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I would like to find the calculated area of a shape, I will have the actual area and perimeter. There is a fixed offset around the shape, for example 3'. What would be the calculated area. The only known variables are actual area and perimeter.

adr0327
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  • Is your shape a polygon? Or is it bounded by a smooth curve? – Intelligenti pauca Nov 07 '16 at 21:15
  • It is a polygon. Sometimes can be a smooth curve – adr0327 Nov 07 '16 at 21:16
  • To really solve this, there must be more information about the shape and the offset. For example, one way to put an "offset" of $3$ units around a square makes a larger square with sides that are $6$ units longer than the original square. But I think you're probably supposed to include only points that are within $3$ units of the original square, in which case you get a "square" with rounded corners. I also think you are probably supposed to require the shape's interior to be convex, or else you are only asked for an upper bound of the area including the offset. – David K Nov 07 '16 at 21:18
  • I think this question might be a variation of http://math.stackexchange.com/questions/1751332 where the "curve" in that question is the boundary of the shape in your question, and you are interested in just the area on one side of the curve (the outside), not both sides (since you already know the total area inside the curve). – David K Nov 07 '16 at 21:27

2 Answers2

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To give you another term to search for: you appear to be looking for the Minkowski sum of your given polygon and a disk of radius $3$ or whatever.

I doubt this can be answered if all you have is area, perimeter and offset. Take for example a square of edge length $10$ (so area $10^2=100$ and perimeter $4\times 10=40$). An offset of $3$ around this will lead to an area of

$$10^2+4\times10\times3+\pi\times3^2\approx248$$

Now take a circle of the same area, i.e. of radius $r=\sqrt{\frac{10^2}\pi}\approx5.64$. It has perimeter $2\pi r\approx35.45$, so that is smaller than the radius of the sphere. Now make a very narrow but deep cut into that circle. That can easily add to the parameter, but if it's narrow enough won't affect the area much. What's more important is that offsetting this resulting shape will look pretty much identical to offsetting the uncut circle, since the boundaries of the cut are so close together that the whole cut vanishes with almost no extra area. So the area of the offset shape would be essentially a circle of radius $r+3$, leading to an area of approximately $235$.

So you'd have two different shapes, with equal area and equal perimeter, but different area for the offset shape. These are not polygons, but they could be approximated by polygons arbitrarily well.

It should be possible to make this work if you only consider convex polygons. There you can consider the outline of the offset shape to be composed of straight runs which form rectangles with the original perimeter, and curved corners which must sum up to one whole disk. So for original perimeter $p$ and area $A$, adding an offset of $w$ will lead to this new area and perimeter:

$$A_w = A + pw + \pi w^2\qquad p_w = p + 2\pi w$$

And since you can approximate other convex shapes by polygons, the above should hold for other convex shapes as well. There should be a proof to that effect somewhere out there, but I don't have a reference at hand.

MvG
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Area = A1; New Area = A2; perimeter = p; offset= w ;

Approximately, i.e., if $w^2<< A1,$ and the offset is outside the convex figure

$$ A2= A1+ pw $$

is if inside the convex figure $$ A2= A1- p w $$ where p*w is the growth or reduction in area.

The co-ordinates or profile of A1 should be known for a more accurate calculation.

Narasimham
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