Problem
As shown in the figure below, two equilateral triangles $ACE$ and $BDF$ are inscribed in a regular hexagon $ABCDEF$, and a great circle with a diameter of $1 \ m$ is inscribed in them.
If a small circle with a diameter of $t \ m$ is inscribed in an isosceles triangle $ABF$, and small circles are also inscribed in other isosceles triangles, find $t$.
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Solution
Since the great circle is inscribed in the equilateral triangle $ACE$ (and $DBF$), its center is the centroid of the equilateral triangle.
At the same time, the length of the perpendicular drawn from one vertex of this equilateral triangle to the opposite side is $\frac{3}{2} \ m$.
Therefore, the length of one side of this equilateral triangle, for example, $CE$, is
$$CE=\frac{3}{2}×\frac{2}{\sqrt{3}}=\sqrt{3}.$$
Next, consider the isosceles triangle $ABF$ at the top.
At the same time, the length of the perpendicular drawn from one vertex of this equilateral triangle to the opposite side is $\frac{3}{2} \ m$.
Therefore, the length of one side of this equilateral triangle, for example, $CE$, is
$$CE=\frac{3}{2}×\frac{2}{\sqrt{3}}=\sqrt{3}.$$
Next, consider the isosceles triangle $ABF$ at the top.
Since $⊿OAG$ in the figure is a right triangle of $30°-60°-90°$ type,
$$AO=\frac{t}{2}×\frac{2}{\sqrt{3}}=\frac{t}{\sqrt{3}},$$
$$∴ \ AH=AO+OH=\frac{t}{\sqrt{3}}+\frac{t}{2}=\frac{2 \sqrt{3}+3}{6} t.$$
However, because the right triangle $∆BAH$ is also a right triangle of $30°-60°-90°$ type,
$$AO=\frac{t}{2}×\frac{2}{\sqrt{3}}=\frac{t}{\sqrt{3}},$$
$$∴ \ AH=AO+OH=\frac{t}{\sqrt{3}}+\frac{t}{2}=\frac{2 \sqrt{3}+3}{6} t.$$
However, because the right triangle $∆BAH$ is also a right triangle of $30°-60°-90°$ type,
$$AH=\frac{\sqrt{3}}{2}×\frac{1}{\sqrt{3}}=\frac{1}{2}.$$
Therefore,
$$\frac{2 \sqrt{3}+3}{6} t=\frac{1}{2},$$
$$(4 \sqrt{3}+6) t=6,$$
$$∴ \ t=\frac{6}{4 \sqrt{3}+6}=\frac{3}{2 \sqrt{3}+3}=2 \sqrt{3}-3≒0.4641.$$
Therefore,
$$\frac{2 \sqrt{3}+3}{6} t=\frac{1}{2},$$
$$(4 \sqrt{3}+6) t=6,$$
$$∴ \ t=\frac{6}{4 \sqrt{3}+6}=\frac{3}{2 \sqrt{3}+3}=2 \sqrt{3}-3≒0.4641.$$
$$(Answer) \quad Approximately \ 0.4641 \ m.$$
Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.38; pp.360-362.