Solution. Let the sides $a$ and $c$ be given, $A F_{1}=x, C F_{2}=y$ (where $F_{1}$ is the midpoint of $B C$ and $F_{2}$ is the midpoint of $A B$).

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Write down the Pythagorean theorem for the right triangles $A S C, A S F_{2}, C S F_{1}$:

$$\begin{aligned} b^{2} & =\left(\frac{2}{3} x\right)^{2}+\left(\frac{2}{3} y\right)^{2} \\ \left(\frac{c}{2}\right)^{2} & =\left(\frac{2}{3} x\right)^{2}+\left(\frac{1}{3} y\right)^{2} \\ \left(\frac{a}{2}\right)^{2} & =\left(\frac{1}{3} x\right)^{2}+\left(\frac{2}{3} y\right)^{2} \end{aligned}$$

Solve the system of equations. Subtract the (3) equation from the 4 times the (2) equation:

$$c^{2}=\frac{16}{9} x^{2}+\frac{4}{9} y^{2}, \quad-\frac{a^{2}}{4}=-\frac{1}{9} x^{2}-\frac{4}{9} y^{2}, \quad \text { hence } \quad x^{2}=\frac{36 c^{2}-9 a^{2}}{60}$$

Similarly, if we multiply the (3) equation by 4 and subtract the (2) equation from it, we get $y^{2}=\frac{36 a^{2}-9 c^{2}}{60}$. Substitute the obtained values into (1):

$$b^{2}=\left(\frac{2}{3}\right)^{2} \frac{36 c^{2}-9 a^{2}}{60}+\left(\frac{2}{3}\right)^{2} \frac{36 a^{2}-9 c^{2}}{60}$$

After rearranging, we get $b=\sqrt{\frac{a^{2}+c^{2}}{5}}$.