Problem Of The Week 1
- In this POW our goal was to help Delilah plant as many flowers in her garden as possible for the least amount of money. The challenge in this problem would be to know how many square feet each part of her garden is and then finding the cheapest way to plant these flowers. To solve this problem we will need the skills of knowing the equations of geometry and how to find the area of a rectangle and the shapes inside of that rectangle. We will also need to know how to make each area as cheap as possible. The final goal here is to help Delilah achieve her goal of spending the least amount of money possible on her flower garden.
- At the current point in time I predict that the answer to this POW will fall between 90 to 100 dollars. To solve this POW originally we tried dividing the numbers lining the squares that were actually meant to solve for the area of each square. Needless to say that didn’t provide us with the answer we were searching for,but it did lead us to realize that we would have to solve for the area before we could go any further. After finding the correct areas for all the squares we started matching the flowers to the spaces that made the most sense. We were on the right track, however we accidentally placed the daisies in a place that would have fit the begonias much better. Upon figuring this out we switched the two and spread the rest of the flowers throughout the garden, leaving us with a final answer.
- Our final answer creates a situation in which Deliliah spends the least money to fill her garden. We paired the cheapest flowers,begonias, to the largest rectangle with an area of twenty one. We left the daisies to group with the second largest space with an area of twenty. This leaves the cannas and roses to occupy the third and second smallest rectangles and the most expensive flowers, cattleya orchids for the smallest space. The prices of flowers go from 1.00 to 3.00 each and by planting one per square feet Delilah would spend a total of one hundred and eight dollars.
- Our conclusion saves Delilah the most money because we ensured that the most expensive flowers were paired with the spaces that would be least expensive to plant them in. For example, the Cattleya Orchids were paired with the 12 square foot rectangle in order to spend the least money for the most expensive flower. The rectangles themselves go from a 21 to the aforementioned 12, we got these figures by multiplying the length by the width for each rectangle. The following describes how we determined were the flowers went:
21ft-daisies for a total of $21
20ft- begonias for a total of $30
15ft-cannas for a total of $30
6ft-roses for a total of $15
4ft-Cattaleya Orchids for a total of $12
Overall cost: $108
Problem Of The Week 2
The second problem of the week tasks students with determining the length between points D and A of a quadrilateral composed of points A,B,C,D. Afterwards, the students must discover the perimeter of the quadrilateral. The real challenge of this problem is identifying the length between point D and point A. The variables of this problem are comprised of the measurements from points A to B, B to C, and C to D which leaves the measurement from D to A to be dicovered.So far it is known that the pythagorean theorem is needed to complete this challenge. The final goal of this POW is to determine the length of point D to point A and then find the perimeter of this quadrilateral.
At the beginning to this product I had no guesses as to what the answer would be, but I estimate that the perimeter of this quadrilateral will be around 40. At first I jumped ahead and began putting numbers to fill the slots of the Pythagorean Theorem. This left me with 27 as the distance between point D and point A. Although at the time this seemed like a valid answer I soon realized my error, I had mistaken the 18 for an 8 therefore botching my entire first attempt. Secondly, I used 18 and 21 to discover the length of the diagonal between point A and C, this turned out to be 27. Using this information I inserted 27 and 14 into the Pythagorean Theorem and got an answer that lead me to unveil the perimeter.
To get the solution to this problem I first used the Pythagorean Theorem, substituting 18 for A squared and 21 for B squared, to determine that the distance between point A and C have a length of 27. Using this knowledge I plugged 27 into the theorem to find that the length between point D and A is 31. This means that the overall perimeter of this quadrilateral is 84.
Although my answer does not correspond with my prediction it is reasonable because I was able to determine the distance between point A and C and in turn found an answer that seems logical in the problems context. A general conclusion I got from completing this problem is that a problem with triangles will most always need the Pythagorean Theorem to be completed.
At the beginning to this product I had no guesses as to what the answer would be, but I estimate that the perimeter of this quadrilateral will be around 40. At first I jumped ahead and began putting numbers to fill the slots of the Pythagorean Theorem. This left me with 27 as the distance between point D and point A. Although at the time this seemed like a valid answer I soon realized my error, I had mistaken the 18 for an 8 therefore botching my entire first attempt. Secondly, I used 18 and 21 to discover the length of the diagonal between point A and C, this turned out to be 27. Using this information I inserted 27 and 14 into the Pythagorean Theorem and got an answer that lead me to unveil the perimeter.
