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In my geometry textbook we have this notation style that I really hate which makes it difficult to understand the proof of a formula for the area of a polygon.

So for a given triangle:

$$A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$$ its area is equal to one half of the magnitude of the vector product $\vec{AB} \times \vec{AC}$. Since we're only working in two dimensions here, and vector product works with 3D vectors, we get:

$$\vec{AB} \times \vec{AC} = \Bigg|\begin{matrix}\vec{e_1} & \vec{e_2} & \vec{e_3} \\ x_2 - x_1 & y_2-y_1 & 0 \\ x_3 - x_1 & y_3-y_1 & 0\end{matrix}\Bigg| = \vec{e_3}\bigg((x_2 - x_1)(y_3-y_1) - (x_3 - x_1)(y_2-y_1) \bigg)$$

Then we introduced $D_{ABC}:= \bigg| \begin{matrix} x_2 - x_1 & y_2-y_1 \\x_3 - x_1 & y_3-y_1 \end{matrix} \bigg|$

So the are of the triangle becomes: $$P_{ABC} = \frac{1}{2}||\vec{AB} \times \vec{AC}|| = \frac{1}{2}||D_{ABC}\vec{e_3}|| = \frac{1}{2}||D_{ABC}||\cdot ||\vec{e_3}|| = \frac{1}{2}|D_{ABC}|$$ We also use the notation: $$P(A, B, C) = \frac{1}{2}D_{ABC}$$ for what we call 'oriented area of a triangle' which would be the number whose absolute value is equal to the are of the triangle, which is positive if the triangle is positively oriented and negative if the triangle is negatively oriented.

Now for the real problem:

Theorem: The oriented area of a polygon $p = A_0A_1 ...A_{n-1}$ is $$P(A_0, A_1, ... A_{n-1}) = P(A, A_0, A_1) + P(A, A_1, A_2) + ... + P(A, A_{n-1}, A_0)$$ for any point $A$

Proof by induction:

(B) For the triangle $A_0A_1A_2$ it's easy to check that $P(A_0, A_1, A_2) = P(A, A_0, A_1) + P(A, A_1, A_2) + P(A, A_2, A_0)$ for any A. This part, even though it's not been proven in my book, is easy to check by simply drawing a triangle and an arbitrary point (note that we're using 'oriented areas' all the way, it has confused me at first).

Now assuming that the formula applies for a polygon with $n$ points, or assuming that:

$$P(A_0, A_1, ... A_{n-1}) = P(A, A_0, A_1) + P(A, A_1, A_2) + ... + P(A, A_{n-1}, A_0)$$ we check what happens when we add another point to it. And this is what it says:

$$P(A_0, A_1, ..., A_n) = P(A, A_0, A_1) + ... + P(A, A_{n-1}, A_n) + P(A, A_n, A_0) =\\ P(A_0, A_1, ..., A_{n-1}) - P(A, A_{n-1}, A_0) + P(A, A_{n-1}, A_n) + P(A, A_n, A_0) =\\ P(A_0, A_1, ..., A_{n-1})+P(A_{n-1}, A_n, A_0)$$

And this was simply thrown in there without any explanation of how they equated those things. The very first equality is where I don't get it. I suppose that they meant to say that the are of the larger polygon (the one with $n+1$ points) is equal to the area of the smaller one which we know from the hypothesis plus the are of the additional triangle that we get. But in that case the $ + P(A, A_n, A0)$ doesn't make sense.

Any ideas?

Thanks.

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  • $\begingroup$ The first equation seems to be what is to be proved. How does the book introduce it? $\endgroup$ – saulspatz May 28 at 16:48
  • $\begingroup$ That's exactly what I thought at first, but I assumed that the book wouldn't make such an enormous mistake. The book does not introduce it at all. As I said, the whole expression is simply given and the proof ends there. $\endgroup$ – Koy May 28 at 16:53
  • $\begingroup$ On further reflection, I think that $P(A_0,A_1,\dots,A_n)$ is just supposed to be an algebraic expression, and that all the $=$ signs are supposed to follow from algebraic manipulations. It's only at the very last that you use the induction hypothesis to conclude that this is the oriented area of the $n$-gon. $\endgroup$ – saulspatz May 28 at 17:13
  • $\begingroup$ I don't think I understand what you mean by that, since I read your first comment I'm still somewhat convinced that after the first equality the theorem itself is used (before being proved_ $\endgroup$ – Koy May 28 at 17:35
  • $\begingroup$ I'm not sure if I'm right my last comment is right, because I don't have the book. Is $P(A_0,A_1,\dots,A_N)$ defined by a formula, or is it just defined as the area of a polygon? The part you quoted makes sense if it was defined by a formula. If not, it seems rather an odd way of writing things. It would make more sense if the first "$P(A_0,A_1,\dots,A_n)$ were moved to the end. $\endgroup$ – saulspatz May 28 at 21:28
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Going by what you stated is

$P(A_0,A_1,A_2) = P(A,A_0,A_1) + P(A,A_1,A_2) + P(A,A_2,A_0)$;

Finding the sign of the area is easy. If we traverse along periphery anticlock-wise then the area is positive, if we traverse clockwise then area is negative. For ex. $P(A,A_0,A_1) = - P(A_0,A,A_1)$

Note that last two points of each term in above equation are adjacent and in order. We can also denote the vertexes of any polygon in such a way that each successive points come when we traverse anticlockwise.

From triangle to quadrilateral the additional point $A_3$ comes after $A_2$ anticlockwise (we can always construct like this as discussed above), we see that the last triangle $(A,A_2,A_0)$ earlier, does not exist in quadrilateral (simply because $A_2, A_0$ are no more adjacent points) instead we have new triangles $(A,A_2,A_3),(A,A_3,A_0)$, ($A_3,A_0$ are now adjacent points) so to proceed from triangle to quadrilateral we need to do $P(A_0,A_1,A_2) - P(A,A_2,A_0) + P(A,A_2,A_3) + P(A,A_3,A_0)$, this can be then modified as $P(A_0,A_1,A_2) + P(A,A_0,A_2) + P(A,A_2,A_3) + P(A,A_3,A_0) = P(A_0,A_1,A_2) + P(A_0,A_2,A_3) = P(A_0,A_1,A_2,A_3)$,

here $(A_0,A_2,A_3)$ is the new triangle added. And $A_0,A_2,A_3$ are in order anticlockwise.

So, $P(A_0,A_1,A_2,A_3) = P(A_0,A_1,A_2) + P(A_0,A_2,A_3) = P(A_0,A_1,A_2) + P(A,A_0,A_2) + P(A,A_2,A_3) + P(A,A_3,A_0) = P(A,A_0,A_1) + P(A,A_1,A_2) + \{P(A,A_2,A_0) + P(A,A_0,A_2)\} + P(A,A_2,A_3) + P(A,A_3,A_0) = P(A,A_0,A_1) + P(A,A_1,A_2) + P(A,A_2,A_3) + P(A,A_3,A_0)$

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