# Is there any theorem about figures of equal area and perimeter being congruent?

I had an idea, that all geometric objects, that are different, as they're not a translation, rotation, and a reflection of one another cannot have the same area AND perimeter, as compared to ONE ANOTHER. They can't be CONGRUENT.

If the shapes are similar there is no similar shapes can contradict this "idea", or that is what I think.

I know that there was some idea on this, is there any theorem, or specific idea, which this is expressed?

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By geometric object do you mean "shapes" as in any sort of shape or is there something more specific you are getting at? –  frogeyedpeas Jul 24 '14 at 20:54
Are you suggesting that two figures with the same area and perimeter have to be congruent? That is certainly false. –  Grumpy Parsnip Jul 24 '14 at 20:58
It can be any shapes that are not congruent. –  Arbuja Jul 24 '14 at 20:59
Okay can this be shown! –  Arbuja Jul 24 '14 at 21:00
Take any shape with area $S$ and perimeter $P$. You can find a rectangle with sides $A$ and $B$ such that $AB=S$ and $2A+2B=P$. –  Yves Daoust Jul 24 '14 at 21:05

Counter-example without words (except for these):

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Note that this can be seen as a special case of a situation described in @hardmath's answer. –  Blue Jul 24 '14 at 22:12
I should've seen the area as a bunch of segments, it was so simple, I just couldn't precieve it. Plus while I was walking I realized the same thing goes for a rhombus and square. Thanks once again @Blue for the truth. –  Arbuja Jul 24 '14 at 22:41
Actually a rhombus and square with the same perimeter and area must be equal. All four sides of a rhombus are equal, and the square is a special case. So for the perimeters to be equal means the sides of the rhombus are the same as the sides of the square. Further, as a regular polygon, the area of the square is a maximum for quadrilaterals of a given perimeter, so if the rhombus area equals the square area, they must both be squares of equal sides, i.e. congruent. –  hardmath Jul 25 '14 at 0:32
I almost agree! However, are the angles the same;)? –  Arbuja Jul 25 '14 at 1:46

This idea is wrong, however attractive.

Let's consider a polygon with $n$ unequal sides. The side lengths determine the perimeter, so any rearrangement of the sides gives the same perimeter.

On the other hand any arrangement of the sides that inscribes the polygon in a circle maximizes the area contained by the polygon. The maximum area is independent of the order of sides, and even given reflection and rotation, there will be $n!/(2n)$ such arrangements, i.e. $(n-1)!/2$ noncongruent polygons, having the same area and perimeter.

On the other hand, restricting ourselves to two figures that are similar is a very narrow restriction. One has a single parameter of scale to vary between the figures. In that setting either equal area OR equal perimeter is enough to compel congruence of the figures.

For an example of two polygons with different numbers of sides, say 4 sides vs. 6 sides, consider a perimeter of six and area of two. We can achieve this with a rectangle of size $2 \times 1$. To match the perimeter we start with a regular hexagon having unit length sides, which will have area $\frac{3\sqrt{3}}{2} \gt 2$. We can maintain the perimeter at six but reduce the area down to two by "squeezing" a pair of opposing (parallel) sides closer together. The exact distance between these two sides to get the area six can be found by solving a cubic equation, but clearly the motion of moving sides closer together could achieve any area between that of the regular hexagon and zero (as the two sides get arbitrarily close).

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Sorry, I need a more specific example. Like how you can arrange them and both are same, like a picture maybe. For some reason I can't envision it. –  Arbuja Jul 24 '14 at 21:03
The smallest example is a quadrilateral. Pick four sides, make them different. I'll give you the example. –  hardmath Jul 24 '14 at 21:04
Never mind I get it now, heron's formula for quadrilaterals stay the same, with both. Thanks! Aplogies:) –  Arbuja Jul 24 '14 at 21:07
No problem, I just wanted to show how to generate examples with a large number of noncongruent variations. –  hardmath Jul 24 '14 at 21:08
Hard Math: What about if the polygons have different number of sides, so one has 4 sides, and the other has 6 sides. Then can they still have the same perimeter or area? Is there some ways this could be improved. –  Arbuja Jul 24 '14 at 23:09

Another counterexample without words:

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Thank you! Now do that when both polygons have different number of sides! –  Arbuja Jul 25 '14 at 0:16
Needed to make an edit! –  Arbuja Jul 25 '14 at 0:17