Using only precalculus mathematics (including that the area of the triangle with vertices at the origin, $(x_1,y_1)$, and $(x_2,y_2)$ is half of the absolute value of the determinant of the $2\times 2$ matrix of the vertices $(x_1,y_1)$ and $(x_2,y_2)$, $\frac{1}{2}\cdot\left|x_1\cdot y_2 – x_2\cdot y_1\right|$) how can one prove that the shoelace method works for all non-self-intersecting polygons?
Geometry – Proving Shoelace Method at Precalculus Level
algebra-precalculusanalytic geometrycontest-mathgeometry
Related Solutions
Consider the fact that if P(x,y) is the unknown point, then when you apply rotation around that point, the point P1(x1, y1) goes to (gets mapped to) P2(x2,y2). From this you can construct equations and solve them to get the x and y. Just lookup the formulas for rotation in the 2D plane. The rotation angles are nice in both cases: 90 and 60 degrees. If you do this, you should get the answers you have posted here.
Say you rotate the point (x1,y1) around the point (a,b) on an angle of $\theta$.
Then the point (x1,y1) gets mapped to (x2, y2) where:
$x_2 = (x_1-a) * cos(\theta) - (y_1-b) * sin(\theta) + a$
$y_2 = (x_1-a) * sin(\theta) + (y_1-b) * cos(\theta) + b$
In your case you don't know the $a$ and $b$, you want to find them. The $\theta$ is $\pm \pi/2$ and $\pm \pi/3$ respectively. The $x_1, y_1, x_2, y_2$ - these you know.
Here is how one can solve the case: $\pi/3$. You have 3 more cases to solve.
See also:
Yes indeed, a direct consequence of the shoelace formula is that a triangle (convex polygon) having all its vertices with rational coordinates has a rational area. It is a nice technique for showing that there is no equilateral triangle in $\mathbb{Z}\times\mathbb{Z}$, too.
Best Answer
One way is to note that $x_1y_2 - x_2y_1$ is a signed area, i.e. it may positive or negative. Adding up all the signed areas of the triangles formed by the points $O$, $P_k$ and $P_{k+1}$ will cancel all the superfluous parts, as can be seen from the sketch:
The same argument also works for trapezoids with the $x$-axis instead of triangles with the origin. (I think this is the most illuminating argument, because the key trick is to give the area a sign depending on orientation.)
One could argue that this is not very rigorous, however. A more rigorous proof is to divide the polygon into two smaller polygons (it's not trivial to show that this is possible) and argue that adding the shoelace sums of the two parts gives the shoelace sum of the whole. That's because the two additional terms for the extra side cancel each other. (This cancellation is well-known from line integrals, we are in essence calculating $\frac12 \oint_{polygon} x\,dy - y\,dx$ here.) By induction, you then only have to verify the formula for a triangle.