CEMC Euclid, 2022

If $r$ is a term in the
sequence and $s$ is the next term,
then $s = 1 + \dfrac{1}{1+r}$.

This means that $s - 1 = \dfrac{1}{1+r}$ and so $\dfrac{1}{s-1} = 1+r$ which gives $r = \dfrac{1}{s-1} - 1$.

Therefore, since $a_3 = \dfrac{41}{29}$, then $$a_2 = \dfrac{1}{a_3-1} - 1 = \dfrac{1}{(41/29)-1} - 1 = \dfrac{1}{12/29} - 1 = \dfrac{29}{12} - 1 = \dfrac{17}{12}$$ Further, since $a_2 = \dfrac{17}{12}$, then $$a_1 = \dfrac{1}{a_2-1} - 1 = \dfrac{1}{(17/12)-1} - 1 = \dfrac{1}{5/12} - 1 = \dfrac{12}{5} - 1 = \dfrac{7}{5}$$
Initially, the water in the hollow tube forms a cylinder with
radius 10 mm and height $h$ mm. Thus,
the volume of the water is $\text{(10 mm ) 2(h mm ) = 100 h mm 3}$.

After the rod is inserted, the level of the water rises to 64 mm. Note
that this does not overflow the tube, since the tube’s height is 100
mm.

Up to the height of the water, the tube is a cylinder with radius 10 mm
and height 64 mm.

Thus, the volume of the tube up to the height of the water is $$\text{(10 mm ) 2(64 mm ) = 6400 mm 3}$$ This volume consists of the water that is
in the tube (whose volume, which has not changed, is $\text{100 h mm 3}$) and the rod up to
a height of 64 mm.

Since the radius of the rod is 2.5 mm, the volume of the rod up to a
height of 64 mm is $\text{(2.5 mm ) 2(64 mm ) = 400 mm 3}$.

Comparing volumes, $\text{6400 mm 3 = 100 h mm 3 + 400 mm 3}$ and so $100h = 6000$ which gives $h = 60$.