[Geometry]
Problem: Let $W=(0,0), A=(7,0), S=(7,1)$, and $H=(0,1)$. Compute the number of ways to tile rectangle $W A S H$ with triangles of area $1 / 2$ and vertices at lattice points on the boundary of WASH.

Solution: Define a fault line to be a side of a tile other than its base. Any tiling of $W A S H$ can be represented as a sequence of tiles $t_{1}, t_{2}, \ldots, t_{14}$, where $t_{1}$ has a fault line of $\overline{W H}, t_{14}$ has a fault line of $\overline{A S}$, and where $t_{k}$ and $t_{k+1}$ share a fault line for $1 \leq k \leq 13$. Also note that to determine the position of tile $t_{k+1}$, it is necessary and sufficient to know the fault line that $t_{k+1}$ shares with $t_{k}$, as well as whether the base of $t_{k+1}$ lies on $\overline{W A}$ (abbreviated " $\mathrm{B}$ " for "bottom") or on $\overline{S H}$ (abbreviated "T" for "top"). Because rectangle $W A S H$ has width 7 , precisely 7 of the 14 tiles must have their bases on $\overline{W A}$. Thus any permutation of 7 B's and 7 T's determines a unique tiling $t_{1}, t_{2}, \ldots, t_{14}$, and conversely, any tiling $t_{1}, t_{2}, \ldots, t_{14}$ corresponds to a unique permutation of 7 B's and 7 T's. Thus the answer is $\left(\begin{array}{c}14 \\ 7\end{array}\right)=\mathbf{3 4 3 2}$.
Let $T(a, b)$ denote the number of ways to triangulate the polygon with vertices at $(0,0),(b, 0),(a, 1),(0,1)$, where each triangle has area $1 / 2$ and vertices at lattice points. The problem is to compute $T(7,7)$. It is easy to see that $T(a, 0)=T(0, b)=1$ for all $a$ and $b$. If $a$ and $b$ are both positive, then either one of the triangles includes the edge from $(a-1,1)$ to $(b, 0)$ or one of the triangles includes the edge from $(a, 1)$ to $(b-1,0)$, but not both. (In fact, as soon as there is an edge from $(a, 1)$ to $(x, 0)$ with $x<b$, there must be edges from $(a, 1)$ to $\left(x^{\prime}, 0\right)$ for all $x \leq x^{\prime}<b$.) If there is an edge from $(a-1,1)$ to $(b, 0)$, then the number of ways to complete the triangulation is $T(a-1, b)$; if there is an edge from $(a, 1)$ to $(b-1,0)$, then the number of ways to complete the triangulation is $T(a, b-1)$; thus $T(a, b)=T(a-1, b)+T(a, b-1)$. The recursion and the initial conditions describe Pascal's triangle, so $T(a, b)=\left(\begin{array}{c}a+b \\ a\end{array}\right)$. In particular, $T(7,7)=\left(\begin{array}{c}14 \\ 7\end{array}\right)=3432$.

Final answer: 3432