Radii of five concentric circles $\omega_0,\omega_1,\omega_2,\omega_3,\omega_4$ form a geometric progression with common ratio $q$ in this order. What is the maximal value of $q$ for which it's possible to draw a broken line $A_0A_1A_2A_3A_4$ consisting of four equal segments such that $A_i$ lies on $\omega_i$ for every $i=\overline{0,4}$?
HIDE: thanks Thanks to the user Vlados021 for translating the problem.Problem
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Tags: geometry, algebra
SAUDITYA
01.05.2019 18:31
Is this correct it looks too simple to be true
Assume the radii to be $\{1,q,q^2,q^3,q^4\}$ with $q \ge 1$ See that $A_0A_1 \le 1+q$ and $A_{3}A_4 \ge q^4 -q^3$.So $1+q \ge q^4-q^3 => q^4-q^3-q-1 \le 0 => q \le \frac{1+ \sqrt{5}}{2}$.For construction just start with any point say $A_0$ on unit circle and draw a circle centered at this point of radius $1+q$ let it intersect the next circle . Label this point $A_1$. Now repeat the process with $A_1$ this time the intersection point maybe two so select any one if them and continue...
JAnatolGT_00
10.01.2023 01:23
We claim the answer $\frac{1+\sqrt 5}{2}.$ WLOG $\omega_i$ has radius $q^i$ and $q\geq 1.$ Observe that $|A_iA_{i+1}|\in [q^i(q-1);q^i(q+1)]$ must hold, so the broken line exists if and only if all such segments have common point, i.e. $$q+1\geq q^3-q^2\iff (q^2+1)(q^2-q-1)\leq 0\iff q\in \left[ \frac{1-\sqrt 5}{2};\frac{1+\sqrt 5}{2}\right] .$$