# Do the given perimeter and area corresponds to many shapes? [closed]

I have a perimeter P and area A of a planar shape. How to prove that there are many shapes that corresponds to those perimeter and area values?

## closed as off-topic by José Carlos Santos, jgon, KReiser, Cesareo, user91500Jan 5 at 8:05

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• Is that the entire information given? Or something more specific? (Eg polygon with fixed number of sides or whatever). – timtfj Jan 4 at 15:34
• Yes, I need a common prove. – Ivan Bunin Jan 4 at 15:44
• That depends on the particular perimeter and area. Note that for all shapes, $4\pi\text{Area}\le\text{Perimeter}^2$, and if $4\pi\text{Area}=\text{Perimeter}^2$, there is only one shape, a circular disk. – robjohn Jan 4 at 15:57

If $$P^2=4A$$, there is only one planar shape (up to translation) that matches: the disc of radius $$\frac1{2\pi}P$$.

If $$P^2<4A$$, there is no such shape.

If $$P^2>4A$$, you may start with a disc of area $$A$$, which has too short perimeter. If you deform a tiny part of the perimeter outside and another part inside, you can achieve that the aea remains unchanged, while the perimeter takes on as large a value as you want. As there is a lot of freedom of choice involved in this process, the claim follows.

• How it is possible to deform a tiny part of disk perimeter outside? Can you draw this? – Ivan Bunin Jan 4 at 15:46
• there should be a $\pi$ in there... $4\pi A\le P^2$. – robjohn Jan 4 at 15:58
• @robjohn So, when $P^2>4 \pi A$ there are infinity set of possible shapes? – Ivan Bunin Jan 4 at 16:07
• @IvanBunin: yes. If more context is added to the question, I can post an answer. – robjohn Jan 4 at 20:50
• @robjohn The Magnus' answer looks enough for me. But if you have a more strong or powerful prove, you are welcome to present it. – Ivan Bunin Jan 4 at 22:04

If you start with a rectangle and have an indent on it, like in shape A, then you can always move that indent to any other place and the perimeter and the area will remain the same, as in shape B (starting with an "outdent" would work too of course.) Perhaps this proof works for what you have in mind.