A set of numbers has “the triple-sum property” (or TSP) if there exist three numbers

in the set whose sum is also in the set. [Repetitions are allowed.]
For example, the set U = {2, 3, 7} has TSP since 2 + 2 + 3 = 7, while V = {2, 3, 10} fails to have TSP.
(a) Suppose the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} is separated into two parts,
forming two subsets A and B.
Prove: Either A or B must have the triple-sum property.
[To begin the proof, suppose that statement is false and there are sets A and B as above, each without TSP.
If 1 lies in A then 3 = 1 + 1 + 1 must be in B. Complete the proof that this situation is impossible.]
(b) Is a similar result true when the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is separated into two parts?

To prove that either subset A or subset B must have the triple-sum property, we will use contradiction.

(a) Suppose we have two subsets A and B of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, and both subsets do not have the triple-sum property. Our goal is to show that this situation is impossible.

Let's assume that 1 lies in subset A. Now, we will examine the possible combinations in subset B:

- If 1 also lies in subset B, then both subsets A and B contain 1. Since repetitions are allowed, we can form the sum 1 + 1 + 1 = 3, which contradicts our assumption that subset B does not have the triple-sum property.

- If 1 does not lie in subset B, then the sum of 1 + 1 + 1 = 3 cannot be formed in subset B. Therefore, subset B does not have the triple-sum property.

In both cases, subset B does not have the triple-sum property. Therefore, if 1 lies in subset A, subset B must have the triple-sum property.

Now, let's consider the case where 1 lies in subset B. Following the same reasoning as above, subset A does not have the triple-sum property.

Therefore, in any separation of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} into two subsets A and B, either A or B must have the triple-sum property.

(b) For the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, we can see that the statement in part (a) does not hold.

For example, we can separate the set into A = {1, 2, 5, 7, 8} and B = {3, 4, 6, 9, 10}. Neither A nor B has the triple-sum property, as no combination of three numbers in either subset sums to a number within the set.

Therefore, the similar result proven in part (a) does not hold for the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

(a) To prove that either subset A or B must have the triple-sum property, we will assume the opposite and show that it leads to a contradiction.

Assume that there exist two subsets A and B, each without the triple-sum property.

Considering the number 1, we can assume that it lies in subset A.

Since 1 lies in A, by the triple-sum property, there exist three numbers in set A whose sum is equal to 1. Let's call these numbers x, y, and z.

So, x + y + z = 1. Now, we have the following possibilities:

1. If x = 1, then y + z = 0. Since all the numbers in the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} are positive, it is not possible for y + z to be equal to 0.

2. If x ≠ 1, then y + z = 1 - x. Since x, y, and z are all positive numbers, the maximum value of y + z is when x is minimized (i.e., x = 2). In this case, y + z = 1 - 2 = -1, which is not possible as all the numbers are positive.

In both cases, we reach a contradiction. Hence, the assumption that there exist subsets A and B without the triple-sum property is false. Therefore, either subset A or B must have the triple-sum property.

(b) In the case of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, we can provide a counterexample to show that the similar result is not true.

Let's separate the set into two subsets A = {1, 3, 5, 6, 10} and B = {2, 4, 7, 8, 9}.

Subset A does not have the triple-sum property since no combination of three numbers from A sums up to any other number in A.

Similarly, subset B does not have the triple-sum property as well.

Therefore, in this case, both subsets A and B do not have the triple-sum property, which contradicts the claim that at least one subset should have the triple-sum property. Hence, the similar result is not true for the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.