It's a Puzzlement
by John P. Robertson
The following puzzlement is harder than it may look. It involves fitting together 30°-60°-90° triangles of various sizes. You may recall that these are right triangles and have sides and hypotenuse proportional to 1, , and 2. The puzzle is to dissect one such triangle with shortest side 10 into one such triangle with shortest side 1, two with shortest side 2, three with shortest side 3, and four with shortest side 4. Otherwise put, arrange ten 30°-60°-90° triangles, of the given sizes, into one 30°-60°-90° triangle.
Because the areas of similar triangles (ones with the same angles) are proportional to the squares of corresponding sides, the above dissection illustrates the relation 13 + 23 + 33 + 43 = 102. In fact, 13 + 23 + 33 + + n3 is a square for any positive integer n. (Actually, 13 + 23 + 33 + + n3 = (1 + 2 + 3 + + n)2.) This is a nonobvious result that may come as a surprise to many of you who thought you were experts in numerical analysis! I don't think there are any geometrical figures known so that one of size 1 and two of size 2 fit together to make one of size 3, or one of size 1, two of size 2, and three of size 3 fit together to form one of size 6.
Thank you to all who submitted palindromes in response to the August 2002 "Puzzlement." Judging on originality, difficulty, and wit has now begun. Winners will be announced and prizes awarded in the February 2003 issue of The Actuarial Review. All entries will be posted on the CAS Web Site.