The cross-sections of a cone of revolution are equilateral triangles. Find 3 points on the surface of the cone such that together with the center of the base, they form the vertices of a regular tetrahedron. What is the edge length of the tetrahedron if the slant height of the cone is one unit?
Let's denote the apex of the cone as \( M \), the center of the base circle as \( O \), the other vertices of the tetrahedron as \( A \), \( B \), and \( C \), the centroid of the \( ABC \) face as \( K \), and the length of the tetrahedron's edge as \( a \). Since the vertices \( A \), \( B \), and \( C \) are at a distance \( a \) from \( O \), these vertices lie on the sphere centered at \( O \) with radius \( a \). If we intersect the cone and the sphere with any plane passing through the axis \( OM \), the intersection will be an equilateral triangle and a circle with radius \( a \), respectively. This circle intersects the triangle's side passing through \( M \) at two points if \( a \leq \frac{1}{2} \) (and naturally \( a \geq \frac{\sqrt{3}}{4} \)), at one point if \( \frac{1}{2} < a \leq \frac{\sqrt{3}}{2} \), and does not intersect if \( a > \frac{\sqrt{3}}{2} \). Therefore, if \( \frac{\sqrt{3}}{4} \leq a \leq \frac{1}{2} \), because then the diameter of the intersection circle would be smaller than \( a \), making it impossible to place three points, each at a distance \( a \) from each other.

![](https://cdn.mathpix.com/cropped/2024_05_02_0dc0ceaabe04dc01cf6bg-1.jpg?height=578&width=1218&top_left_y=627&top_left_x=431)

# Figure 1

According to this, \( \frac{\sqrt{3}}{4} \leq a \leq \frac{1}{2} \), and the sphere with radius \( a \) intersects the cone's lateral surface in two circles, whose planes are parallel to the base. Since the problem does not ask for all solutions, it is practical to determine \( a \) in the simplest arrangement, where the vertices \( A \), \( B \), and \( C \) lie on the same circle. In this case, this circle is the circumcircle of the equilateral triangle \( ABC \) with side length \( a \), so its radius is \( \frac{a}{\sqrt{3}} \) (Figure 1).

The centroid \( K \) lies on the axis \( OM \), and in the right triangle \( AOK \),

\[
K O^{2}=A O^{2}-A K^{2}=a^{2}-\frac{a^{2}}{3}=\frac{2}{3} a^{2}
\]

Furthermore, since \( AK \) is parallel to the base, the ratio \( \frac{KM}{AK} \) is \( \sqrt{3} \), so \( MK = \sqrt{3} AK = a \). Thus, the segments \( MK \) and \( KO \) of the \( \frac{\sqrt{3}}{2} \) long segment \( MO \) are expressed in terms of \( a \), from which \( a \) can be determined:

\[
\begin{gathered}
a \frac{\sqrt{2}}{\sqrt{3}} + a = \frac{\sqrt{3}}{2} \\
a = \frac{3(\sqrt{3} - \sqrt{2})}{2} = 0.4768
\end{gathered}
\]

Remarks. 1. We also examine the arrangements where two of the vertices \( A \), \( B \), and \( C \) lie on one circle, and one on the other.

I. First, assume that the circle containing the two vertices has a larger radius, and let the vertices \( B \) and \( C \) be on this circle. Let the midpoint of the edge \( BC \) be \( F \), and the point where the circle through \( B \) and \( C \) intersects the generator \( MA \) be \( D \) ( \( F \) is in the plane \( AMO \), Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_0dc0ceaabe04dc01cf6bg-1.jpg?height=529&width=655&top_left_y=2186&top_left_x=716)

## Figure 2

The triangle \( AFO \) is isosceles, with base \( a \) and legs of length \( \frac{a \sqrt{3}}{2} \) (in-plane height). Since \( DF \) is parallel to the base of the cone, \( D \) sees the segment \( AF \) at an angle of \( 60^\circ \), so \( D \) lies on the \( 60^\circ \) circle above \( AF \) and on the side of the line \( AF \) not containing \( O \). The center \( E \) of this circle is at a distance \( \frac{a}{2} \) from the points \( A \) and \( F \), from which the distance \( EO \) can be determined. Using the cosine rule in the triangle \( AEO \),

\[
\begin{gathered}
E O^{2}=a^{2}+\frac{a^{2}}{4}-a^{2} \cos \angle EAO = a^{2} \left\{ \frac{5}{4} - \cos \left( \angle F AO + 30^\circ \right) \right\} = \\
= a^{2} \left\{ \frac{5}{4} - \cos \angle F AO \cdot \cos 30^\circ + \sin \angle F AO \cdot \sin 30^\circ \right\} = a^{2} \left\{ \frac{5}{4} - \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{1}{2} \right\} = \frac{9 + 2 \sqrt{6}}{12} a^{2}.
\end{gathered}
\]

In the isosceles triangle \( ADO \), the legs are of length \( a \), and the height \( m \) to the base \( AD \) is

\[
m = a \cos \angle AOE = a \frac{A O^{2} + E O^{2} - A E^{2}}{2 A O \cdot E O} = a \frac{\frac{3}{4} + \lambda^{2}}{2 \lambda},
\]

where

\[
\lambda = \sqrt{\frac{9 + 2 \sqrt{6}}{12}}.
\]

In an equilateral triangle with side length 1, the distance from the midpoint of the base \( O \) to the other two sides is \( m = \frac{\sqrt{3}}{4} \), so

\[
\begin{aligned}
a = \frac{2 \sqrt{3} \lambda}{3 + 4 \lambda^{2}} = & \sqrt{\frac{9 + 2 \sqrt{6}}{\left[3 + \frac{1}{3}(9 + 2 \sqrt{6})\right]^{2}}} = \sqrt{\frac{81 + 18 \sqrt{6}}{(18 + 2 \sqrt{6})^{2}}} = \\
& = \frac{\sqrt{567 + 12 \sqrt{6}}}{50} = 0.4884.
\end{aligned}
\]

II. If \( A \) is on the larger circle and \( B \) and \( C \) are on the smaller one, then the angle \( \angle ADF \) is \( 120^\circ \), \( E \) is on the side of the line \( AD \) containing \( O \) (Figure 3), and similar calculations yield

\[
\begin{gathered}
E O^{2} = \frac{9 - 2 \sqrt{6}}{12} a^{2} \\
a = \frac{\sqrt{567 - 12 \sqrt{6}}}{50} = 0.4637.
\end{gathered}
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_0dc0ceaabe04dc01cf6bg-2.jpg?height=485&width=550&top_left_y=1953&top_left_x=771)

Figure 3

The values of \( a \) obtained in the three arrangements are very close to each other. For better visualization, Figures 2 and 3 are not to scale.

The approximate value of the edge length of the skewed tetrahedron was given by Katalin Párkány (Budapest, Berzsenyi D. Gymnasium, 2nd grade) +