Polygon Coordinates given the Area and Center point

[PhysicsInept]

TL;DR

Polygon

I’m wondering if there is a formula for calculating the coordinate points of a polygon given the following
- Center point is known
- area is known
- Point A is known
- Points B, C, and D are UNKNOWN

I am NOT a math pro - this is for a puzzle I’m trying to solve and I can’t remember if this was even covered in high school math!
Thanks

 

[jedishrfu]

So the polygon has 4 points meaning it's a quadrilateral of some sort like a square, rectangle, rhombus, parallelogram, trapezoid or some arbitrary shape.

One could start with a right triangle contain the center point and point A choose B such that the triangle has one quarter of the area given.

 

[PhysicsInept]

Hmmm okay so bear with me as I try to understand ..
So I would use the Center point as Point A of a right triangle, then I know Point B already, and I can calculate the last Point since I have area. Why would the area be 1/4 of what I have been given? A polygon isn’t 4 equal triangles right?

 

[pbuk]

I’m wondering if there is a formula for calculating the coordinate points of a polygon given the following
- Center point is known

As you only mention four points we need to assume the polygon is a quadrilateral - this is important. It is also important to define what you mean by the "center point": if you mean what is usually called the centroid (also sometimes geometric centre/center or even barycentre/barycenter) then yes this should be solvable with some work; it will probably involve Varignon's parallelogram.

One could start with a right triangle contain the center point and point A choose B such that the triangle has one quarter of the area given.

I assume you mean that the "centre point" is the right angled vertex of the triangle? That will construct a quadrilateral with the "centre point" at the intersection of the diagonals which is not in general the centroid.

 

[jedishrfu]

Yes I didn't want to solve it for the OP.

 

[mfb]

A polygon with four sides has 8 degrees of freedom (e.g. the 4 corners), we only have 5 values given. This can only be solved if we know it's a special type of polygon.

 

[pbuk]

A polygon with four sides has 8 degrees of freedom (e.g. the 4 corners), we only have 5 values given. This can only be solved if we know it's a special type of polygon.

It can only be solved uniquely if we have more information, but I'm not sure if uniqueness is required (the OP doesn't ask for "the polygon").

 

[mfb]

If a question asks for the coordinates of points without saying anything else it's usually a unique solution (only dependent on given parameters). A general polygon doesn't have a single unambiguous center either, that makes me think there is some information missing in the first post. If you require some special shapes and define their center then you can find solutions of course.

 

[jedishrfu]

To my mind it implies a square or a rhombus.

 

[Mark44]

I’m wondering if there is a formula for calculating the coordinate points of a polygon

To my mind it implies a square or a rhombus.

I don't agree, based on the OP. If someone asks about a polygon, without additional information, it's a stretch to infer that a more specific geometric figure is meant. Even if we take the mention of points A, B, C, and D to mean that these are vertices, and that there are no other vertices, the figure could still be a rectangle, trapezoid, or just an irregular quadrilateral.
Seeing the exact problem statement would clear up these uncertainties.

 

[Maarten Havinga]

If you take a polygon with 4 points, there is always an inner diagonal. Take as origin one of the endpoints of the diagonal and name the 3 other endpoints u, v and w as vectors where v is the other endpoint of the inner diagonal.

Then the formula for the surface of the polygon equals $|(Det(u\: v) + Det(v\: w)/2|$

Thus if this area a is known, you can take angle phi and take 2 matrices from $SL_2(R)$ where the 1st times $cos(\phi)$ has as 2nd column the 1st column of the 2nd matrix times $sin(\phi)$. These matrices multiplied with $a\sqrt(1/2)cos(\phi)$ or $sin(\phi)$ for the 2nd have as columns the vectors we wanted to find.