227. $\left\{\begin{array}{l}x+y+z=10 \\ x z=y^{2} \\ z^{2}+y^{2}=x^{2}\end{array}\right.$

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Problems from "Practica geometriae" by Leonardo Fibonacci.
227. Leonardo's solution (using false position). He assumes $z=1$. Then

$$
\begin{aligned}
& x_{1}+y_{1}=9 ; x_{1}=y_{1}^{2} ; 1+y_{1}^{2}=y_{1}^{4} \text { and } x_{1}=\sqrt{\frac{5}{4}}+\frac{1}{2} ; \\
& y_{1}=\sqrt{\sqrt{\frac{5}{4}}+\frac{1}{2}}
\end{aligned}
$$

Instead of 10, the sum of the unknowns is now

$$
S=\sqrt{\sqrt{\frac{5}{4}+\frac{1}{2}}}+\sqrt{\frac{5}{4}}+\frac{1}{2}+1
$$

Then he takes the proportion: $S: 10=1: z$, from which $S z=10$

Substituting $S$ gives: $z=5-\sqrt{\sqrt{3125}-50}$ and so on.