11.5. The radii of five concentric circles $\omega_{0}, \omega_{1}, \omega_{2}, \omega_{3}$, $\omega_{4}$ form a geometric progression with common ratio $q$ in the given order. For what largest $q$ can a non-closed broken line $A_{0} A_{1} A_{2} A_{3} A_{4}$, consisting of four segments of equal length, be drawn such that $A_{i}$ lies on $\omega_{i}$ for all $i=0,1,2,3,4$?

(I. Bogdanov)
Answer. When $q=\frac{\sqrt{5}+1}{2}$.

Solution. We can assume that $q \geqslant 1$. Let the radius of $\omega_{i}$ be $R_{i}=R q^{i}$.

Choose some positive $\ell$ and try to construct the required broken line with segments of length $\ell$, starting from an arbitrary point $A_{0} \in \omega_{0}$. Suppose the point $A_{i} \in \omega_{i}$ has already been constructed. The distances from it to the points of the circle $\omega_{i+1}$ range over the interval $\left[R_{i+1}-R_{i}, R_{i+1}+R_{i}\right]$, that is, $\left[R q^{i}(q-1), R q^{i}(q+1)\right]$. The point $A_{i+1}$ can be constructed if and only if $\ell$ belongs to this interval. Therefore, the broken line can be constructed if and only if $R q^{i}(q-1) \leqslant \ell \leqslant R q^{i}(q+1)$ for all $i=0,1,2,3$.

Since $q \geqslant 1$, this system of inequalities is equivalent to the inequalities $R q^{3}(q-1) \leqslant \ell \leqslant R(q+1)$. A length $\ell$ satisfying them exists if and only if $q^{3}(q-1) \leqslant q+1$, that is, $q^{4}-q^{3}-q-1 \leqslant 0$, or $\left(q^{2}-q-1\right)\left(q^{2}+1\right) \leqslant 0$. The largest value of $q$ satisfying this inequality is $q=\frac{\sqrt{5}+1}{2}$.