# Polygons with equal area and perimeter but different number of sides?

Let's say we have two polygons with different numbers of sides. They can be any sort of shape, but they have to have the same area, and perimeter.

There could be such possibilities, but can someone show me with pictures? I just need visualize it.

Sometimes in life you just have to know it, and sometimes we need a picture shown in our faces :).

Just worked out a quick example, so the numbers may not be optimal: take a triangle with side lengths 2,3,4 - this has perimeter 9 and area $3\sqrt{15}/4$. It's easy enough to construct a rectangle with this data as well, by solving the equations $st = 3\sqrt{15}/4$ and $2s+2t = 9$. In fact, the sides lengths $s,t$ of the rectangle work out to be $\frac{1}{4}(9 \pm \sqrt{81-12\sqrt{15}})$.

• A 3-4-5 triangle has perimeter 12 and area 6, the corresponding rectangle satisfies $x(6-x)=6$ or $x^2 - 6x + 6 = 0$. By the quadratic formula, $x = 3 \pm \sqrt{3}$, so a $3 + \sqrt{3}$ by $3 - \sqrt{3}$ rectangle will work. – NovaDenizen Jul 25 '14 at 4:53
• love the way you solve it! – mattecapu Jul 27 '14 at 12:39
┌─┐   ┌┐
│ │   │└─┐
│ └─┐ │  └┐
└───┘ └───┘


Edit: I like the above figures because they're easy to generalize to many sides. But if it's unclear that they have the same area, here's another pair: the L and T tetrominoes.

You can imagine sliding the square on the right side up and down relative to the 1x3 bar on the left side; this operation preserves both area and perimeter. Explicitly, both tetrominoes have area 4 and perimeter 10. The L has 6 sides, and the T has 8 sides.

• How do I know those have the same area? – oconnor0 Jul 25 '14 at 16:51
• @oconnor0 Chop a 2x2 square from the top of the first figure and put in into the "notch" that remains. You can do the same with the 2x1 rectangle from the top of the second figure. – templatetypedef Jul 25 '14 at 17:01
• The Tetris pieces make it much more obvious. And they're prettier. :) – cHao Jul 27 '14 at 4:46

For any triangle there is a rectangle with the same area and perimeter.

proof: by Heron's formula the area of the triangle with semiperimeter $S$ and triangle sides $a,b,c$ is $$\sqrt{S(S-a)(S-b)(S-c)}\leq \sqrt{\frac{8S^4}{27}}$$

A square with semiperimeter $S$ has area $\frac{S^2}{4}\geq\sqrt{\frac{8S^4}{27}}$.

We can then proceed to make the square into a rectangle, making the area smaller and smaller preserving the perimeter. Until the area is as small as the triangle's.

• In fact I am sure for any n<m and any n-gon we can find an m-gon with the same area and perimeter as the n-gon, but I can't prove this – Jorge Fernández Hidalgo Jul 25 '14 at 0:37
• See my answer "cut and flip" – DavidButlerUofA Aug 4 '14 at 20:01
• are you sure that works for any arbitrary n,n and an y arbitrary n-gon? – Jorge Fernández Hidalgo Aug 5 '14 at 2:58
• actually, upon thinking about it, there are some cases I am not sure about. I think it deserves its own question. math.stackexchange.com/questions/888011/… – DavidButlerUofA Aug 5 '14 at 10:58

This the first example that came to my mind, using four identical $45-45-90$ triangles.

• Might be better to clone the first shape, but flip one of the sub-triangles (horizontally in this case) so that the dotted-line sides still entirely touch. That should make it more obvious that the perimeter is exactly the same, without requiring any math. – cHao Jul 27 '14 at 4:38
• @chao actually, that was my original construction, but I somehow convinced myself they had the same number of sides! Thanks for rescuing the idea from my lapse :) – rschwieb Jul 27 '14 at 12:51
• @chao rolled back to revision 1. It's minimal and requires basically no explanation. – rschwieb Aug 9 '14 at 1:56
• You already had my upvote :) But yeah, that does seem a lot cleaner. – cHao Aug 9 '14 at 3:23

Cut-(flip)-join method

Take a polygon and choose a line segment between two points on the perimeter that is fully contained inside the polygon.

