Existence of the Tetrahedron

We would like to show that a certain Platonic solid exists. In this case, we take equilateral triangles for the faces, and we have $3$ intersect at each vertex. As we all know, this produces a tetrahedron. But how can we show that it exists?

First, take a single point $D$. Now, put three of the equilateral triangles around point $D$ such that each triangle has a vertex at $D$, and that each pair of triangles shares exactly one edge.

Denote the three other vertices $A$, $B$, and $C$. Then our triangles are $DAB$, $DBC$, and $DCA$.

Now, we make an important conclusion. Notice that $AB$ is a side of the equilateral $\triangle DAB$. A similar conclusion holds for $BC$ and $CA$. But since all of the equilateral triangles are congruent, we have that $\triangle ABC$ is also an equilateral triangle, with the same size as the first three!

Place a fourth triangular face over $\triangle ABC$. Now, let us examine what we have made.

Surrounding point $A$ are equilateral triangles $ABC$, $ACD$, and $ADB$. This is, indeed, exactly the same situation as point $D$. You can check for yourself that it also works for $B$ and $C$.

To be even more sure, we can even look at Euler's formula. There are four vertices: $A$, $B$, $C$, and $D$. There are six edges: $AB$, $AC$, $AD$, $BC$, $BD$, and $CD$. Last, there are four faces: $ABC$, $ABD$, $ACD$, and $BCD$. It is easy to see that $V-E+F=2$, so we can be extra sure that this is indeed a valid three-dimensional solid.

Thus, we have satisfied the conditions for our Platonic solid, and the tetrahedron exists!