hajimbrak wrote: Starting with the triple $(1007\sqrt{2},2014\sqrt{2},1007\sqrt{14})$, define a sequence of triples $(x_{n},y_{n},z_{n})$ by $x_{n+1}=\sqrt{x_{n}(y_{n}+z_{n}-x_{n})}, y_{n+1}=\sqrt{y_{n}(z_{n}+x_{n}-y_{n})}, z_{n+1}=\sqrt{z_{n}(x_{n}+y_{n}-z_{n})}$, for $n\geq 0$.Show that each of these sequences converges to a limit and find these limits. Sketch of proof : Considering that $(x_n,y_n,z_n)$ are the three positive sidelengths or a triangle, we can rather easily show that : 1) $(x_{n+1},y_{n+1},z_{n+1})$ are the sides of a new triangle with same area 2) $|x_n^2-y_n^2|$, $|x_n^2-z_n^2|$ and $|y_n^2-z_n^2|$ are decreasing sequences 3) $x_n,y_n,z_n$ have same limits, sides of an equilateral triangle of same area Area of starting triangle is $\frac{\sqrt{(x_0+y_0+z_0)(x_0+y_0-z_0)(x_0-y_0+z_0)(-x_0+y_0+z_0)}}4$ Area of limit equilateral triangle is $\frac{\sqrt 3}4l^2$ Hence required limit is $l=\sqrt[4]{\frac{(x_0+y_0+z_0)(x_0+y_0-z_0)(x_0-y_0+z_0)(-x_0+y_0+z_0)}3}$ With $(x_0,y_0,z_0)=(1007\sqrt 2,2014\sqrt 2,1007\sqrt{14})$, this formula gives $\boxed{l=2014}$

Any motivation for that invariant ?

scale down by 1007 we get sides of a triangle

Adding some motivation to my 'cryptic' solution from more than 2 years ago We basically want to remove radicals from this expression, so squaring is very natural. The existence of the square root implicitly assumes that we have $y_n + z_n \ge x_n$. Also, the thing under the radical suspiciously looks like the product of length of tangent from a vertex and some other thing. This motivates the introduction of a triangle with sides $x_n, y_n, z_n$. Usually we need some kind of invariant in multivariable limits. So, we try the squared area of the triangle since it is a polynomial in $x_n, y_n, z_n$, more precisely, $16 \Delta^2 = 2\sum_{cyc} b^2c^2 - \sum_{cyc} a^4$. This surprisingly comes out to be invariant. Also to show the convergence, it seems better to choose the difference between squares. Also this problem can be done using this paper.

Exceptionally easy for P3; Idea only since I am in hurry. Central observation is if we treat the 3 values as lengths of a triangle then the ares is an invariant ! Another important observation is the subtended angles converges to 60 deg. Now bad calculations but a smart ans. !