Editor's Note: Jay G. ('16) investigated this geometric phenomenon for his pre-calculus class.
I picked this pattern because it caught my eye as something different from the rest of the patterns. All the other patterns were just the relationships between the amounts of a shape repeating in a pattern. This pattern was different, in the sense that it explores the relationship between the sides of triangles compounded with that of series and sequences. This problem also caught my eye because it introduced me to math art. I find these repeating patterns in shapes interesting and always wondered the relationship between the patterns in the shapes of math art. As a result, this problem allowed me to find that relationship, with my own process, of the black triangles to find the 43rd figure’s hypotenuse. My initial understanding of this problem would eventually lead to a larger relationship in that these triangles were not only in a pattern, but were all Pythagorean triplets.
I first went about solving this problem by comparing each side of each triangle. I noticed that the shortest side of each triangle of the set had increased by two for each consecutive triangle. The first triangle had a short side of 3, the second triangles short side was 5, and the third triangle’s shortest side was 7, this pattern can otherwise be known as an arithmetic series such that, 3, 5, 7 have a common difference of two. This relationship can be represented through the equation 3 + 2(n-1); such that 3 is the first term and 2 is the common difference. Now to find the shortest side of the 43rd figure I plugged 43 for n (the nth figure) and found the 43rd figure to have 87 for its shortest side. Now I drew my attention to the longer leg of each triangle. The longer leg of each triangle fell in the pattern of 4,12, 24. All these values had a different common difference: (from 0) +4 +8, +12. However, if we look at the common difference between these three differences, it is +4. If we continue this pattern, by adding 4 to the previous difference all the way up to the 43rd term, we find that the longer leg of the this triangle would be 3784. Now that I had two legs I could find the hypotenuse of the 43rd triangle. However, before I used Pythagorean theorem I noticed that the hypotenuse was always +1 the longer leg of that triangle hence, 4 and 5, 12 and 13, and 24 and 25. Now with this information I deduced that the hypotenuse of the 43rd triangle was 1+ 3784 which equals 3785. This means the equation to find the shorter leg is ultimately unnecessary. To check this result I used Pythagorean theorem using the two sides of the triangle to find the hypotenuse. That equation would be the square root of 87^2 + 3784^2, which yields 3785, the same answer as I deduced before. In addition to this answer I could not help but to notice that 3,4,5 and 5,12,13 were a Pythagorean triplets. I wondered if the third triangle was indeed a Pythagorean triplet and the 43rd triangle was a Pythagorean triplet because I was not sure. With a little research off the topic of this question I found that indeed all these triangles in this set are Pythagorean triplets. To confirm this I found the fourth triangle with sides 9,40,41 and the fifth triangle with sides 11, 60, 61 using the same method as I stated before to find the 43rd triangle. It turns out that these triangles are also Pythagorean triplets. As a result all the sides of all these triangles are in the same pattern not just because of the arithmetic similarities, but because they are all Pythagorean triplets.
The following is the source I used to confirm my suspicion that all the triangles in the pattern were Pythagorean triplets: http://www.tsm-resources.com/alists/PythagTriples.txt