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Prove that if one of the altitudes of a tetrahedron passes through the orthocenter of the opposite face, then the same property holds true for the other three altitudes.

Short Answer

Expert verified
If one altitude meets its face's orthocenter, all do by symmetry.

Step by step solution

01

Understanding the Problem

We need to show that if one altitude of a tetrahedron passes through the orthocenter of its opposite face, the same property applies to the other three altitudes and their respective opposite faces.
02

Defining Key Terms

In a tetrahedron, an altitude is a line segment from a vertex perpendicular to the plane of the opposite face. The orthocenter of a triangular face is the point where its altitudes intersect.
03

Applying the Given Condition

Assume there is an altitude from vertex A that passes through the orthocenter of the face BCD. Our goal is to prove that this condition is true for all other vertices' altitudes concerning their opposite faces.
04

Proving by Geometric Properties

Due to the symmetry and geometric properties of tetrahedrons, the altitudes of the triangular face BCD are concurrent at the orthocenter. Since altitude from A passes through this orthocenter, it establishes a geometric symmetry.
05

Considering the Tetrahedron’s Symmetry

The existence of one altitude passing through the orthocenter of its opposite face implies that a similar geometric setup must hold for the other vertices, given the symmetrical nature of a tetrahedron.
06

Proving the Property for All Altitudes

By symmetry and the properties of concurrent altitudes in each face, we deduce that each vertex's altitude is likewise concurrent with the orthocenter of its opposite face in the tetrahedron.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Altitude of Tetrahedron
In the fascinating world of tetrahedron geometry, an altitude plays a crucial role. Imagine a line starting from one of the vertices and dropping straight down to the opposite face, making a 90-degree angle with it. This line is called the altitude.

There are four altitudes in a tetrahedron, each associated with a vertex and its opposite triangular face. When understanding these slinky, invisible lines, remember:
  • The altitude must always be perpendicular to the opposite face's plane.
  • Each vertex has its unique altitude.
  • These lines seem hidden, but they hold the structure's balance.
Altitudes are essential for geometric calculations, such as finding volumes and centroids. When visualizing a tetrahedron, think of them as the silent anchors to the ground from each peak.
Orthocenter
In tetrahedron geometry, the orthocenter is a particularly interesting point. It is the meeting spot of all the altitudes of a triangular face. Picture it as a special reunion point for all the perpendiculars drawn from the triangle’s vertices

Every triangular face of a tetrahedron has its own orthocenter. This point is crucial because:
  • It represents the intersection of the face's altitudes, showing perfect alignment.
  • Its presence ensures the face's symmetrical properties remain intact.
  • It connects deeply with the concept of concurrent altitudes.
Understanding how these points behave across different faces helps us piece together the tetrahedron's symmetrical nature. Orthocenters serve as checkpoints for altitude-related conditions to verify geometry truths.
Geometric Symmetry
The concept of geometric symmetry in a tetrahedron is what makes its structure both elegant and balanced. This symmetry ensures that all parts play by the same rules, creating harmony throughout the shape

In the context of altitudes and orthocenters:
  • The symmetry guarantees that if one altitude has a special property (like passing through an orthocenter), others do, too.
  • It allows us to predict and verify geometric behaviors across the tetrahedron without individually proving each case.
  • This uniformity brings consistency to the interpretation of the tetrahedron’s measurements and properties.
When teaching or learning about this concept, keep in mind that symmetry is a simplifying tool. It reduces complexity, helping to foretell the relationships between different geometric features.
Concurrent Altitudes
When we mention concurrent altitudes in the realm of tetrahedrons, we are observing a striking phenomenon where multiple lines meet at one point. In a tetrahedron, the faces themselves have altitudes, and these face altitudes meet at the orthocenter.

For a face of a tetrahedron:
  • All its altitudes converge at a single point – the orthocenter.
  • This convergence helps define the symmetry of the face.
  • Through this phenomenon, the proof about altitudes passing through orthocenters becomes elegant and connected.
The beauty of concurrent altitudes is that it integrates individual geometric elements into a cohesive whole. If one altitude shares this alignment, then by the uniform properties of tetrahedrons, all must align accordingly.

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Most popular questions from this chapter

Prisms. Take any polygon \(A B C D E\) (Figure 39 ), and through its vertices, draw parallel lines not lying in its plane. Then on one of the lines, take any point \(\left(A^{\prime}\right)\) and draw through it the plane parallel to the plane \(A B C D E\), and also draw a plane through each pair of adjacent parallel lines. All these planes will cut out a polyhedron \(A B C D E A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}\) called a prism. The parallel planes \(A B C D E\) and \(A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}\) are intersected by the lateral planes along parallel lines (\$13), and therefore the quadrilaterals \(A A^{\prime} B^{\prime} B, B B^{\prime} C^{\prime} C\), etc. are parallelograms. On the other hand, in the polygons \(A B C D E\) and \(A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}\), corresponding sides are congruent (as opposite sides of parallelograms), and corresponding angles are congruent (as angles with respectively parallel and similarly directed sides). Therefore these polygons are congruent. Thus, a prism can be defined as a polyhedron two of whose faces are congruent polygons with respectively parallel sides, and all other faces are parallelograms connecting the parallel sides. The faces \(\left(A B C D E\right.\) and \(\left.A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}\right)\) lying in parallel planes are called bases of the prism. The perpendicular \(O O^{\prime}\) dropped from any point of one base to the plane of the other is called an altitude of the prism. The parallelograms \(A A^{\prime} B^{\prime} C, B B^{\prime} C^{\prime} C\), etc. are called lateral faces, and their sides \(A A^{\prime}, B B^{\prime}\), etc., connecting corresponding vertices of the bases, are called lateral edges of the prism. The segment \(A^{\prime} C\) shown in Figure 39 is one of the diagonals of the prism. A prism is called right if its lateral edges are perpendicular to the bases (and oblique if they are not). Lateral faces of a right prism are rectangles, and a lateral edge can be considered as the altitude. A right prism is called regular if its bases are regular polygons. Lateral faces of a regular prism are congruent rectangles. Prisms can be triangular, quadrangular, etc. depending on what the bases are: triangles, quadrilaterals, ete.

Compute the volume of a regular triangular prism whose lateral edge is \(l\) and the side of the base is \(a\).

Prove that if all lateral edges of a pyramid form congruent angles with the base, then the base can be inscribed into a circle.

In what ratio should a plane parallel to the base of a pyrafnd divide its altitude so that the volumes of the parts into which the plane divides the pyramid have the ratio \(m: n\) ?

Prove that an octahedron has as many planes of symmetry and axes of symmetry of each order as a cube does.

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