Problem:
Legs L1​,L2​,L3​,L4​ of a square table each have length n, where n is a positive integer. For how many ordered 4-tuples (k1​,k2​,k3​,k4​) of nonnegative integers can we cut a piece of length ki​ from the end of leg Li​(i=1,2,3,4) and still have a stable table? (The table is stable if it can be placed so that all four of the leg ends touch the floor. Note that a cut leg of length 0 is permitted.)
Solution:
Turn the table upside down so its surface lies in the xy-plane. We may assume that the corner with leg L1​ is at (1,0), and the corners with legs L2​,L3​,L4​ are at (0,1),(−1,0),(0,−1), respectively. (We may do this because rescaling the x and y coordinates does not affect the stability of the cut table.) For i=1,2,3,4, let ℓi​ be the length of leg Li​ after it is cut. Thus 0≤ℓi​≤n for each i. The table will be stable if and only if the four points F1​(1,0,ℓ1​),F2​(0,1,ℓ2​),F3​(−1,0,ℓ3​), and F4​(0,−1,ℓ4​) are coplanar. This will be the case if and only if F1​F3​​ intersects F2​F4​​, and this will happen if and only if the midpoints of the two segments coincide, that is,
(0,0,(ℓ1​+ℓ3​)/2)=(0,0,(ℓ2​+ℓ4​)/2)(*)
Because each ℓi​ is an integer satisfying 0≤ℓi​≤n, the third coordinate for each of these midpoints can be any of the numbers 0,21​,1,23​,…,n.
For each nonnegative integer k≤n, let Sk​ be the number of solutions of x+y=k where x,y are integers satisfying 0≤x,y≤n. The number of stable tables (in other words, the number of solutions of (∗)) is N=∑k=0n​Sk2​.
Next we determine Sk​. For 0≤k≤n, the solutions to x+y=k are described by the ordered pairs (j,k−j),0≤j≤k. Thus Sk​=k+1 in this case. For each n+1≤k≤2n, the solutions to x+y=k are given by (x,y)=(j,k−j),k−n≤j≤n. Thus Sk​=2n−k+1 in this case. The number of stable tables is therefore
N​=12+22+⋯n2+(n+1)2+n2+⋯+12=26n(n+1)(2n+1)​+(n+1)2=31​(n+1)(2n2+4n+3)​
This problem was proposed by Elgin Johnston.
The problems on this page are the property of the MAA's American Mathematics Competitions