A circle of radius 100 cm is given. The goal is to place 7 circles of unequal radius inside the initial circle, so that each 3 of circles don't overlap each other and all of them stay inside the initial circle. The first smaller circle is of radius 50 cm and so on.
I thought it has to do something with Ford circles, but answer r = 14.36 cm for the 7th circle is incorrect.
Hint:
One way is to consider this construction as a part of the (infinite) Steiner’s chain of circles defined by two circles, $\bigcirc O((0,0), R)$ and $\bigcirc o((-r,0),r)$, with the distance between their centers $d=r$, where $R=100,\ r=50$. The other $6$ circles $\bigcirc O_0,\dots,\bigcirc O_5$ are sandwiched between the two, starting with $\bigcirc O_0((r,0),r)$.
All tangent points of the kissing circles in the chain are located on the circle (shown green in the picture) centered at $X$ with radius $r_x=|XO_i|$, $i=0,\dots,5$,
\begin{align} r_x&= \sqrt{rR-\frac{rRd^2}{(r+R)^2}} =\frac{200}3\approx 66.7 , \end{align}
while all the centers $O_i$, $i=0,\dots,5$, belong to the ellipse (shown yellow in the picture) focused at $o,\,O$.
Edit:
Using Descartes' theorem, for the curvatures $k_1,k_2,k_3,k_k$ of four mutually tangent circles, \begin{align} k_4&=k_1+k_2+k_3\pm 2\sqrt{k_1 k_2+k_2 k_3+k_3 k_1} \tag{1}\label{1} . \end{align}
In this particular arrangement of circles, \begin{align} k_1&=-\tfrac1R ,\quad k_2=\tfrac2R ,\quad k_3=k_2=\tfrac2R \end{align}
since each next circle in the chain is smaller, the sign in \eqref{1} would be always positive and we have
\begin{align} k_4&=\tfrac3R ,\\ k_5&=\tfrac{6}R ,\\ k_6&=\tfrac{11}R ,\\ k_7&=\tfrac{18}R ,\\ k_8&=\tfrac{27}R , \end{align}
so the radii of $\bigcirc O_i$, $i=1,\dots,5$ can be found as
\begin{align} r_i&=\frac R{i^2+2} . \end{align}
For example, the seventh inner circle, the smallest one, must have the radius
\begin{align} r_5&=\frac R{27} \approx 3.7037 . \end{align}