<s>[INST] Elementary Geometry The side of the square measures $1\  \mathrm{cm}$ , and $AC = 1\  \mathrm{cm}$, find the value of $AB$

 [/INST] First try:
There are 3 rectangular triangles here, where we get 3 Pythagoras from:
$$
\begin{eqnarray*}
1^2+(1-d)^2&=&c^2\\
a^2+d^2&=&1^2\\
(1+a)^2+1^2&=&(1+c)^2,
\end{eqnarray*}
$$
with $a=\overline{AB}$,$d=\overline{BC}$ and $c=\overline{0C}$, where $0$ is the upper left point. With this, we ask Wolfram to get 
$$
d=\overline{AB}\approx 0.883204 = \frac{1}{2}\left( \sqrt{2}-1+\sqrt{2\sqrt{2}-1} \right)
$$

Second try:
We have $\sin \alpha= \frac{a}{1}= \frac{1+a}{1+c}$ and $1+(1-d)^2=c^2$. 
Further, since $d=\cos( \arcsin (a) )=\sqrt{1-a^2}$, we get
$$
a=\frac{1+a}{1+\sqrt{1+(1-d)^2}}=\frac{1+a}{1+\sqrt{1+(1-\sqrt{1-a^2})^2}}, \tag{*}
$$
with the same solution as above.
$(*)$ can be reformulated to 
$$
a^8-2a^6+7a^4-6a^2+1=A^4-2A^3+7A^2-6A+1=0,
$$
where the roots of polynomial of $4$th order can be found here. But because it works out nice, let's go one step further and substitute $a=u+1/2$, so we get:
$$
u^4+\frac{11}{2}u^2-\frac{7}{16}=0.
$$
From here on, the rest is left to you. Nice Question!
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