\section*{Problem 6 - 331036}

Let \(K\) be a right circular cone. The base of \(K\) has the center \(M\) and the radius \(r\), the apex of \(K\) is \(S\), and the height \(h = MS\) of \(K\) is \(h = r \cdot \sqrt{3}\).

On the circumference of the base, there are two points \(A, B\) such that the angle \(\angle AMB\) is \(120^\circ\).

An ant, located at point \(A\), wants to reach a point \(C\) on the generatrix \(BS\) and choose this target point \(C\) such that it travels the shortest possible path without leaving the lateral surface.

Determine the length of such a path, expressed in terms of \(r\)!
}

The main idea is to unroll the cone. The question is then simply about the shortest distance from a point to a line segment. This shortest distance is known to be the perpendicular.
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1633.jpg?height=374&width=1004&top_left_y=1669&top_left_x=524)

What we are looking for is thus the height from the side \(\overline{A S}\) in the triangle \(A S B\) (denoted as \(d\) hereafter). To do this, we first calculate the length of the side \(\overline{A B}\) using the Law of Sines in \(\triangle A B M\). Since this triangle is isosceles, the two unknown angles are \(30^{\circ}\) (Base Angle Theorem). Now we have:

\[
\frac{|\overline{A B}|}{\sin \left(120^{\circ}\right)}=\frac{r}{\sin \left(30^{\circ}\right)} \Longleftrightarrow|\overline{A B}|=\frac{\sin \left(120^{\circ}\right) r}{\sin \left(30^{\circ}\right)}
\]

It is known that \(\sin \left(30^{\circ}\right)=\frac{1}{2}\) and

\[
\sin \left(120^{\circ}\right)=\sin \left(90^{\circ}+30^{\circ}\right)=\sin \left(90^{\circ}\right) \cos \left(30^{\circ}\right)+\cos \left(90^{\circ}\right) \sin \left(30^{\circ}\right)=\frac{1}{2} \sqrt{3}
\]

Thus, we have:

\[
|\overline{A B}|=\frac{\frac{1}{2} \sqrt{3} \cdot r}{\frac{1}{2}}=\sqrt{3} r
\]

Next, we calculate the sides \(\overline{A S}\) and \(\overline{B S}\), which are the slant heights of the body. By the Pythagorean theorem, we have:

\[
s^{2}=h^{2}+r^{2}=(\sqrt{3} r)^{2}+r^{2}=4 r^{2}
\]

and thus \(s=2 r\).

Let \(\alpha=\angle B S A\). By the Law of Cosines in triangle \(A S B\), we have:

\[
\begin{aligned}
(\sqrt{3} r)^{2} & =(2 r)^{2}+(2 r)^{2}-2 \cdot 2 r \cdot 2 r \cdot \cos (\alpha) \\
3 r^{2} & =4 r^{2}+4 r^{2}-8 r^{2} \cdot \cos (\alpha) \\
\alpha & =\arccos \left(\frac{5}{8}\right)
\end{aligned}
\]

Furthermore, \(\sin (\alpha)=\frac{d}{2 r}\). Substituting and solving, we get:

\[
d=\sin \left(\arccos \left(\frac{5}{8}\right)\right) \cdot 2 r
\]

This can be simplified further. By the Pythagorean identity \(\sin ^{2} x+\cos ^{2} x=1 \Longleftrightarrow \sin ^{2} x=1-\cos ^{2} x\),

we get for \(\sin (\arccos x)\):

\[
\sin (\arccos x)=\sqrt{1-(\cos (\arccos (x)))^{2}}=\sqrt{1-x^{2}}
\]

Thus:

\[
d=\sin \left(\arccos \left(\frac{5}{8}\right)\right) \cdot 2 r=\sqrt{1-\left(\frac{5}{8}\right)^{2}} \cdot 2 r=\sqrt{\frac{39}{64}} \cdot 2 r=\frac{\sqrt{39}}{4} r
\]