Question

Let \(\triangle A B C\) be an equilateral triangle with center \(O .\) Let \(\ell\) be the line through \(O\) perpendicular to the plane of \(\triangle A B C .\) Prove that there is a point \(D\) on \(\ell\) such that the triangles \(\triangle D A B, \triangle D B C, \triangle D C A,\) and \(\triangle A B C\) are all congruent. Calculate the length \(O D\) in terms of \(A B\)

Short answer

The length \( OD = \frac{s\sqrt{6}}{3} \), where \( s \) is the length of side \( AB \).

Step-by-step solution

Understanding the Problem

We are given an equilateral triangle \( \triangle ABC \) with center \( O \). Line \( \ell \) is perpendicular to the plane of the triangle at \( O \). We need to find a point \( D \) on \( \ell \) such that \( \triangle DAB, \triangle DBC, \triangle DCA, \) and \( \triangle ABC \) are congruent. Finally, we calculate the distance \( OD \) in terms of side \( AB \).

Find the Center of the Equilateral Triangle

The center \( O \) of equilateral triangle \( \triangle ABC \) is equidistant from all the vertices. Since all the sides are equal, the centroid and the circumcenter coincide at point \( O \). Given \( AB = BC = CA = s \), we know the distance from \( O \) to each vertex is \( \frac{s}{\sqrt{3}} \).

Define Congruent Triangles in 3D Space

For \( \triangle DAB, \triangle DBC, \triangle DCA \) to be congruent to \( \triangle ABC \), \( D \) must be at an equal perpendicular distance from all vertices of \( \triangle ABC \) due to symmetry. Thus, \( D \) should lie on line \( \ell \) above \( O \), such that the altitude from \( D \) to the plane of \( \triangle ABC \) is equal to the circum-radius of \( \triangle ABC \).

Calculate OD using Symmetry

The circum-radius of an equilateral triangle \( \triangle ABC\) with side \( s \) is \( \frac{s}{\sqrt{3}} \). For triangles \( \triangle DAB, \triangle DBC, \triangle DCA \) to be congruent to \( \triangle ABC \), \( D \) must form a regular tetrahedron with the vertices of \( \triangle ABC \). Hence, the height \( OD \) from \( O \) to \( D \) is given by: \[ OD = \sqrt{R^2 - \left(\frac{s}{2}\right)^2} = \sqrt{\left(\frac{s}{\sqrt{3}}\right)^2 - \left(\frac{s}{2}\right)^2} = \frac{s\sqrt{2}}{2\sqrt{3}} = \frac{s\sqrt{6}}{3} \].

Answer Verification

Re-check to ensure \( D \) satisfies the requirement that \( \triangle DAB, \triangle DBC, \triangle DCA, \triangle ABC \) are congruent. Since all side lengths of the tetrahedron formed by connecting \( D \) to \( A, B, \) and \( C \) are equal, \( D \) and all the described congruent triangles are correctly defined in 3D space.

Key concepts

Equilateral Triangle

An equilateral triangle is a special type of triangle where all three sides are of equal length. This means that if a triangle has sides labeled as AB, BC, and CA, then AB = BC = CA. One fascinating property of an equilateral triangle is that all the internal angles are also equal and measure 60 degrees each. This symmetry makes equilateral triangles a fundamental shape in geometry because they represent perfect balance.

• Due to its equal sides and angles, an equilateral triangle is highly symmetrical and it serves as a basic building block in constructing complex geometric shapes, such as a regular tetrahedron.
• The centroid, circumcenter, and orthocenter of an equilateral triangle all coincide at the same point, which is unique to this type of triangle.

This center point is crucial in forming other geometric structures, like the regular tetrahedron, by creating congruent figures in three-dimensional space.

Congruence

Congruence in geometry indicates that two figures have the exact same shape and size. For triangles specifically, congruence means that all corresponding sides and angles are equal. In the context of this problem, we explore the concept of congruence in three-dimensional space.

• For triangles to be congruent, such as \(\triangle DAB\) and \(\triangle ABC\), each corresponding side and angle must match perfectly.
• This can be visualized in three dimensions by ensuring that a point, like \(D\), is equidistant from all sides of the triangle \(\triangle ABC\), forming a tetrahedral shape where each "face" of the tetrahedron is congruent.

The congruence guarantees that the spatial relations are symmetrical and helps mathematicians form regular tetrahedrons efficiently.

Regular Tetrahedron

A regular tetrahedron is a three-dimensional shape composed of four equilateral triangles. In terms of vertices, it means there are four points with equal edges connecting them. This creates a solid tetrahedron that is perfectly symmetrical. In this exercise, we're essentially building a regular tetrahedron using the equilateral triangle \(\triangle ABC\) and a new point \(D\).

• Since each face of a regular tetrahedron is an equilateral triangle, it ensures that connecting \(D\) to \(A, B,\) and \(C\) forms congruent triangles.
• Positioning \(D\) correctly ensures the tetrahedron is regular, meaning all its edges have equal lengths, thus maintaining symmetry.

Visualizing a regular tetrahedron helps in understanding complex three-dimensional arrangements and their geometric properties, such as the circumradius.

Circumradius

The circumradius of a geometric figure is the radius of a circle (or sphere in three dimensions) that passes through all its vertices. For an equilateral triangle \(\triangle ABC\), the circumradius is a crucial measure, defined as the distance from the center to any of the vertices.

• For our equilateral triangle \(\triangle ABC\) with side length \(s\), the circumradius \(R\) is \(\frac{s}{\sqrt{3}}\).
• This radius is not only a critical component in two-dimensional settings but also serves as the foundation for establishing spatial relationships in three-dimensional shapes such as tetrahedrons.

When expanding into three dimensions, the circumradius from the triangle helps determine appropriate distances and measurements to form the congruent structures needed for a regular tetrahedron, displaying the elegance of geometric relationships across dimensions.