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+### Statement
+
+$7.4.37.$ [Insert the problem statement]
+
+### Solution
+
+![For problem $7.4.37$ |630x670, 31%](../../img/7.4.37/Savchenko.png)
+
+Since the center of mass $G$ of the system remains stationary, the total force $\vec F=\vec F_1+\vec F_2$ on a charge $q$ must be in the same direction as the line joining $G$ and the charge. As the masses of the three charges are the same, $G$ is the geometric centroid of the triangle; that is, $G$ is on the median $AM$ such that $AG=2\cdot MG$. The magnitudes of $\vec F_1$ and $\vec F_2$ are
+
+\[F_1=\frac{kQq}{(AB^2}\qquad\mbox{and}\qquad F_2=\frac{kq^2}{(2\cdot AB\sin(\alpha/2))^2}.\]
+
+Let $\angle GBM=\angle BDE=\phi$ (shown in pink) and $\angle GBA=\angle BDC=\theta$ (shown in yellow). Using the law of sines in $\triangle BDE$ and $\triangle BDC$, we have
+
+\[\frac{F_1}{\sin\phi}=\frac{BD}{\sin(180^\circ-(\theta+\phi))}=\frac{F_2}{\sin\theta}.\]
+
+Consequently,
+
+\[\frac{\sin\theta}{\sin\phi}=\frac{F_2}{F_1}=\frac{q}{Q}\cdot\frac{1}{(2\sin(\alpha/2))^2}.\]
+
+Using the law of sines in $\triangle GBM$ and $\triangle GBA$, we have
+
+\[\frac{MG}{\sin\phi}=\frac{GB}{\sin90^\circ}\qquad\mbox{and}\qquad\frac{AG}{\sin\theta}=\frac{GB}{\sin(\alpha/2)}.\]
+
+As a result,
+
+\[\frac{\sin\theta}{\sin\phi}=\frac{AG}{MG}\cdot\sin(\alpha/2)=2\sin(\alpha/2).\]
+
+Equating the two expressions we have for $\sin\theta/\sin\phi$, we obtain
+
+\[\frac{q}{Q}\cdot\frac{1}{(2\sin(\alpha/2))^2}=2\sin(\alpha/2)\qquad\Rightarrow\qquad\frac{Q}{q}=\frac{1}{(2\sin(\alpha/2))^3}.\]
+
+#### Answer
+
+[Insert a concise answer or boxed result]