In the left Figure, consider a right triangle $OPA$ with $\angle {AOP} = 90^0$. Let $\ell$ be the reflection of $PO$ in $PA$ and $\ell$ meets $OA$ at $A_1$. Let $O_1$ be the center of the circle $(PAA_1)$, the line $PO_1$ meets $OA$ at $A_2$. Let $O_2$ be the center of the circle $(PA_1A_2)$, the line $PO_2$ meets $OA$ at $A_3$,.....Let $O_n$ be the center of the circle $(PA_{n-1}A_{n})$, the line $PO_n$ meets $OA$ at $A_{n+1}$ for $n=3, 4, 5....$. Denote $\angle {OPA} = \alpha_1$, $ \angle A_{i}PA_{i+1}=\alpha_{i+1}$ for $i=\overline{1,n}$

Easily to show that $\alpha_2=\alpha_1$ and $\alpha_{n+1}=\alpha_{n}+\alpha_{n-1}$ for $n=2,3,4,...$ this is Fibonacci sequence.

Now in Cartesian coordinates, let $P=(0,1)$, $O=(0,0)$, $A=(x,0)$. What is the locus equation of some circumcenters $O_1$, $O_2$, $O_3$...? if we can find these locus equations, we can divide any angles by coefficients in the Fibonacci sequence.

Example: In right Figure, let $AOB$ be a right triangle with $A(0,1)$, $O(0,0)$, $B(x,0)$. The locus of $O_1$ (blue curve) meets $AB$ at $O_1$, the circle $(O_1, O_1B)$ meets $OB$ at $A_1, A_2$ then $\angle OAA_1 = \angle A_1AA_2 = \angle A_2AB = \frac{\angle OAB}{3}$.