Here we start with a triangle `ABC` where each vertex is labeled a different color (red, blue, gold). Triangulate this triangle in such a way that:
Sperner's lemma claims there will always be a panchromatic triangle, namely a triangle with vertices of all three colors. In fact, there will always be an odd number of panchromatic triangles!
(In the example above there are 9 such panchromatic triangles.)
An amazing proof that uses determinantal signed area goes as follows.
For each vertex `P` in the triangulation, move it linearly in time `t in [0,1]` to one of the vertices `A`, `B`, or `C` that matches its own color, by linear interpolation. Namely, consider linear interpolation `P(t) = (1-t) P + t V`, where `V` is `A`, `B`, or `C` if `P` is colored red, blue, or gold respectively. Note that as we deform, all the triangles that are non-panchromatic become a degenerate triangle with zero area at `t=1`, while all the panchromatic triangles maps onto the original triangle `ABC`.
For sufficiently small `t` in some interval `[0, epsilon)`, where `epsilon > 0`, we get a small perturbation of the vertices `{P(t)}` that still gives a valid triangultaion of `ABC`. In this example above, for `t in [0,0.1)` we still have a valid triangulation. For larger `t` it need not remain a triangulation (self-intersections may occur), however that will be ok as we are going to investigate the signed areas of each triangle for all time `t`.
Say at `t=0`, this triangulation gives some `m` triangles, indexed by `1, 2, ..., m`, each with some initial positive area. Using determinant, define `q_i(t)` to be the signed area of triangle `i` of vertices `PQR`, where `PQR` are positively oriented at `t=0`, namely `q_i(t) = 1/2 * det(Q(t)-P(t) | R(t)-P(t))`, and since we chose a positive orientation, we have `q_i(0) > 0`. As `t` varies, the sign of 'q_i(t)' may change since the vertices are moving around. And since the points are moving linearly in `t`, each `q_i(t)` is a polynomial function in `t`. This can be deduced easily as we are computing determinants!
We note that initially the total area is `sum q_i(0) = "area"(ABC)`.
We also note that on the small interval `[0,epsilon)`, the deformed triangulation remains a triangulation, so for all `t in [0,epsilon)` , `sum q_i(t) = "area"(ABC)`, a constant.
But a polynomial that is constant on a nontrivial interval is globally a constant! Hence we have `sum q_i(t) = "area"(ABC)` for all `t`!
Note well that as the points move, `q_i(t)` could change signs due to the orientation of the vertices of the triangles changes.
So at time `t = 1`,
Finally, observe that if we have a finite sequence of `+1` and `-1` that sums to `+1`, we must have odd many terms in the sequence. Hence we have odd many panchromatic triangles!
Some topological remark. Sperner's lemma of course is true on topological simplices, but such a straightline embedding is always possible (at least for planar graphs, this is Fary's theorem). And this signed volume argument extends to higher dimensional simplices. Also, Sperner's lemma is of importance as it is the combinatorial equivalent of Brouwer fixed point theorem and KKM covering theorem. This signed volume argument is given by McLennan and Tourky. What is more interesting is that this argument can be adapted to prove Tucker's theorem, which is equivalent to Borsuk-Ulam and LSB covering theorem, given by Su et. al. (But it seems a bit more involved in the Tucker's case, one has to embed a cross polytope into a larger polytope first, and then argue using degree of simplicial maps of triangulations of spheres...)