14.21. Find the equation of the circumscribed sphere of the tetrahedron $A_{1} A_{2} A_{3} A_{4}$ in barycentric coordinates associated with it.
14.21. Let $X$ be an arbitrary point, $O$ the center of the circumscribed sphere of a given tetrahedron, $\mathbf{e}_{i}=\overrightarrow{O A}_{i}$, and $\mathbf{a}=\overrightarrow{X O}$. If the point $\delta$ has barycentric coordinates $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$, then $\sum x_{i} \left( \mathbf{e}_{i} + \mathbf{a} \right) = \sum x_{i} \overrightarrow{X A}_{i} = 0$, since $X$ is the center of mass of the points $A_{1}, \ldots, A_{4}$ with masses $x_{1}, \ldots, x_{4}$. Therefore, $\left(\sum x_{i}\right) \mathbf{a} = -\sum x_{i} \mathbf{e}_{i}$. The point $X$ lies on the circumscribed sphere of the tetrahedron if and only if $|\mathbf{a}| = |X O| = R$, where $R$ is the radius of this sphere. Thus, the circumscribed sphere of the tetrahedron is given in barycentric coordinates by the equation $R^{2}\left(\sum x_{i}\right)^{2} = \left(\sum x_{i} \mathbf{e}_{i}\right)^{2}$, i.e., $R^{2} \sum x_{i}^{2} + 2 R^{2} \sum_{i<j} x_{i} x_{j} = R^{2} \sum x_{i}^{2} + 2 \sum_{i<j} x_{i} x_{j} \left(\mathbf{e}_{i}, \mathbf{e}_{j}\right)$, since $|\mathbf{e}_{i}| = R$. This equation can be rewritten as $\sum_{i<j} x_{i} x_{j} \left(R^{2} - \left(\mathbf{e}_{i}, \mathbf{e}_{j}\right)\right) = 0$. Note now that $2\left(R^{2} - \left(\mathbf{e}_{i}, \mathbf{e}_{j}\right)\right) = a_{i j}^{2}$, where $a_{i j}$ is the length of the edge $A_{i} A_{j}$. Indeed, $a_{i j}^{2} = \left|\mathbf{e}_{i} - \mathbf{e}_{j}\right|^{2} = \left|\mathbf{e}_{i}\right|^{2} + \left|\mathbf{e}_{j}\right|^{2} - 2\left(\mathbf{e}_{i}, \mathbf{e}_{j}\right) = 2\left(R^{2} - \left(\mathbf{e}_{i}, \mathbf{e}_{j}\right)\right)$. In the end, we obtain that the circumscribed sphere of the tetrahedron $A_{1} A_{2} A_{3} A_{4}$ is given in barycentric coordinates by the equation $\sum_{i<j} x_{i} x_{j} a_{i j} = 0$, where $a_{i j}$ is the length of the edge $A_{i} A_{j}$.