Question

Let \(A B C D\) be a square with center \(O .\) Let \(\ell\) be the line through \(O\) perpendicular to the plane of \(A B C D .\) Prove that there is a point \(E\) on \(\ell\) such that the four triangles \(\triangle E A B, \triangle E B C, \triangle E C D,\) and \(\triangle E D A\) are all equilateral. Calculate the length \(O D\) in terms of \(A B .\) How can you use this construction to construct a regular octohedron?

Short answer

Point \(E\) lies \(\frac{\sqrt{3}}{2}s\) above or below \(O\). \(OD = \frac{s}{\sqrt{2}}\). Use \(E\) to form two pyramids for a regular octahedron.

Step-by-step solution

Understanding the square and its properties

Let the square \(ABCD\) be centered at \(O\), with each side having length \(s\). Since \(\ell\) is perpendicular to the plane of the square, it is a vertical line going through \(O\). The line \(\ell\) can be imagined as standing up from the center of \(ABCD\).

Identifying the equilateral triangle condition

We want to find a point \(E\) on \(\ell\) such that the triangles \(\triangle EAB, \triangle EBC, \triangle ECD,\) and \(\triangle EDA\) are equilateral. This requires each triangle to have all sides equal, specifically each equal to \(AB = s\).

Calculate the height \(OE\) required for equilaterality

In triangle \(\triangle EAB\), since \(EA = EB = AB = s\), \(E\) must be directly above \(O\) at a height \(OE\). From geometry of equilateral triangles, the altitude \(h\) from vertex \(E\) to side \(AB\) is given by \(h = \frac{\sqrt{3}}{2}s\). Therefore, point \(E\) should be at a distance \(\frac{\sqrt{3}}{2}s\) from \(O\) on line \(\ell\).

Calculation of OD

Since \(O\) is the center of square \(ABCD\), \(\triangle OAD\) is right-angled at \(O\), and \(OA = OD = \frac{AB}{\sqrt{2}}\) because \(O\) is the intersection of the diagonals, each one being \(s\sqrt{2}\) (diagonal of a square formula). Thus, \(OD = \frac{s}{\sqrt{2}}\).

Constructing a regular octahedron

Using point \(E\) as the apex directly above (or below) the center\(O\), form two pyramids with square bases \(ABCD\). Thus, the square becomes the joint base of two congruent square pyramids. This creates a regular octahedron, with equilateral triangular faces, incorporating \(\triangle EAB, \triangle EBC, \triangle ECD,\) and \(\triangle EDA\).

Key concepts

Equilateral Triangles

An equilateral triangle is a triangle where all three sides are of equal length. Not only are the sides equal, but all interior angles of an equilateral triangle are the same too, each being 60 degrees.
When constructing equilateral triangles on the vertices of a square, as seen in the exercise, we aim for the sides of each triangle to equal the side length of the square. Suppose we have a square named ABCD with side length 's'. To form triangles like \( \triangle EAB \), with point E positioned over the center, the challenge is ensuring that \( EA = EB = AB = s \).
This requires setting the height \( OE \), from the centroid of the square to point E, such that it maintains these equal lengths and complements the geometry with the square's sides.

Square Geometry

Square geometry revolves around understanding the properties of squares. A square is a four-sided polygon, with equal sides, called quadrilateral. All angles in a square are right angles, making each angle 90 degrees. In the exercise, square ABCD is central, with its defining property that its diagonals are congruent and bisect each other at right angles. This results in the center point, O, being simultaneously a midpoint of every diagonal. When constructing a regular octahedron, we leverage the square as the base with line \( \ell \) perpendicular to its plane, aimed at forming equilateral triangles with its vertices and an apex point on \( \ell \) such as point E. Understanding how to manipulate the distances within these structural elements helps in shaping the desired geometry.

Altitude in Triangles

The altitude in a triangle is a crucial concept, especially in equilateral triangles. It is a straight line drawn from a vertex perpendicular to the opposite side, also known as the base. In an equilateral triangle, all altitudes have a particular relationship with the sides: they are each \( \frac{\sqrt{3}}{2} \) times the side length. In our exercise, the altitude from point E to any side such as AB is instrumental in keeping each triangle equilateral. This relationship determines both the height at which E exists above the square and how these triangles consistently maintain equal side lengths relative to the square's dimension. The set altitude effectively turns theoretical geometric construction into spatial reality.

3D Shapes in Geometry

Three-dimensional shapes, or 3D shapes, have depth, width, and height, offering a more complex spatial experience compared to 2D. In this problem, a regular octahedron is the 3D shape in focus. The regular octahedron comprises eight faces, each an equilateral triangle. Here, constructing two identical square pyramids with a shared base is key. These pyramids are bounded by the square ABCD and the apex point E, either above or below the base. Understanding how 3D shapes result from the interaction of 2D shapes, like squares and triangles, is invaluable in geometry. It deepens comprehension of spatial relations and geometric transformations necessary to visualize and execute structures like the octahedron.