Three rays originating from a single point form pairwise angles of $60^{\circ}$. A unit radius sphere is tangent to all three rays. Calculate the distance from the center of the sphere to the starting point of the rays.
I. solution. Let the rays be $e_{1}, e_{2}, e_{3}$, their common point $C$, the center of the tangent sphere $O$, and the points of tangency $E_{1}, E_{2}, E_{3}$, respectively, so $O E_{1} \perp C E_{1}, O E_{2} \perp C E_{2}, O E_{3} \perp C E_{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_6b0e04896fa66be11e1ag-1.jpg?height=569&width=550&top_left_y=263&top_left_x=771)

Figure 1

Tangents drawn from an external point to a sphere are equal: $C E_{1}=C E_{2}=C E_{3}$, therefore the triangles $C E_{1} E_{2}, C E_{2} E_{3}, C E_{3} E_{1}$ are congruent equilateral triangles, and the pyramid $C E_{1} E_{2} E_{3}$ is a regular tetrahedron. $O$ and $C$ are equidistant from $E_{1}$ and $E_{2}$, so they lie in the perpendicular bisector plane of the segment $E_{1} E_{2}$. Therefore, the line $C O$ is perpendicular to $E_{1} E_{2}$, and similarly to $E_{1} E_{3}$, so it is perpendicular to the plane $E_{1} E_{2} E_{3}$. Thus, $O$ lies on the altitude of the tetrahedron from $C$. Denoting the foot of this altitude, the center of the equilateral triangle $E_{1} E_{2} E_{3}$, by $K$ and the segment $C E_{1}$ by $c$, $K E_{1}$ is equal to the $\frac{2}{3}$ of the height of the equilateral triangle with side $c$, which is $c \sqrt{3 / 2}$, so its length is $c \sqrt{3}$. From the similarity of the right triangles $C O E_{1}$ and $C E_{1} K$,

$$
C O: O E_{1}=C E_{1}: E_{1} K=c: c / \sqrt{3}
$$

and from this, since $O E_{1}=1$, the sought distance is $C O=\sqrt{3}$.

Ghihor Zoltán (Kaposvár, Táncsics M. g. III. o. t.)

Remark. With the same reasoning, we can also obtain the distance between the center of the tangent sphere and the common point of the rays, even if the angle between any two rays is $\alpha$, and $\alpha<120^{\circ}$. In this case,

$$
C O=\frac{\sqrt{3}}{2 \sin \frac{\alpha}{2}}
$$

Zalán Péter (Aszód, Petőfi S. g. III. o. t.)

II. solution. Notice that the three face diagonals starting from a vertex of a unit cube and the circumscribed unit sphere around the opposite vertex form a figure congruent to the given one.

![](https://cdn.mathpix.com/cropped/2024_05_02_6b0e04896fa66be11e1ag-1.jpg?height=485&width=480&top_left_y=1869&top_left_x=798)

Figure 2

Indeed, the angle between any two face diagonals of a cube that share a common endpoint is $60^{\circ}$, because the endpoints of the three edges starting from any vertex are at face-diagonal distance from each other, forming an equilateral triangle. Furthermore, the sphere touches the three cube faces that do not pass through its center, and the points of tangency are the vertices of the cube adjacent to the center. These vertices are precisely the non-common endpoints of the mentioned face diagonals, so the sphere also touches these diagonals.

Thus, the sought distance is equal to the space diagonal of the cube, which has a length of $\sqrt{3}$.