Problem 10. Let's call a mountain a truncated right circular cone with the circumference of the lower base being 10, and the upper base - 9. The slope of the mountain is inclined at an angle of $60^{\circ}$ to the base plane. On the circumference of the lower base lies point $A$. A tourist starts climbing the slope from point $A$ to the nearest point on the upper base, and then continues his journey along the edge of the upper base, covering a distance of 3 (see figure). After that, he returns to point $A$ by the shortest route. What is the length of the return path?

![](https://cdn.mathpix.com/cropped/2024_05_06_c901cfc70ca56cbce9b9g-23.jpg?height=472&width=469&top_left_y=1403&top_left_x=1453)
Answer: $\frac{5 \sqrt{3}}{\pi}$

Solution. Let's denote (see the left figure) the vertex of the cone by $Q$, the center of the smaller base by $T$, the point on the upper base closest to $A-B$ as the starting point of the return path $-C$. Let $D$ be the last point of the return path on the circumference of the upper base (possibly $D$ coincides with $C$ or $B$). The angle $\angle D T C$ is denoted by $\alpha$.
![](https://cdn.mathpix.com/cropped/2024_05_06_c901cfc70ca56cbce9b9g-23.jpg?height=512&width=916&top_left_y=2156&top_left_x=524)

Consider the development of the lateral surface of the cone (see the right figure), and place the upper base at point $D$ so that it touches the development of the lateral surface. Keep the notation of the points, and in case of duplication, use indices (for example, points $C_{1}$ and $C_{2}$ on the right figure correspond to point $C$ on the left figure). Any arc of the upper base circle transitions to an arc of a circle with twice the radius, since $Q B=\frac{T B}{\cos 60^{\circ}}=2 T B$. Therefore, the angular measure will be halved. In particular, the upper base circle will transition to a semicircle, i.e., the lateral surface will transition to a part of the plane bounded by two concentric semicircles and a diameter passing through their ends, $Q D$ will be the diameter of the upper base.

$U_{\text {hol }} \angle D Q C_{1}=\frac{\angle D T C_{1}}{2}=\frac{\alpha}{2}$. As mentioned above, the angular measures of arcs $\widetilde{D C_{1}}$ and $\widetilde{D C}_{2}$ are in the ratio $2: 1$. This means that $\angle D Q C_{2}=\frac{\angle D T C_{1}}{2}=\frac{\alpha}{2}$. Then $\angle D Q C_{1}=\angle D Q C_{2}$, and therefore point $C_{1}$ lies on the line $Q C_{2}$. Considering that $Q D$ is a diameter, we get that $C_{1}$ is the foot of the perpendicular dropped from point $D$ to the line $Q C_{2}$. Then the length of the path going from $C$ to $D$ on the upper base (left figure) and from $D$ to $A$ on the lateral surface will be no less than the length of the broken line (right figure) $C_{1} D A_{1}$ and no less than the length of the perpendicular dropped from $A_{1}$ to the line $Q C_{2}$.

We will prove that there exists a path equal to the length of this perpendicular. For this, it is sufficient to show that the projection of point $A_{1}$ lies on the segment $Q C_{2}$, since in this case the perpendicular intersects the semicircle $B_{1} C_{2} B_{2}$. The point of intersection gives the point $D$ for the shortest path, and the perpendicular represents the shortest path on the development. Since the path along the edge of the upper base constitutes one-third of the length of the circumference of the upper base, then $\angle B_{1} Q C_{2}=60^{\circ}$. Then the ratio of $Q C_{2}$ to $Q A_{1}$ is equal to the ratio of the lengths of the circumferences of the lower and upper bases and is equal to $\frac{9}{10}$, which is greater than $\cos 60^{\circ}$. Therefore, the projection of $Q A_{1}$ on the line $Q C_{2}$ is less than the length of the segment $Q C_{2}$, which is what we needed.

The radius of the lower base $R=\frac{10}{2 \pi}$. Then $Q A=\frac{R}{\cos 60^{\circ}}=\frac{10}{\pi}$. From the right triangle (figure 2) the length of the perpendicular is

$$
Q A_{1} \cdot \sin \angle B_{1} Q C_{2}=Q A_{1} \cdot \sin 60^{\circ}=Q A \cdot \frac{\sqrt{3}}{2}=\frac{5 \sqrt{3}}{\pi}
$$