dr. wiggins' pythagorean challenge

     In February 2013, Dr. Wiggins challenged the Algebra 1 students of East Tipp Middle School to determine if the Pythagorean Theorem could be generalized to shapes other than squares built on the sides of a right triangle.

     After some time to think, Zach broke the ice with his semicircle proof below. Soon afterward, however, so many kids were weighing in that they began to become a bit territorial. To make sure we knew who got dibs on what, we started an ownership board.





     (You can click the photo to make it bigger, but don't read too closely ... not all of our wishes came true.) The board not only reserved shapes for specific kids, but it also helped to accelerate a sort of arms race between the more competitive of the group.

     And speaking of competitive, two of the kids who took Algebra 1 last year as seventh graders have been getting pretty jealous of this year's class as the kids worked their way through 42b and the Rock Problem. So if you see new faces below, it's because the class finally took pity on Ben and Carl.

     Without further ado, here are the findings ...



Equilateral Triangles (Meredith)
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Isosceles Right Triangles (Zach)
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Regular Pentagons (Drew)
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Regular N-Gons (Jon)
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Semicircles (Zach)
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Parabolic Arches (Zach)
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Infinte Squares (Carl)
click photo to enlarge / click here for pdf


Cubes Surface Area (Zach)
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Regular Dodecahedra Surface Area (Drew)
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Tetrahedra Volume (Ben)
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     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!





UPDATE: The oddballs have been identified!






     When looking for the two photos that tell a different story, kids didn't find it immediately apparent.





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     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, ...





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     ... 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)
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Parabolic Arches (Zach)
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     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:












     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.





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     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:





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click photo to enlarge


     And thus, the Pythagorean Theorem can be restated as follows:

In any right triangle, the length of the hypotenuse Wigginsed
is equal to the sum of the Wigginses of the two legs.


     Thanks again for all the fun, Dr. Wiggins!