Two rods of equal cross-section but different densities and different thermal expansion coefficients are welded together so that the center of gravity of the system remains unchanged despite the heat-induced expansion. How do the lengths of the rods relate to each other?
Let the length of one rod at $0^{\circ}$ be $l_{1}$, its density (mass per unit length) $d_{1}$, and its linear expansion coefficient $\alpha_{1}$; for the other rod, these data are $l_{2}, d_{2}, \alpha_{2}$. Let the cross-sectional area of each rod be $q$. The center of gravity $S_{1}$, and $S_{2}$ of each rod is $\frac{1}{2} l_{1}$, and $\frac{1}{2} l_{2}$ from the welding point $O$ of the two rods, respectively. Let the distance of the common center of gravity $S$ from $O$ be $x$. The moment of the fictitious total weight at the $S$ point with respect to $O$ is equal to the algebraic sum of the moments of the weights acting at the $S_{1}$ and $S_{2}$ points, that is, if we consider $S_{2}$ to be positive, then

$$
q\left(l_{1} d_{1}+l_{2} d_{2}\right) x=\left(l_{2} d_{2} \frac{l_{2}}{2}-l_{1} d_{1} \frac{l_{1}}{2}\right) q
$$

and

$$
x=\frac{l_{2}^{2} d_{2}-l_{1}^{2} d_{1}}{2\left(l_{1} d_{1}+l_{2} d_{2}\right)}
$$

(If $l_{2}^{2} d_{2}-l_{1}^{2} d_{1}>0$, then $S$ is between $O$ and $S_{2}$, and if $l_{2}^{2} d_{2}-l_{1}^{2} d_{1}<0$, then $S$ is between $O$ and $S_{1}$).

If the cross-sectional area $q$ of the rod is small enough that we can neglect its extension, then the denominator of $x$ is constant, because $\left(l_{1} d_{1}+l_{2} d_{2}\right) q$ represents the total mass. Therefore, we only need to examine the condition for the numerator to be independent of the temperature $t$.

At temperature $t$, the lengths of the rods are $l_{1}\left(1+\alpha_{1} t\right)$ and $l_{2}\left(1+\alpha_{2} t\right)$, and their densities are $\frac{d_{1}}{1+\alpha_{1} t}$ and $\frac{d_{2}}{1+\alpha_{2} t}$. Thus, the value of the numerator is independent of $t$ if

$$
l_{2}^{2} d_{2}-l_{1}^{2} d_{1}=l_{2}^{2}\left(1+\alpha_{2} t\right)^{2} \frac{d_{2}}{1+\alpha_{2} t}-l_{1}^{2}\left(1+\alpha_{1} t\right)^{2} \frac{d_{1}}{1+\alpha_{1} t}
$$

After performing the indicated operations and simplifying, we get:

$$
\left(l_{2}^{2} d_{2} \alpha_{2}-l_{1}^{2} d_{1} \alpha_{1}\right) t=0
$$

This condition must hold for any value of $t$, that is,

$$
l_{2}^{2} d_{2} \alpha_{2}-l_{1}^{2} d_{1} \alpha_{1}=0, \quad \text { or } \quad \frac{l_{1}}{l_{2}}=\sqrt{\frac{d_{2} \alpha_{2}}{d_{1} \alpha_{1}}}
$$

Béla Schreiber (Izr. rg. VIII. o. Bp.).