Example 80. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$, with an edge length of $a$. A sphere $\Omega$ is drawn through the vertices $A$ and $C$ and through the midpoints of the edges $B_{1} C_{1}$ and $C_{1} D_{1}$. Let's find the radius of this sphere.

Construction of the image. Let the figure $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 89) be the image of the given cube, and let points $F_{1}$ and $E_{1}$ be the midpoints of its edges $B_{1} C_{1}$ and $C_{1} D_{1}$, respectively. This image is complete and metrically defined (verify this independently). As in the previous example, constructing the image of the given sphere is quite difficult, since the length of the sphere's radius is unknown. However, the image of the sphere is not so necessary: if the center of the sphere and any point on its surface (we already have four such points shown) are shown in the image, then perhaps it will be possible to calculate the length of the radius.

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-141.jpg?height=646&width=671&top_left_y=1713&top_left_x=358)

Fig. 89
The solution is as follows. First of all, let's find the point $M$ - the center of the sphere $\Omega$. Since $A \in \Omega$ and $C \in \Omega$, the point $M$ belongs to the set of points equidistant from points $A$ and $C$. It is not difficult to guess that this set of points is the diagonal plane $B B_{1} D_{1} D$. Similarly, the point $M$ belongs to the diagonal plane $A A_{1} C_{1} C$. Thus, the point $M$ belongs to $\left(O O_{1}\right)=\left(A C C_{1}\right) \cap$ $\left(B D D_{1}\right)$. Finally, the point $M$ belongs to the plane $\alpha$, which intersects the segment $C E_{1}$ at point $L$ - the midpoint of the segment $C E_{1}$, and $\alpha \perp [C E_{1}]$.

Since the plane $\alpha$ has a common point $L$ with the plane $D C C_{1}$, these planes intersect along a line $l$ passing through point $L$, and since $[C E_{1}] \perp \alpha$, then $[C E_{1}] \perp l$.

Construct the point $E$ - the midpoint of the segment $C D$ and construct the segment $E E_{1}$. Since the image on which we are making additional constructions is metrically defined, we cannot arbitrarily draw a line $l$ through point $L$ and say that let $l \perp [C E_{1}]$. Nevertheless, the line $l$ will intersect the segment $E E_{1}$. Let $l \cap [E E_{1}]=K$. Then $\triangle K L E_{1}$ will be a right triangle and similar to triangle $C E E_{1}$. From the similarity of these triangles, it follows that $|K E_{1}|:|C E_{1}|=|L E_{1}|:|E E_{1}|$. And since $|E E_{1}|=a$, $|C E|=\frac{a}{2}$, then

$$
|C E_{1}|=\frac{a \sqrt{5}}{2}, \text{ and } |L E_{1}|=\frac{a \sqrt{5}}{4}
$$

Thus, we find that if $l \perp [C E_{1}]$, then $|K E_{1}|=\frac{5}{8} a$. Using this equality, we will construct the point $K$.

Next, construct $[O E]$ and draw a line $k \parallel [O E]$ through point $K$. It is easy to see that $k \subset (O E K)$ and that $k \cap (O O_{1})$. Let $M=k \cap (O O_{1})$. Then, since $(M K) \parallel [O E]$ and $[O E] \perp (D C C_{1})$, then $(M K) \perp (D C C_{1})$.

Thus, $[C E_{1}] \perp (L K)$ and $[C E_{1}] \perp (M K)$. Then $[C E_{1}] \perp (L K M)$. In other words, $(L K M)=\alpha$, and therefore, $M \in \alpha$, and thus $M$ is the center of $\Omega$.

Let's calculate $|M E_{1}|$ - the radius of the sphere $\Omega$.

From the right triangle $\triangle M K E_{1}$, we get:

$$
|M E_{1}|=\sqrt{|M K|^{2}+|K E_{1}|^{2}}=\frac{a \sqrt{41}}{8}
$$

(We could, of course, find $|M C|$ - the radius of the sphere $\Omega$ from the right triangle $M C O$, where $|O M|=\frac{3}{8} a$, $|O C|=\frac{a \sqrt{2}}{2}.$)

Investigation. By the meaning of the problem, $|M E_{1}| > |M O_{1}| = |K E_{1}| = \frac{5}{8} a$. But the inequality $\frac{a \sqrt{41}}{8} > \frac{5}{8} a$ is true. Thus, the radius of the sphere $\Omega$ is $\frac{a \sqrt{41}}{8}$.