Equilateral Triangles (Meredith) click photo to enlarge |
Isosceles Right Triangles (Zach) click photo to enlarge |
Regular Pentagons (Drew) click photo to enlarge |
Regular N-Gons (Jon) click photo to enlarge |
Semicircles (Zach) click photo to enlarge |
Parabolic Arches (Zach) click photo to enlarge |
Infinte Squares (Carl) click photo to enlarge / click here for pdf |
Cubes Surface Area (Zach) click photo to enlarge |
Regular Dodecahedra Surface Area (Drew) click photo to enlarge |
Tetrahedra Volume (Ben) click photo to enlarge |
     There will be one more investigation uploaded, but not until the kids at East Tipp determine which two of the photos above are telling a slightly different story than the other eight. It's not necessarily a search for those that look the coolest or scariest (meaning the diagrams ... not the kids). When they come up with the answer to that, we'll be ready for our last upload.     
Thanks again for all the fun, Dr. Wiggins!
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     When looking for the two photos that tell a different story, kids didn't find it immediately apparent. |
click photo to enlarge |
     As a result, we starting grouping photos that looked to be telling the same story. For example, the equilateral triangles and the regular pentagons are just sub-cases of the regular n-gons. Since those three go together, they must be a part of the eight that are not as interesting as the mystery two. Furthermore, the dodecahedron proof seems to fit with the pentagon proof; the semicircles can be classified as an n-gon with regular sides. As a matter of fact, when we moved them around, ... |
click photo to enlarge |
click photo to enlarge |
     ... the idea of 'regular' came up a lot. When we separated out those that were of regular figures, we were left with just two. |
Isosceles Right Triangles (Zach) click photo to enlarge |
Parabolic Arches (Zach) click photo to enlarge |
     While these trios of figures aren't regular, they are similar (or proportional) to each other. Would the Pythagorean relationship hold for any group of proportional figures that are sized according to a Pythagorean triple? To find out, we decided on this irregular figure: |
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     Dr. Wiggins, please accept being the face of 'irregular' as the compliment for which it was intended. After all, none of the 'regular' people answered our email about Problem 42b!      It's easy enough to use a computer to make three of these figures in proportion; it's also easy enough to build them such that their bottom edges hold a Pythagorean relationship. But now ... how do we calculate the area of such a thing?      An early suggestion was to use a string to find the perimeter of each figure and then translate that into an area; however, a set perimeter doesn't define a certain area. Another suggestion was to cut Dr. Wiggins into familiar and manageable shapes ... but that felt icky even typing it, let alone doing it!      Finally, Drew suggested that we relate each printed head's area to its mass. Without the availabity of a scale capable of weighing paper accurately, we began constructing three right Wiggins prisms, as seen below. |
click photo to enlarge |
click photo to enlarge |
click photo to enlarge |
     Each right Wiggins prism has the same height; therefore, when we compare their volumes as related to their weights, a constant height will put the comparison squarely on the base area. The payoff: |
click photo to enlarge |
click photo to enlarge |
     And thus, the Pythagorean Theorem can be restated as follows: is equal to the sum of the Wigginses of the two legs.> |
    
Thanks again for all the fun, Dr. Wiggins!
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