Week Of Inspirational Maths
In this week we experimented with various maths problems and learned about how maths and learning about maths effects the brain.
What was the purpose of this week of investigation and video-watching? |
This week served as both an introduction into what we would be learning in 10th grade maths, as well as a way to get us excited about learning math. A big theme that was frequently mentioned was how messing up on a problem doesn't mean you're a failure, it means you're learning new things. We also talked about how nobody was a "Maths-Person" and how everyone can get good at maths. These ideas were used to help get us in a growth mindset so that we won't give up on tough problems we encounter in class throughout the year.
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During the week we investigated four main maths problems. We attempted to find the minimum amount of squares needed to completely fill a 11*13 rectangle, explored a pattern regarding the total amount of squared in a group of squares forming a staircase shape, hypothesized about the yet unsolved sequence of increasing and decreasing numbers known as the Hailstone Sequence and discovered the composition of cubes if they were put in a 3*3*3 cube and dipped in paint. We also watched a series of videos made by Stanford professor Jo Boaler that talked about math and it's effect on the brain.
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Give an overview of all the activities and videos. |
Choose two messages from the five videos we watched. Explain the personal significance of those videos |
One of the videos we watched talked about how, when it comes to math, speed doesn't matter. This is significant to me because throughout elementary school every math test we took was timed, and while I usually got all the problems correct I would always doubt how good I actually was because other people could finish faster than me. This also affects me currently because in my Trigonometry class the quizzes all have extremely harsh time limits. Another concept covered in the videos that has personal significance for me was about how, whether you do so consciously or not, your brain uses the image of fingers to perform calculations. This was interesting to me because when I think about numbers I never consciously visualizing anything, I have always preferred to view numbers more abstractly so I can more easily apply concepts to different things but I suppose that subconsciously, my brain pust be adding in a sort of placeholder image, and it's really cool that that image is what most people use to initially count when they are younger.
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Out of the 4 problems we explored in class during our week of inspirational maths, the one I explored the most and the one I decided to extend here was the painted cube problem. The problem is this: if a 3*3*3 cube constructed out of 1*1*1 cubes was lowered into a vat of paint, removed and then disassembled into its 1*1*1 cube components, how many 1*1*1 cubes would be painted on 3 sides, 2 sides, 1 side or no sides?
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To find cubes with 3 sides painted, I realized that the only cubes exposed on three of their sides would be the ones on the corners of the cube. A cube has 8 vertices, so there were 8 cubes painted on 3 times.
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To find the cubes with one side painted, I realized that the only curves that could be exposed on only one face would be located in the center of the faces on the 3*3*3 cubes. On a 3*3*3 cube only one square on each side isn't on an edge, and because there are six sides to a cube there are 6 1*1*1 cubes with only one side painted.
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The cubes that were painted on two faces are all the painted cubes that weren't painted once or thrice. The cubes on the center of each face were painted once, so the twice painted cubes must be on the edge. The cubes on the vertices were painted thrice, leaving the twice painted cubes to be on all edges that are not also vertices. For a 3*3*3 cube, that means 12 1*1*1 cubes were painted on two sides.
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The only cube that was not painted at all is the one in the very center. It was protected by a "shell" of other cubes. To verify that all my values were correct, I added them up. 8+6+12+1=27 and 3*3*3=27 so I know all cubes were accounted for and I didn't count anything more than once.
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After I solved the 3*3*3 cube I was challenged to solve the problem for a 4*4*4 cube. Using a similar method, I found that the 4*4*4 cube had 8 cubes painted on 3 sides, 24 cubes painted on 2 sides, 24 cubes painted on 1 side and 8 cubes painted on no sides. At this point I started to recognise a pattern. For a larger cube of any size, there would always be 8 cubes painted on 3 sides (y=8) because all cubes have 8 vertices. The number of cubes painted on 2 side would be equal to the number of edges (always 12) multiplied by their length subtracted by 2 to account for the vertices (y=12(x-2)). The number of cubes with one side painted is equal to the length of each side squared, but you have to subtract 2 again from that original number because you can't count any cubes on an edge. There are 6 sides to a cube so (y=6(x-2)^2). Finally the number of cubes painted on no sides is equal to all the cubes protected by the outer "shell" of cubes ((x-2)^3).
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Above: the work I wrote down to solve this problem
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