To get the solution to this problem I first used the Pythagorean Theorem, substituting 18 for A squared and 21 for B squared, to determine that the distance between point A and C have a length of 27. Using this knowledge I plugged 27 into the theorem to find that the length between point D and A is 31. This means that the overall perimeter of this quadrilateral is 84.
Although my answer does not correspond with my prediction it is reasonable because I was able to determine the distance between point A and C and in turn found an answer that seems logical in the problems context. A general conclusion I got from completing this problem is that a problem with triangles will most always need the Pythagorean Theorem to be completed.
Problem Of The Week 3
In this scenario students are tasked with finding the length of a tarp used to set up a tent. The real challenge of this problem is to use PQ,QR,and RS in such a manner that will lead to a solution. I know that QU and RT are placed 12 m apart and that their lengths are 4 meters and 7.5 meters. Now that the problem has been laid out, we can identify the final goal. Which in this case is to find the length of the tarp using the height and base figures that have been presented.
This problem began with me sketching out a rough trapezoid that would act as the tent. As I started to get a feel for this problem I estimated that the answer will be 18.5. My process for testing this estimate started with my first failure when I attempted to get an answer for the first right angle and then square it to get a final answer. I used the equation:
A= ½ (b)(h)
I plugged my numbers in and got
A=½ (4)(7.5)
This left me stumped with an answer of 900 for the length of the tarp. I tried the same thing on the other right angle,
A=½ (3)(4)
This supplied me with yet another radically imprecise answer. A few days later I tried again this time using the Pythagorean Theorem and found something more accurate to my prediction.
My answer to this Problem of The Week is that the tent is 26 meters long.
I reached this conclusion by squaring 4 and 3 to get 16+9 to get 25. Then I determined that QU2+PU2 must equal 25. My thought process during this problem:
QU2 = 42
PU2=32
Which means that the distance between P and Q is 5. I went through the same process with R and T and got 8.5. At this point I was stuck, luckily Lyric Hildner could offer some assistance giving my the lineup needed to figure out the length of the tarp. This was:
P Q + QR + RS= length of the tarp
I was then able to plug in my numbers and got:
5m+12.5m+8.5m=26 meters or the length of the tarp.
This problem began with me sketching out a rough trapezoid that would act as the tent. As I started to get a feel for this problem I estimated that the answer will be 18.5. My process for testing this estimate started with my first failure when I attempted to get an answer for the first right angle and then square it to get a final answer. I used the equation:
A= ½ (b)(h)
I plugged my numbers in and got
A=½ (4)(7.5)
This left me stumped with an answer of 900 for the length of the tarp. I tried the same thing on the other right angle,
A=½ (3)(4)
This supplied me with yet another radically imprecise answer. A few days later I tried again this time using the Pythagorean Theorem and found something more accurate to my prediction.
My answer to this Problem of The Week is that the tent is 26 meters long.
I reached this conclusion by squaring 4 and 3 to get 16+9 to get 25. Then I determined that QU2+PU2 must equal 25. My thought process during this problem:
QU2 = 42
PU2=32
Which means that the distance between P and Q is 5. I went through the same process with R and T and got 8.5. At this point I was stuck, luckily Lyric Hildner could offer some assistance giving my the lineup needed to figure out the length of the tarp. This was:
P Q + QR + RS= length of the tarp
I was then able to plug in my numbers and got:
5m+12.5m+8.5m=26 meters or the length of the tarp.
Problem Of The Week 4
After Maya and Jason went to Cream Bean Berry they mounted their bikes and began the venture back to their respective homes. May travels north at 32 mph while Jason travels east at 24 mph, students must determine the time when Maya and Jason will be 130 miles apart and how far each of them have travelled. The real challenge of this problem will be determining how far each of them traveled. I know that Maya is traveling north at 32 miles per hour and Jason is traveling east at 24 miles per hour. I know to get to this challenge I need to accomplish the first part of the problem or figure out the time that it took before Maya and Jason were 130 miles apart. For me the final goal of this problem will be determining how far each of them traveled to get 130 miles away from each other.
