Determine if a Point is inside a polygon without coordinates

Question

I have a polygon $ABCD$. I do not know the coordinates of the corners but I know the length of its sides (i.e. I know length of $AB$, $BC$, $CD$ & $DA$).

I have a point $P$. I do not know the coordinates of this point either but I know the distances $AP$, $BP$, $CP$ & $DP$.

How do I determine if point $P$ lies inside the polygon $ABCD$?

In the euclidean plane? Besides, note that the shape of $ABCD$ is not determined from the side lengths alone---however, the distances to $P$ would add information. — ccorn, Dec 20 '16 at 08:19
Consider a square and a rhombus that both have the same side length. How do you hope to distinguish between the two? — J. M. ain't a mathematician, Dec 20 '16 at 08:27
The Cayley-Menger determinants for any four points out of $A,B,C,D,P$ should be zero (in the plane). This would help determining the (squared) lengths of $AC$ and $BD$, if there is a need for that. — ccorn, Dec 20 '16 at 09:29
??? For polygons other than the triangle, you would need at least some angular information, or diagonal lengths, as well, to flesh out the shape. — Senex Ægypti Parvi, Dec 20 '16 at 10:37

score 2 · Answer 1 · answered Dec 20 '16 at 09:30

With $P$ as center draw four circles $\gamma_i$ $(1\leq i\leq4)$ with radii $|PA|$, $\ldots$, $|PD|$. Choose $A_1$ on $\gamma_1$ arbitrarily and construct a point $A_2 \in \gamma_2$ such that $|A_1A_2|=|AB|$. There are two such points, by symmetry you can discard one of them. Then construct the two points $A_{3i}\in\gamma_3$ having distance $|BC|$ from $A_2$, and keep both of them. Finally construct the four points $A_{4ik}\in\gamma_4$ having distance $|CD|$ from $A_{3i}$. At least one of these four points should have distance $|DA|$ from $A_1$. This allows you to draw at least one quadrangle $Q$ satisfying the given conditions. Now check whether $P\in Q$.

Of course you can do all of this numerically as well: Put $P=(0,0)$, $A_1=\bigl(|PA|,0\bigr)$, and proceed in terms of analytic geometry: intersection of circles, etc.

score 0 · Answer 2 · answered Dec 20 '16 at 08:16

0

Try this: You have the Distance AP. Now you take AB & DA to get to the same distance ($x*AB +y*DA = AP$). Repeat that for every corner point. x and y have to be both smaller than 1 to have the point in the polygon.

answered Dec 20 '16 at 08:16

WhoAmI

181

Could you elaborate on this. I don't understand how it works. Do you have any links that I could refer? – Nirumal Thomas Dec 20 '16 at 09:06

Determine if a Point is inside a polygon without coordinates

2 Answers2