Prove: In an equilateral tetrahedron, all faces are equilateral triangles.?

If you prove the second part, you're done with the rest, since the definition of "equilateral" takes care of all the triangular faces. If every edge of the solid is the same length, then every edge of the (triangular) faces are the same length.

Now, since a tetrahedron has four faces, each face can have at most 3 edges, since there are only three other faces with which it can share an edge. However, it must have more than two edges (or it's not going to be a face), and so each face will be a triangle.

Hope that's enough for you!

First of all, a tetrahedron must have triangular faces by definition, so there's nothing to prove there. (Are you using a non-standard definition of tetrahedron? I assume you're saying a tetrahedron is any polyhedron with four faces. I tend to prefer to give "tetrahedron" the usual definition, and then prove that it is the unique polyhedron with 4 faces, but admittedly that's partly an aesthetic preference. One problem with naming a polyhedron by the number of faces is that the number of faces does not uniquely determine a polyhedron up to isomorphism. For example, the dodecahedron and the octahemioctahedron both have 12 faces, but are not isomorphic, so calling them both dodecahedra would be unsatisfying. Compare the case of n-gons, where it is much more reasonable to use "n-gon" to refer to a polygon with n sides; in this case, any two n-gons are isomorphic.)

Regardless, I assume the answer you want is this: the number of edges of any face is >=3 because a polygon must have 3 sides, and <= 3 because the face can share at most one edge with each of the other 3 faces. Therefore each face has exactly three edges. Of course, if you want a really, really rigorous proof, you'll have to specify what exactly you mean by a polyhedron; defining polyhedron completely rigorously turns out to be harder than you might think.

Secondly, what do you mean by "equilateral tetrahedron"? If you mean that all edges have the same length L, then it's obvious that each face is an equilateral triangle. Each face consists of three edges; therefore it is a triangle with three sides of length L, so it is equilateral.

2 tetrahedrons glued jointly supply yet another occasion. So an equilateral triangle will do, besides as any isosceles triangle. you will see it by making use of dilating the unique hexahedron alongside the path orthogonal to the gluing airplane. honestly, you may style a tetrahedron whose 4 faces would be congruent to any triangle having 3 acute angles. From this you besides could get an hexaedron by making use of gluing 2 of them. by making use of Euler's formula Vertices + Faces = Edges + 2. If we call n the form of aspects of P, then V <= 6n / 3, on account that each and each vertex belongs to a minimum of three faces. F = 6, and E = 6n/2 = 3n, on account that each and each ingredient belongs to precisely 2 faces. We get 2 n + 6 >= 3n + 2, so n <= 4. with the aid of fact the case n = 3 is solved, it in ordinary terms maintains to be to establish the attainable quadrilaterals which artwork. it is real, yet needs somewhat checking, that in case you dilate a cube in the path of huge diagonal, then you somewhat get carry of an hexahedron whose faces are a rhombus. So P could be any rhombus. Now by making use of doing the floor plan of your hexahedron, like in case you wanted to construct it by making use of reducing folding and gluing, you notice that, each and all the perimeters could desire to have a similar length. basic to establish yet no longer basic to describe. only initiate with a,b,c,d the lengths of the aspects. placed 4 faces around one, and the only way you will get them up and glue them is that all and sundry edges be same length. besides, it is your answer: P must be the two a rhombus, or any triangle with 3 acute angles. Edit: appropriate, I missed a relatives of quadrilaterals that have one diagonal as an axis of symmetry. Edit: Oh nicely. they could could desire to be rhombuses besides. i think Scythian's shape supplies all quadrilaterals. to establish which of them you get, initiate with any triangle ABC isosceles in B with terrific attitude decrease than 2pi/3. finished it with D to make a rhombus. enable E be the ellipse based at B and containing ACD. Then a quadrilateral P having ABCD' as vertices in that order, would be a answer if D' is on the ellipse E. This ellipse is the intersection between the airplane containing ABC and a cylinder which includes 6 of the vertices of the hexahedron, the two final ones being on its axis.