# Cutting three sticks

Cosmo plays the following (single-player) game with three wooden sticks:

• He first checks whether he can form a triangle from the three sticks (which means: whether the longest stick is at most as long as the combined length of the two shorter sticks).
• If Cosmo can form such a triangle: Cosmo has won the game.
• If Cosmo cannot form such a triangle: Cosmo takes the longest stick, cuts off from it a piece as long as the combined length of the two shorter sticks, throws away this piece, and keeps the remaining shortened stick.
• Then Cosmo repeats these steps with the two shorter sticks and the shortened stick.

Will Cosmo always win the game after a finite number of steps? For any choice of wooden sticks to start with?

• So a winning state is a + b >= c, rather than a + b > c?
– dmg
Feb 5, 2015 at 9:48
• @dmg: Yes, such degenerate triangles are winning states. Feb 5, 2015 at 9:49

Consider sticks of length $1, a, a^2$ - each stick is a times bigger than the previous one. Then the next step we will have sticks of length $a^2 - a - 1, 1, a$. If we choose a such that $a^2 - a - 1 = \frac{1}{a}$, then the sticks will retain the same ratios at each step, and we will never terminate. The required a is a root of $a^3 - a^2 - a - 1 = 0$. This turns out to have exactly one root at a = 1.8393, and we can check that $a^2 = 3.383 > a + 1$, so we don't terminate at step 1.
• It's notable that this solution relies only on the idea of finding a tuple which, under Cosmo's process, iterates to a multiple of itself - and this can yield other more complex examples. It happens that the solution to the system: $$b-2a-2=k$$ $$1=ka$$ $$a=kb$$ has that $1,\,a,\,b$ will, after two steps, iterate to $k$ times itself and hence never be solved. (Also, it appears that such solutions are not stable; numbers near your desired $a$ do not iterate forever - I would venture to conjecture that your $a$ is, in fact, the only one such that $1,a,a^2$ will never let Cosmo win) Feb 6, 2015 at 4:30