I roughly predict that the solution to how long it takes for 130 miles to separate Maya and Jason is 4 hours and I believe that the distance traveled would be between 96 and 128 miles. I began this problem knowing that the answer should be in between 1 and 3 hours and that they should travel anywhere from 100 to 200 miles. I started this problem by listing the variables and figuring out how far apart they would be after one hour of riding. The solution to this mini problem is 40 miles that seperate Jason and Maya. After I had a general idea of the math needed to complete this problem I moved on to tackling the problem as a whole. At first I hit a snag when trying to find a way to get the time in units. I had misplaced a squared symbol which left me with a drastically wrong answer. Once I realized this I moved on to get my final answer.
My answer to the fourth problem of the week is that it took three hours and twenty five minutes for Jason and Maya to be 130 miles apart. Using the time presented in the problem (1:15) we can establish that it is 4:40 when Maya and Jason are 130 miles apart. Using this we can calculate how far each of them travelled, that being:
Jason-78 miles travelled
Maya-104 miles travelled
My final answer somewhat resembles the guesses that I previously made which deems the solution reasonable. A general conclusion that I can draw from this problem is that the extent to which Pythagoras Theorem can be used is near limitless and can be applied to many different problems. My solution is correct because I was able to use the time I calculated (3 hours and twenty five minutes) and multiply it by 24 and 32 to get 78 and 104. I then plugged these into Pythagoreans Theorem as such:
1042 + 782 = C2
10,816+6,084=C2
16,900=C2
130=C
By proving this in such a way I am positive that I provide a correct answer to the fourth problem of the week.
Problem of the Week 5: Fractals!
By Dylan King
10/23/15
Stage 1
Fractals demonstrate a repeating pattern that can encompass geometry. Fractals can range from the detail injected into an image on a computer monitor to the beat of a human heart and deal with rough shapes or, “Monsters” as referred to by the mathematical community until the late twentieth century. The Sierpinski triangle is a subdivision of many iterations of triangles that all seem to be built inside of a large equilateral triangle. The Sierpinski triangle is a fractal because it demonstrates the idea of a repeating pattern which, in itself, is the simple definition of a fractal. Now that we know that the Sierpinski triangle is a fractal we can touch on the vocabulary that will be used to describe it. One of the keywords being, iterations. An iteration is a repetition of a process this is very clear in the aforementioned Sierpinski triangle. Another useful vocabulary word is recursion which is the repeated application of a procedure and is featured in many fractals. Two more useful words that will come in handy are initiator and generator. Initiator being the starting shape and generator being a collection of scaled copies of the initiator. The final keyword is self similarity which in geometry means that no matter how far you zoom in or out of an image it appears to be still or the same shape. Now that we have covered the basics, let’s move on to examining my own distinct Sierpinski triangle.
Stage 2
Through the strategic use of a ruler and protractor I created a number of drafts representing a Sierpinski triangle. My final choice came down to the fianite details present in the measure of the triangle's angles. During this time I was challenged by my inability to provide detail and refinement to the length of the lines present in my equilateral triangle. My Sierpinski triangle ended up with four iterations. This is so because the more iterations I added the more I was challenged with the process of creating truly equilateral triangles. In the end I decided that having less iterations was worth it if it meant my triangle would look more professional and accurately put together. My motivation for the design of my Sierpinski triangle was based around the three major themes that I use most when drawing. The bottom left being black and white and the bottom right representing the use of pattern drawn sporadically to fill in an area. On the top triangle I went with a different method and left some of the triangles blank only to go back and shade them in an effort to create something relatively three dimensional. These three different methods represent the way I enjoy drawing and formed together nicely to create my Sierpinski triangle.
Stage 3
My fractal would be identifiable as a Sierpinski triangle because it takes the overall shape of an equilateral triangle, the initiator, and is subdivided into four iterations of smaller self similar equilateral triangles, generator. All of these elements compiled into one make my triangle easily identifiable as a Sierpinski triangle. The recursion present in my triangle showcases its ties to fractals as a whole with it’s clear repetition of a pattern throughout my piece. The concept of self similarity also dominates my triangle and shows many aspects of fractals that appear in my Sierpinski triangle.
