## Task B-4.5.

The length of the height of a right circular cone is equal to the diameter of the base of the cone. What is the ratio of the radius of the sphere inscribed in the cone to the radius of the sphere circumscribed about the cone?
## Solution.

We will consider the axial section of the cone, i.e., the isosceles triangle $ABC$ whose base is the diameter of the cone's base, and the legs are the generators of the cone. Then, the radius of the sphere inscribed in the cone is equal to the radius of the circle inscribed in this triangle, and the radius of the sphere circumscribed around the cone is equal to the radius of the circle circumscribed around this triangle. (A sketch of the axial section with the inscribed and circumscribed circles is sufficient)

Let the radius of the inscribed circle be denoted as $r_{u}$, the radius of the circumscribed circle as $r_{o}$, and the radius of the base as $r$.
![](https://cdn.mathpix.com/cropped/2024_05_30_d3836afd55956f7fff9cg-25.jpg?height=692&width=1266&top_left_y=861&top_left_x=270)

The length of the generator of the cone is $|AC| = \sqrt{r^2 + (2r)^2} = r\sqrt{5}$.

The area of triangle $ABC$ is $P = \frac{1}{2} \cdot 2r \cdot 2r = 2r^2$,

and its semiperimeter is $s = \frac{1}{2}(2r + 2 \cdot r\sqrt{5}) = r(1 + \sqrt{5})$.

The radius of the circle inscribed in triangle $ABC$ is calculated from $P = r_{u} \cdot s$:

$$
r_{u} = \frac{P}{s} = \frac{2r^2}{r(1 + \sqrt{5})} = \frac{2r}{1 + \sqrt{5}}
$$

The radius of the circle circumscribed around triangle $ABC$ is calculated using the formula $P = \frac{abc}{4r_{o}}$:

$$
r_{o} = \frac{abc}{4P} = \frac{2r \cdot r\sqrt{5} \cdot r\sqrt{5}}{4 \cdot 2r^2} = \frac{10r^3}{8r^2} = \frac{5}{4}r
$$

Finally, the ratio of the radii of the sphere inscribed in the cone and the sphere circumscribed around the cone is

$$
\frac{r_{u}}{r_{o}} = \frac{\frac{2r}{1 + \sqrt{5}}}{\frac{5}{4}r} = \frac{8}{5(1 + \sqrt{5})} = \frac{2(\sqrt{5} - 1)}{5}
$$

Note: The radii of the inscribed and circumscribed spheres can also be obtained using similarity.

Observe in the first sketch that triangle $ANC$ is similar to triangle $CD_1S_u$. Therefore, $\frac{r}{r\sqrt{5}} = \frac{r_{u}}{2r - r_{u}}$, from which it follows that $r_{u} = \frac{2r}{1 + \sqrt{5}}$. (2 points)

From the right triangle $ANS_o$ in the second sketch, it follows that $r_{o}^2 = r^2 + (2r - r_{o})^2$, from which it follows that $r_{o} = \frac{5}{4}r$. (2 points)

The remaining 2 points are for the sketch and the calculation of the desired ratio as in the previously detailed solution.