A plane passes at a distance $a$ from the center of a unit sphere. Find the edge of the cube, one face of which lies in this plane, and the vertices of the opposite face are on the sphere.
${ }_{1} C_{1} D_{1} \mathrm{c}$ edge $x$ lies in the given plane, and the vertices $A_{1}, B_{1}, C_{1}$ and $D_{1}$ are on the given sphere with center $O$

(Fig. 1-4)

Fig. 1

. Draw a line through point $O$ perpendicular to the given plane. Let this line intersect the planes of the faces $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$

at points $P$ Fig. 2

and $P_{1}$

respectively. Since points $A_{1}, B_{1}, C_{1}$ and $D_{1}$ are equidistant from point $O$, point $P_{1}$ is the center of the square $A_{1} B_{1} C_{1} D_{1}$. Then $P$ is the center of the square $A B C D$.

Fig. 3
![](https://cdn.mathpix.com/cropped/2024_05_06_35476b28e4766e84fc81g-39.jpg?height=1820&width=1090&top_left_y=906&top_left_x=198)

Fig. 5

![](https://cdn.mathpix.com/cropped/2024_05_06_35476b28e4766e84fc81g-39.jpg?height=780&width=783&top_left_y=902&top_left_x=1255)

Fig. 6

![](https://cdn.mathpix.com/cropped/2024_05_06_35476b28e4766e84fc81g-39.jpg?height=930&width=776&top_left_y=1776&top_left_x=1267)

passing

through the line $OP$ and vertex $A_{1}$. We get a circle of unit radius with center at point $O$ and

rectangle

$A A_{1} C_{1} C$ one side $A C$ of which lies on the chord of the circle,

![](https://cdn.mathpix.com/cropped/2024_05_06_35476b28e4766e84fc81g-40.jpg?height=1058&width=1086&top_left_y=80&top_left_x=200)

Fig. 4

![](https://cdn.mathpix.com/cropped/2024_05_06_35476b28e4766e84fc81g-40.jpg?height=1012&width=786&top_left_y=86&top_left_x=1253)

Fig. 8

located

at a

distance $a$ from the center of the circle, and the vertices $A_{1}$ and $C_{1}$ - on the circle, with $A A_{1}=x$,

and $A_{1} C_{1}=x \sqrt{ } 2$

Let points $O$ and $P_{1}$ lie on opposite sides of the line $A C$ (Fig. 5). Then in the right triangle $O P_{1} A_{1}$ it is known that

$O A_{1}=1, A_{1} P_{1}=\frac{1}{2} A_{1} C_{1}=\begin{aligned} & x \sqrt{ } 2 \\ & 2\end{aligned}$,

$O P_{1}=O P+P P_{1}=a+x$

By the Pythagorean theorem $O A_{1}{ }^{2}=$

$=A_{1} P_{1}{ }^{2}+O P_{1}{ }^{2}$, or

$1=\frac{x^{2}}{2}+(a+x)^{2}$.

From this equation, we find that

$$
x=\begin{aligned}
& \sqrt{ } 6-2 a^{2}-2 a \\
& 3
\end{aligned}
$$

(the second root is obviously negative).

Let points $O$ and $P_{1}$ lie on the same side of the line $A C$ (Figs. 6-8). Then in the right triangle $O P_{1} A_{1}$ it is known that

$O A_{1}=1, A_{1} P_{1}=1 A_{1} C_{1}=$
$O P_{1}=\left|P P_{1}-O P\right|=|x-a|$.

By the Pythagorean theorem $O A^{2}=$

$=A_{1} P_{1}{ }^{2}+O P_{1}{ }^{2}$, or

$1=\frac{x^{2}}{2}+(x-a)^{2}$

From this equation, we find that

$x=$

$\pm \sqrt{ } 6-2 a^{2}+2 a$

3

Send a comment