Problem of the Week 6
Dylan King
The diagram below shows the graph of 3x + 4y = 12. The shaded figure is a square, three of whose vertices are on the coordinate axes. The fourth vertex is on the line. Find a) the x and y intercepts of the line, b) the coordinates of the vertex that falls on the line, c) the length of a side of the square, and d) the area of the square.
The featured diagram shows the graph of 3x + 4y = 12 with a square beside it. Three of the square’s vertices are on the coordinate axis leaving the fourth on the line graphed above. Students were tasked with finding the x and y intercept and using that to determine the vertex that falls on the line. This would then allow them to uncover the length of a side of the square and the area of the square. I know that I will only need to find the length of one of the sides of a square because all of the sides in a square are of equal length. I also know that will need to break this problem into multiple pieces to solve it. The final goal for this Problem Of The Week is to use the length of the square's sides to discover the area of the square.
I roughly predicted that the answer to this problem would be under four because of the size of the square compared to the x and y intercepts. The first step I took to achieving my final goal was finding the x and y intercepts. I then looked to the square immediately and frustratingly found that I could not resolve the problem that simply. After some time I realized that I had to think larger and drew the graph as a large triangle. This meant I could find the area of this triangle and then use it to find the area of the square. I labeled the graph as PQD and the square as ABCD, this let me quickly try different methods to find an answer. I began to see that there were two other triangles on the opposite sides of the square. I was able to return to ABCD and quickly found an answer.
The x intercept is 4 leaving the y intercept to be 3. The coordinate of the vertex that falls on the line is (1.7,1.7). The length of one side of the square is 1.7 and the area of the square is 2.9.
This answer loosely corresponds with my prediction and is more reasonable than a guess of anything under four. A general conclusion I can draw from this problem is that any graphed problem that includes a square will always provide more than one triangle that will aid in finding a solution. I know my answer is correct because I was able to use the equation to prove it:
(3-a)(4-a)=a2
(3)(4)=12
12-4a-3a+a2=a2
4a+3a=7a
12-7a+a2=a2
12=7a
12/7=1.7
1.7=a
This answer also conforms to the x and y intercept and does not go farther than (4,0) and(0,3). All in all, this was my process towards understanding and solving the sixth Problem Of The Week.
I roughly predict that the solution to how long it takes for 130 miles to separate Maya and Jason is 4 hours and I believe that the distance traveled would be between 96 and 128 miles. I began this problem knowing that the answer should be in between 1 and 3 hours and that they should travel anywhere from 100 to 200 miles. I started this problem by listing the variables and figuring out how far apart they would be after one hour of riding. The solution to this mini problem is 40 miles that seperate Jason and Maya. After I had a general idea of the math needed to complete this problem I moved on to tackling the problem as a whole. At first I hit a snag when trying to find a way to get the time in units. I had misplaced a squared symbol which left me with a drastically wrong answer. Once I realized this I moved on to get my final answer.
My answer to the fourth problem of the week is that it took three hours and twenty five minutes for Jason and Maya to be 130 miles apart. Using the time presented in the problem (1:15) we can establish that it is 4:40 when Maya and Jason are 130 miles apart. Using this we can calculate how far each of them travelled, that being:
Jason-78 miles travelled
Maya-104 miles travelled
My final answer somewhat resembles the guesses that I previously made which deems the solution reasonable. A general conclusion that I can draw from this problem is that the extent to which Pythagoras Theorem can be used is near limitless and can be applied to many different problems. My solution is correct because I was able to use the time I calculated (3 hours and twenty five minutes) and multiply it by 24 and 32 to get 78 and 104. I then plugged these into Pythagoreans Theorem as such:
1042 + 782 = C2
10,816+6,084=C2
16,900=C2
130=C
By proving this in such a way I am positive that I provide a correct answer to the fourth problem of the week.
Problem of the Week 5: Fractals!
