<s>[INST] Calculating the limit of a geometrical construction Suppose I have a rectangle ABCD, with AD larger than AB. I trace its diagonal AC, then the perpendicular to AC that goes through B. It intersects AC in $B_1$. I then trace the perpendicular to AD that goes through $B_1$, which intersects AD in $A_1$, and the perpendicular to CD that goes through $B_1$, which intersects CD in $C_1$.

We iterate the process to make $B_2$, $A_2$, $C_2$...

So on, so forth.
Clearly, $\lim_{n \to +\infty}$$C_n$ is D. But $\lim_{n \to +\infty}A_n$ is not D, and is instead somewhere on AD.
Question : Given the initial size AD and AB, with AD>AB, what's the length of $\lim_{x \to +\infty}A_n$D?
I think that if the initial lengths are $\{a,b\}$, the lengths on the next step are $$\left\{\frac{a^3}{a^2+b^2},\frac{b^3}{a^2+b^2}\right\},$$ but I have no idea where to go from there...
 [/INST] The lengths of the next step you obtained are correct.
Now consider the ratio between the initial height and width, i.e. write $b = ka, k>1$.
The new lengths are given by:
$$\frac {a^3}{a^2+ b^2} = \frac {a^3}{a^2 + k^2a^2} = \frac a{1+k^2}$$
$$\frac {b^3}{a^2+b^2} = \frac {k^3a^3}{a^2+b^2}$$
hence the new ratio of height and width is $k^3$. We can also write
$$\frac {b^3}{a^2 + b^2} = \frac {bk^2a^2}{a^2+k^2a^2} = \frac b{1+k^{-2}}$$
and so a recurrence relation becomes apparent. For the height, we are looking for:
$$\frac a{(1+k^2)(1+(k^3)^2)(1+((k^3)^3)^2)\dots} = \frac a{\prod_{n=0}^\infty (1+k^{2\cdot 3^n})}$$
Since $k > 1$, the infinite product diverges, so the height tends to zero as expected.
As for the width, the opposite occurs: we are looking for
$$\frac b{(1+k^{-2})(1+(k^3)^{-2})(1+((k^3)^3)^{-2})\dots} = \frac b{\prod_{n=0}^\infty (1+k^{-2\cdot 3^n})}$$
To show that the infinite product converges, we can check the equivalent series
$$\sum_{n=0}^\infty k^{-2\cdot 3^n} < \sum_{n=0}^\infty k^{-n} = \frac1{1-k^{-1}}$$
which converges as $0 < k^{-1} < 1$. It can also be expressed as the infinite sum:
$$\prod_{n=0}^\infty (1+k^{-2\cdot 3^n}) = 1 + k^{-2} + k^{-6} + k^{-8} + k^{-18} + \dots = \sum_{n \in S} k^{-n}$$
where $S$ is the set of all nonnegative integers that does not contain $1$ in their base $3$ representation (https://oeis.org/A005823).
A trivial upper bound is $\displaystyle \sum_{n = 0}^\infty k^{-2n} = \frac 1{1-k^{-2}} = \frac {k^2}{k^2-1}$, which shows that the limit of the width cannot be less than $\dfrac {b(k^2-1)}{k^2} = b\left(1-\dfrac1{k^2}\right)$. I am however unable to think of a closed form for the infinite product or the infinite sum.
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