# Find $a,b,c$ such that if $(−a + b + c, a - b + c, a + b - c)$ is all positive the process is repeated indefinitely

Given a tern of real numbers $$(a, b, c)$$, you build a new tern $$(−a + b + c, a - b + c, a + b - c)$$. If all the elements are positive, then the process is repeated; otherwise it stops. Determine all triples of positive Reals (a, b, c) for which the process never stops.

I think the answer is that for any picked element $$a$$ cannot be larger than $$b+c$$

• or equivalently $a,b,c$ are the sides of triangle – Albus Dumbledore Sep 29 '20 at 2:35

## 3 Answers

Hint: if any element is larger than the sum of the other two, you immediately get a negative member, and that's the only way you can get a negative member. On the other hand, the property that each member is at most the sum of the other two is preserved under the transformation.

• The property is not preserved: $(2,3,4)\rightarrow(5,3,1)\rightarrow(-1,3,7)$. – Jaap Scherphuis Sep 29 '20 at 6:40
• Oops, you're right. OP, please un-accept so I can delete this. – Robert Israel Sep 29 '20 at 22:20

Note that the sum of elements is not changing over transformations. Denote it by $$S = a + b + c$$, then transformations are taking following form: $$(a, b, c)\to (S - 2a, S - 2b, S - 2c) \to (4a - S, 4b - S, 4c - S) \to \dots$$

Let's consider this recursion more precise for $$a$$: $$a_0 = a, \quad a_{n+1} = S - 2a_n.$$ Using a simple induction it's easy to get closed forms for even and odd elements: $$a_{2n} = \sum_{k=0}^{2n-1}(-2)^{k}S + (-2)^{2n}a = -\sum_{k =0}^{n-1}2^{2k}S + 2^{2n}a = -\frac{2^{2n} - 1}{3}S + 2^{2n}a > 0,$$ which gives us following bound: $$\frac{3a}{S} > 1 - \frac{1}{2^{2n}} \quad \forall n\in\mathbb{N}; \tag1$$ By the same way, for odd elements $$a_{2n+1} = \sum_{k=0}^{2n}(-2)^{k}S +(-2)^{2n+1}a = -\sum_{k =0}^{n}2^{2k}S + 2^{2n}S -2^{2n+1}a = \frac{2^{2n + 1} + 1}{3}S - 2^{2n+1}a > 0,$$ we obtain that $$\frac{3a}{S} < 1 + \frac{1}{2^{2n}} \quad\forall n\in\mathbb{N}. \tag2$$

Now the rest is only to see that taking limit $$n\to\infty$$ in $$(1)$$ and $$(2)$$ yields the only possible value of $$a = \frac{S}{3}$$. The same for $$b$$ and $$c$$.

So the answer is $$a = b = c$$.

Here is an easier way to see what is going on.

Write the triple as $$(A-x, A, A+y)$$, so $$A$$ is the middle value and $$x$$, $$y$$ are the differences with the outliers.

After the transformation we get $$(A+y+x, A+y-x, A-x-y)=(B+2x, B, B-2y)$$, where $$B=A+y-x$$ is the new middle value, and the differences are now $$2x$$ and $$2y$$. Clearly the differences have doubled. This happens every time, and since their sum remains constant eventually the smallest will become negative.

The only way to avoid this happening is to have no differences, i.e. start with three equal numbers.