By Dylan King
10/23/15
Stage 1
Fractals demonstrate a repeating pattern that can encompass geometry. Fractals can range from the detail injected into an image on a computer monitor to the beat of a human heart and deal with rough shapes or, “Monsters” as referred to by the mathematical community until the late twentieth century. The Sierpinski triangle is a subdivision of many iterations of triangles that all seem to be built inside of a large equilateral triangle. The Sierpinski triangle is a fractal because it demonstrates the idea of a repeating pattern which, in itself, is the simple definition of a fractal. Now that we know that the Sierpinski triangle is a fractal we can touch on the vocabulary that will be used to describe it. One of the keywords being, iterations. An iteration is a repetition of a process this is very clear in the aforementioned Sierpinski triangle. Another useful vocabulary word is recursion which is the repeated application of a procedure and is featured in many fractals. Two more useful words that will come in handy are initiator and generator. Initiator being the starting shape and generator being a collection of scaled copies of the initiator. The final keyword is self similarity which in geometry means that no matter how far you zoom in or out of an image it appears to be still or the same shape. Now that we have covered the basics, let’s move on to examining my own distinct Sierpinski triangle.
Stage 2
Through the strategic use of a ruler and protractor I created a number of drafts representing a Sierpinski triangle. My final choice came down to the fianite details present in the measure of the triangle's angles. During this time I was challenged by my inability to provide detail and refinement to the length of the lines present in my equilateral triangle. My Sierpinski triangle ended up with four iterations. This is so because the more iterations I added the more I was challenged with the process of creating truly equilateral triangles. In the end I decided that having less iterations was worth it if it meant my triangle would look more professional and accurately put together. My motivation for the design of my Sierpinski triangle was based around the three major themes that I use most when drawing. The bottom left being black and white and the bottom right representing the use of pattern drawn sporadically to fill in an area. On the top triangle I went with a different method and left some of the triangles blank only to go back and shade them in an effort to create something relatively three dimensional. These three different methods represent the way I enjoy drawing and formed together nicely to create my Sierpinski triangle.
Stage 3
My fractal would be identifiable as a Sierpinski triangle because it takes the overall shape of an equilateral triangle, the initiator, and is subdivided into four iterations of smaller self similar equilateral triangles, generator. All of these elements compiled into one make my triangle easily identifiable as a Sierpinski triangle. The recursion present in my triangle showcases its ties to fractals as a whole with it’s clear repetition of a pattern throughout my piece. The concept of self similarity also dominates my triangle and shows many aspects of fractals that appear in my Sierpinski triangle.
Problem of the Week 6
Dylan King
The diagram below shows the graph of 3x + 4y = 12. The shaded figure is a square, three of whose vertices are on the coordinate axes. The fourth vertex is on the line. Find a) the x and y intercepts of the line, b) the coordinates of the vertex that falls on the line, c) the length of a side of the square, and d) the area of the square.
The featured diagram shows the graph of 3x + 4y = 12 with a square beside it. Three of the square’s vertices are on the coordinate axis leaving the fourth on the line graphed above. Students were tasked with finding the x and y intercept and using that to determine the vertex that falls on the line. This would then allow them to uncover the length of a side of the square and the area of the square. I know that I will only need to find the length of one of the sides of a square because all of the sides in a square are of equal length. I also know that will need to break this problem into multiple pieces to solve it. The final goal for this Problem Of The Week is to use the length of the square's sides to discover the area of the square.
I roughly predicted that the answer to this problem would be under four because of the size of the square compared to the x and y intercepts. The first step I took to achieving my final goal was finding the x and y intercepts. I then looked to the square immediately and frustratingly found that I could not resolve the problem that simply. After some time I realized that I had to think larger and drew the graph as a large triangle. This meant I could find the area of this triangle and then use it to find the area of the square. I labeled the graph as PQD and the square as ABCD, this let me quickly try different methods to find an answer. I began to see that there were two other triangles on the opposite sides of the square. I was able to return to ABCD and quickly found an answer.
The x intercept is 4 leaving the y intercept to be 3. The coordinate of the vertex that falls on the line is (1.7,1.7). The length of one side of the square is 1.7 and the area of the square is 2.9.
