The proof of the day is: $V-E+F=2$ for every convex polyhedron
Theorem: For every convex polyhedron $P$ with $F$ faces, $E$ edges and $V$ vertices, $V-E+F=2$.
Proof (proven by
Zoroastro Azambuja Filho):
First of all, a subset $C$ of the plane or space is called convex when every segment that
connects two points of $C$ is a subset of $C$. And a finite union of convex polygons called
faces, such that the intersection of whichever two faces be whether a common edge of these two faces,
or a common vertex or null, is called a polyhedron.
Define a line $r$ that is not parallel to none of the faces of $P$, and a plane $H$ that
does not intersect $P$ and is perpendicular to $r$. $H$ will be called the horizontal plane
and the lines parallel to $r$ will be called vertical lines. $H$ divides the space into two
half-spaces, of which one contain $P$. This one will be called the upper half-space, and
we'll say that its points are above $H$.
Now, consider the sun shining at its peak over the upper half-space, considering that its
light-rays are vertical lines. To each point of the upper half-space corresponds a point
$Sh(x)$ in $H$, that we'll call the shadow of $x$, obtained as the intersection of the $H$
plane with the vertical line that passes through $x$. The shadow of every set $X$, contained
on the upper half-space is, by definition, the set formed by the shadows of the points of
$X$.
The intersection of a vertical line with the convex set limited by $P$ is a convex subset of
this line, therefore, if it's not null, it is a segment whose extrema belong to $P$, or it's
a single point of $P$. So, an arbitrary vertical line can only have $0$, $1$ or $2$ points in
common with $P$. That is, each points of the shadow of $P$ is the shadow of one or two points
of $P.$
But the shadow of $P$ is a convex polygon on the horizontal plane, whose boundary $\gamma'$ is
the shadow of a closed polygonal $\gamma$, made of edges of $P$. Each point of $\gamma'$ is the
shadow of a single point of $P$ (belonging to $\gamma$). The polygonal $\gamma$ is called the
apparent boundary of $P.$ Each interior point of the shadow of $P$ (i.e., it doesn't belong to
$\gamma'$) is the shadow of $2$ points of $P$. Given two points of $P$ that have the same shadow,
the highest (most distant to $H$) we'll call a light point; the lowest one will be called a dark
point.
Therefore, $P$ is the union of $3$ disjoint parts: the set of all light points, the set of all
dark points and the apparent boundary $\gamma$.
Define $Li(P)$ as the union of the set of light points of $P$ and the apparent boundary of $P$,
and $Sh(P)$ the shadow of $P$. Each point of $Sh(P)$ is the shadow of a single point of $Li(P)$.
That is, the map that takes each point $x \in Li(P)$ to its shadow $x' \in Sh(P)$ is a bijection
between $Li(P)$ and $Sh(P)$. We'll use the notation $Li(P)$ to represent the polygon $Sh(P)$
decomposed as the union of overlapped polygons that are shadows of the faces contained in $Li(P)$,
i.e.: the light faces.
Evidently, we can also consider the set $\overline{Li}(P)$, made by the union of the set of the
dark points of $p$ and the apparent boundary of $P$. The map that takes each point $y \in \overline{Li}(P)$
to its shadow $y' \in Sh(P)$ is also a bijection between $\overline{Li}(P)$ and $Sh(P)$. We'll write
$\overline{Li}(P)$ to indicate the shadow of $\overline{Li}(P)$ expressed as the union of the shadows
of the dark faces of $p$, i.e., contained in $\overline{Li}(P)$.
Finally, if we decompose each face of $P$ in triangles and draw diagonals in each one of them,
we'll change the values of $F$, $E$ and $V$ individually, but the expression $V-E+F$ remains with
the same value, since every time we draw a diagonal, the numbers $F$ and $E$ increase equally, so
on the expression $V-E+F$ they cancel out. Therefore, without loss of generality, suppose that
each faces of $P$ are triangles.
As each face have $3$ edges and each edge belong to $2$ faces, we have that $3F=2E$. There are $F$
triangles and the sum of the inner angles of each one is equal to $\pi$ radians, so, the sum
$\Sigma$ of the inner angles of the triangles that make up $P$ is given by $\Sigma=2\pi E - 2\pi F$.
At the same time, we have that $\Sigma = \Sigma_1 + \Sigma_2$, where $\Sigma_1$ is the sum of the
inner angles of the light triangles and $\Sigma_2$ is the sum of the inner angles of the dark triangles.
Since the sum of the inner angles of a triangle $\Delta$ is equal to the sum of the inner angles of its
shadow $Sh(\Delta)$, we know that $\Sigma_1$ is equal to the sum of the inner angles of the triangles on which
is decomposed the shadow of $Li(P)$. To calculate this sum, we can sum the angles vertex by vertex.
Let $V_1$ be the number of light vertices, $V_2$ be the number of dark vertices and $V_0$ be the
number of vertices of the apparent boundary $\gamma$. So $V=V_0+V_1+V_2$. Note that $V_0$ is also
the number of vertices (and of sides) of the polygonal $\gamma'$, boundary of $Sh(P)$.
In $Li(P)$, we have $V_1$ interior vertices (shadows of light vertices) plus $V_0$ vertices from the
boundary $\gamma'$. The sum of the angles that have a given interior vertex as its vertex is equal to
$2\pi$ radians. The sum of all angles that have some vertex on the boundary $\gamma'$ is equal to
$\pi(V_0-2)$, according to the well-known expression of the sum of inner angles of a polygon with $V_0$ sides.
So we end up with $\Sigma_1 = 2\pi V_1+\pi(V_0-2)$, and similarly, $\Sigma_2 = 2\pi V_2+\pi(V_0-2)$,
so we can sum up this equations, obtaining $\Sigma = \Sigma_1+\Sigma_2 =$
$2\pi\left(V_0+V_1+V_2\right)-4\pi=2\pi V-4\pi$.
Comparing with the equation $\Sigma = 2\pi E-2\pi F$ and dividing by $2\pi$, it results that
$E-F=V-2 \iff V-E+F=2$.