got it. I moved all the knights around in a circle until they came to a point where they had switched places. I moved each piece four times to come to a total of 16 moves. I know that this is correct because I cross-referenced my answer with my classmates. Several of them were in their spots at only about six moves, but the way they were placed made it impossible to put their counterparts in their spots. If this is still confusing, see the paper attached.

An interesting extension to this problem would be to make the board 4 by 4 and see if it was still possible. I think it still would be, but it would be much more difficult, since there are no borders. I think the borders helped a bit in the original, since they made it less confusing.

I learned from this project that the problem was slightly more complicated than I thought, but still not too difficult. I learned also that while I figured out the problem quickly, it was much harder for me to describe how I solved it than I thought. This made it difficult for me to get it done on time. I think I deserve somewhere from 20 to 25 points out of 30 on this problem, since, while I understood and solved the problem well and easily, I was not the most punctual on getting it written down.