This answer loosely corresponds with my prediction and is more reasonable than a guess of anything under four. A general conclusion I can draw from this problem is that any graphed problem that includes a square will always provide more than one triangle that will aid in finding a solution. I know my answer is correct because I was able to use the equation to prove it:
(3-a)(4-a)=a2
(3)(4)=12
12-4a-3a+a2=a2
4a+3a=7a
12-7a+a2=a2
12=7a
12/7=1.7
1.7=a
This answer also conforms to the x and y intercept and does not go farther than (4,0) and(0,3). All in all, this was my process towards understanding and solving the sixth Problem Of The Week.
Problem Of The Week 7
Find the order of 10 playing cards such that if you flip one card over and put the next one under the deck, and repeat this process, the flipped cards will be A, 2, 3, 4, 5, 6, 7, 8, 9, 10 in order.
Understanding the problem:
This Problem of The Week challenges students to use ten playing cards ( A-10) and align them in such a way that if you flip one card over and put the next one under the deck they would be laid down in order from A-10. The real challenge of this problem is determining which cards would partner well in order to play them from A through 10. Firstly, we need to create a process in which we can flip the cards that will give us the solution. Once we have this, we have accomplished the goal of the POW and can move on to describing our results. The final goal of this POW is not only to get the answer, but to use critical thinking to avoid simple trial and error, when solving this problem.
Process:
One of the predictions we came up with during the process of figuring out the POW was that the numbers on the cards that were flipped did not matter, they could be anything, the only number that mattered was the fact that the last card you flipped had to be before the number 10, so when you flipped it over, it would go in order, from 9 to 10. After we had a general prediction we moved on to attempting to get an answer. We started this process by playing with a deck of cards in order to get a feel of how the problem would be resolved. Once we had established a feel for the playing cards we moved on to testing certain formulas in order to achieve an answer. At first we tried stacking twenty cards in this order: 1, 10, 2, 9, 3, 8, 4, 7, 5, 6, 6, 5, 7, 4, 8, 3, 9, 2, 10. We figured out how to the cards to go from A-10, but we knew that there was still a more efficient way to solve this and use less cards. I (Dylan) then looked at aligning the first five cards which, after much contemplation, I realized would simply be A-5. This approach made me look at the POW in a way in which I saw the cards divided into to two stacks (A-5 and 6-10) This perspective quickly lead me to a feasible answer. Solution: Our solution for this problem is to order the cards in this fashion:
Cards you flip: A 2 3 4 5
Cards you tuck: 6 7 8 9 10
This means that you would first lay down A-5 and then be presented with the opportunity to lay down 6-10 creating a perfect answer for this POW.
Justification:
At first glance at this POW I had a hard time justifying my answer, but after deeply writing my process I have realized the precise steps I took that not only gave me the answer, but justified it. The first thing to note is that this problem should be looked at as if it was split in half. This way you only focus on the first five cards. After this there are five empty spaces in between the previous cards to be filled if the cards would be laid down from A-10. This would have to be 6-10 in that order because the cards will all be tucked in the first, “round” and would then be laid down from 6-10 later.
Understanding the problem:
This Problem of The Week challenges students to use ten playing cards ( A-10) and align them in such a way that if you flip one card over and put the next one under the deck they would be laid down in order from A-10. The real challenge of this problem is determining which cards would partner well in order to play them from A through 10. Firstly, we need to create a process in which we can flip the cards that will give us the solution. Once we have this, we have accomplished the goal of the POW and can move on to describing our results. The final goal of this POW is not only to get the answer, but to use critical thinking to avoid simple trial and error, when solving this problem.