Separate the polygon along this line segment to produce two polygons. Be sure to distinguish the "exposed cut edge".

Choose one of the two parts and join it to the other so that the exposed cut edge is touching an edge of the other part along its full length. It does not have to be joined to the other exposed edge, as long as it is touching some edge along its full length. You may choose to flip one of the parts before rejoining.

This will produce a polygon with the same area and same perimeter as the original polygon.

rschwieb's answer is an example of this method.

Special case: cut-flip-join-in-same-place

Cut the polygon in two as described above. Flip one part. Join the two exposed cut edges.

Unless the part flipped was symmetric, you will have made a new polygon (as opposed to simply restoring the original polygon). If at least one end of the line segment is not at a vertex, then you will also change the number of sides.

The simplest example is an isosceles triangle and a parallelogram:

However you can start with any triangle and progressively make shapes with more and more sides:

• ...and this one looks like an example of your 'flipping' method in my question, iterated repeatedly. You really like these two problems :) – Semiclassical Aug 4 '14 at 11:22
• @Semiclassical I'm a mathematician; sometimes I find it hard to let things go. :) – DavidButlerUofA Aug 4 '14 at 19:45

Here are Gordon–Webb–Wolpert polygons, they have more than just common area and perimeter, they are isospectral! Drums shaped as them would sound the same. It's a fun exercise to construct more examples using tangram pieces.

More isospectral polygons can be found here, including ones with different numbers of sides.

• I wonder if the common area & perimeter property is sufficient as well as necessary. Probably not, but a counter-example to that would be interesting as well. – Semiclassical Jul 25 '14 at 0:28
• Unfortunately this does not answer the original question, as the two polygons given both have 8 sides. – vadim123 Jul 25 '14 at 0:29

You can make a hexagon and a pentagon with the same area and perimeter by gluing triangles to a square:

The area of both is $6$ units$^2$ and the perimeter of both is $4 + 4\sqrt2$ units. It's particularly pleasant that both shapes are convex.

(Note also that these two are related by a "cut and flip" as described in my other answer.)

• That's also nice as an example in the spirit of the bump case you did for my problem. – Semiclassical Aug 3 '14 at 23:30
• The idea of adding "bumps" inspired both of them actually -- good example of multiple problems feeding into each other. – DavidButlerUofA Aug 3 '14 at 23:32

This may be cheating a bit, but why not just create two polygons that "look the same" but have a different graph? For example, we may take a triangle and insert a new vertex somewhere in one of its edges. This produces a new polygon that looks like that triangle, but is actually a rectangle quadrilateral with three collinear vertices.

• that's not a rectangle, and not even a 4-sided polygon because its sides must be non-adjacent – mattecapu Jul 27 '14 at 12:48
• @mattecapu According to which definition? – Sebastian Negraszus Jul 27 '14 at 12:55
• A rectangle has four right interior angles. Your figure is not a rectangle. – Joel Reyes Noche 2 days ago
• @JoelReyesNoche Yes, it's a quadrilateral, not a rectangle. That was a slip or mistranslation. My question concerning the definition was about the claim that it's "not even a 4-sided polygon because its sides must be non-adjacent", not about the definition of rectangles. Sorry for the confusion. – Sebastian Negraszus yesterday

For the first shape, take a rectangle with sides $1,x$. For the second shape, take a right triangle with sides $4,\frac{x}{2}, \sqrt{(\frac{x}{2})^2+4}$. They both have area $x$, for every $x$. Making them have the same perimeter reduces to solving a quadratic equation; a solution is $x=\frac{3+\sqrt{33}}{2}$. Then both shapes have perimeter $5+\sqrt{33}$.

Let $A$ be a triangle with corners at the points $(0,0)$, $(2,0)$, and $(0,1)$, with area $1$ and perimeter a little larger than 5. Let $B$ be a parallellogram with corners at $(0,0)$, $(1,0)$, $(x,1)$, and $(x+1,1)$. For any $x$, the areas of $A$ and $B$ are the same, namely $1$. If $x=0$, $A$ has largest perimeter, while if $x$ is large then $B$ has largest perimeter. So by continuity, there is a value of $x$ such that the perimeters are equal.