Coin Splitting Invariant

Start with 1,000 coins. Split piles any way you like. Each split n → x + (n−x) adds x(n−x). Why is the final total always the same?

Final total
499,500
= C(1000, 2)
Invariant
running total + Σ C(pile size, 2) = C(1000, 2)
Every split transfers exactly x(n−x) from “remaining pair potential” into the running total.

Interactive example with 8 coins

Click any pile larger than 1, choose where to split it, and watch the invariant stay fixed at 28.
Running total 0
Remaining pair potential 28
Their sum 28
Current piles
Select a pile larger than 1.
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Algebraic proof

C(n,2) − C(x,2) − C(n−x,2)
= x(n−x)

Before splitting, a pile of size n contributes C(n,2) to the remaining potential. After splitting, its two children contribute C(x,2)+C(n−x,2). The drop is exactly x(n−x), the amount added to the running total.

Combinatorial proof

Consider every unordered pair of coins. A pair stays together until one particular split separates its two coins for the first time.

A split into x and n−x separates exactly x(n−x) pairs. Therefore every pair of coins is counted exactly once.

C(1000,2) = 1000×999/2
Answer
499,500
The total is independent of the splitting strategy because each unordered pair of coins contributes exactly once—at the moment the pair is first separated.