Process:
One of the predictions we came up with during the process of figuring out the POW was that the numbers on the cards that were flipped did not matter, they could be anything, the only number that mattered was the fact that the last card you flipped had to be before the number 10, so when you flipped it over, it would go in order, from 9 to 10. After we had a general prediction we moved on to attempting to get an answer. We started this process by playing with a deck of cards in order to get a feel of how the problem would be resolved. Once we had established a feel for the playing cards we moved on to testing certain formulas in order to achieve an answer. At first we tried stacking twenty cards in this order: 1, 10, 2, 9, 3, 8, 4, 7, 5, 6, 6, 5, 7, 4, 8, 3, 9, 2, 10. We figured out how to the cards to go from A-10, but we knew that there was still a more efficient way to solve this and use less cards. I (Dylan) then looked at aligning the first five cards which, after much contemplation, I realized would simply be A-5. This approach made me look at the POW in a way in which I saw the cards divided into to two stacks (A-5 and 6-10) This perspective quickly lead me to a feasible answer. Solution: Our solution for this problem is to order the cards in this fashion:
Cards you flip: A 2 3 4 5
Cards you tuck: 6 7 8 9 10
This means that you would first lay down A-5 and then be presented with the opportunity to lay down 6-10 creating a perfect answer for this POW.
Justification:
At first glance at this POW I had a hard time justifying my answer, but after deeply writing my process I have realized the precise steps I took that not only gave me the answer, but justified it. The first thing to note is that this problem should be looked at as if it was split in half. This way you only focus on the first five cards. After this there are five empty spaces in between the previous cards to be filled if the cards would be laid down from A-10. This would have to be 6-10 in that order because the cards will all be tucked in the first, “round” and would then be laid down from 6-10 later.
Problem Of The Week 8
Understanding the problem: In this problem of the week (POW) students were tasked with solving a problem through reasoning. This problem was about four students,Brianna, Andy, Courtney and Brandon, who entered a competition at their school. The prizes for the competition are:
First place: An anniversary Math Faculty Pink Tie.
Second Place: A book of Brain Teasers.
Third Place: “I Love Math” T-shirt.
Teachers at this school make predictions about who of the four students will win a prize. These predictions are:
Lauren predicted that Brandon would win the pink tie, Andy would win the book and Courtney would win the T-shirt.
Hannah predicted that Brianna would win the pink tie, Brandon would win the book and Courtney would win the T-shirt.
Aliza predicted that Andy would win the pink tie, Brianna would win the book and Brandon would win the T-shirt.
It turns out that each teacher predicted one student correctly, leaving students to precisely determine whom. The real challenge of this problem is keeping track of the information and separating the useful facts from the problem’s, “ Filler.” I already know that each teacher was right about one student, I need to determine a way to figure out who they were right about. This will lead me to accomplishing this POW. My final goal for this POW is to use reasoning to weigh the predictions of each teacher and figure out the final three students.
Process:
To begin this POW I carefully read each teacher's prediction and made a list of what place certain students would receive. The only progress that this gave me is that Courtney is voted for third place twice. Using this information I saw that Andy and Brandon would be first and second place because brianna was only voted for twice. Unfortunately I learned that this would not work because both Lauren and Hannah predicted Courtney and her third place stance would contradict seeing as each teacher only guessed one correct student. I then went back to the first steps of this problem and re wrote a list. Unlike the original I organized mine into categories that showed each teacher's guesses from 1st-3rd place next to each other. I then crossed out Courtney, who I already knew was wrong, and discovered that there was only one more option for third place which was Brandon. Knowing this I crossed Brandon's name out any other place it appeared on the chart. I then looked at Hannah's guesses; and because she predicted Courtney would be third and brandon would be second ,which were both wrong, I realized her guess for first place was the correct one. Hannah guessed that Brianna would come in at first place, this left me with only one student for second place and an answer to the POW.
Solution:
The three students that placed in this competition are:
1st place, Brianna- guessed by Hannah
2nd place, Andy-guessed by Lauren
3rd place, Brandon, guessed by Aliza
Justification:
I know that my answer is correct because of the way that I used reasoning to weigh the predictions of each teacher and figure out the final three students. Even though I did not predict the answer to this POW I believe that it would have been different because of the way that I first viewed this Problem. A general conclusion that I pulled from this problem is that a question like this can be solved by rereading the problem. By this I mean that the problem clearly states that each staff member predicted exactly one prize winner correctly which would have lead me to an answer sooner than my method. Even though I did not reread the problem I eventually came up with an answer, I know that this answer is correct because the process I took crossed off each unfit or impossible guess until I was left with the final students which was exemplified above. All in all, this POW was a good mix of a word problem and creative thinking and pushed me to go above and beyond my usual work.
First place: An anniversary Math Faculty Pink Tie.
Second Place: A book of Brain Teasers.
Third Place: “I Love Math” T-shirt.
Teachers at this school make predictions about who of the four students will win a prize. These predictions are:
Lauren predicted that Brandon would win the pink tie, Andy would win the book and Courtney would win the T-shirt.
Hannah predicted that Brianna would win the pink tie, Brandon would win the book and Courtney would win the T-shirt.
Aliza predicted that Andy would win the pink tie, Brianna would win the book and Brandon would win the T-shirt.
It turns out that each teacher predicted one student correctly, leaving students to precisely determine whom. The real challenge of this problem is keeping track of the information and separating the useful facts from the problem’s, “ Filler.” I already know that each teacher was right about one student, I need to determine a way to figure out who they were right about. This will lead me to accomplishing this POW. My final goal for this POW is to use reasoning to weigh the predictions of each teacher and figure out the final three students.
Process:
To begin this POW I carefully read each teacher's prediction and made a list of what place certain students would receive. The only progress that this gave me is that Courtney is voted for third place twice. Using this information I saw that Andy and Brandon would be first and second place because brianna was only voted for twice. Unfortunately I learned that this would not work because both Lauren and Hannah predicted Courtney and her third place stance would contradict seeing as each teacher only guessed one correct student. I then went back to the first steps of this problem and re wrote a list. Unlike the original I organized mine into categories that showed each teacher's guesses from 1st-3rd place next to each other. I then crossed out Courtney, who I already knew was wrong, and discovered that there was only one more option for third place which was Brandon. Knowing this I crossed Brandon's name out any other place it appeared on the chart. I then looked at Hannah's guesses; and because she predicted Courtney would be third and brandon would be second ,which were both wrong, I realized her guess for first place was the correct one. Hannah guessed that Brianna would come in at first place, this left me with only one student for second place and an answer to the POW.
Solution:
The three students that placed in this competition are:
1st place, Brianna- guessed by Hannah
2nd place, Andy-guessed by Lauren
3rd place, Brandon, guessed by Aliza
Justification:
I know that my answer is correct because of the way that I used reasoning to weigh the predictions of each teacher and figure out the final three students. Even though I did not predict the answer to this POW I believe that it would have been different because of the way that I first viewed this Problem. A general conclusion that I pulled from this problem is that a question like this can be solved by rereading the problem. By this I mean that the problem clearly states that each staff member predicted exactly one prize winner correctly which would have lead me to an answer sooner than my method. Even though I did not reread the problem I eventually came up with an answer, I know that this answer is correct because the process I took crossed off each unfit or impossible guess until I was left with the final students which was exemplified above. All in all, this POW was a good mix of a word problem and creative thinking and pushed me to go above and beyond my usual work.
Problem Of The Week 9
This Problem Of The Week was different from the regular goings on of Geometry. For this POW students were sorted into groups of four and had to produce four different puzzles that had to do with math. The following is mine.
The puzzle that I envision is a combination of a measurement puzzle, a formulaic puzzle, and a lock and key puzzle. This puzzle would begin by forcing the escapee’s to convert the length and width of a rug or piece of furniture from the imperial system to the metric system. They would do this via a guide that was discovered on a table at the start of the escape room. After this they would use the metric length to determine the area of the rug/ furniture using the formula length*width=area. Finally, they would use the area of the rug as a code to unlock a box containing a sheet of paper from a biology class, which leads to Haley's puzzle. To carry this out I need one rug, a ruler, a paper describing the metric system, a box, and a lock with a number code.
The puzzle that I envision is a combination of a measurement puzzle, a formulaic puzzle, and a lock and key puzzle. This puzzle would begin by forcing the escapee’s to convert the length and width of a rug or piece of furniture from the imperial system to the metric system. They would do this via a guide that was discovered on a table at the start of the escape room. After this they would use the metric length to determine the area of the rug/ furniture using the formula length*width=area. Finally, they would use the area of the rug as a code to unlock a box containing a sheet of paper from a biology class, which leads to Haley's puzzle. To carry this out I need one rug, a ruler, a paper describing the metric system, a box, and a lock with a number code.