5 distinct points in a plane ---> max. number of pentagons that can be formed

[lookagain]

Given 5 distinct points in one plane, what is the maximum number of distinct pentagons (convex and non-convex) possible that can be formed where the 5 points are vertices of a pentagon, and you get to decide what the 5 distinct points are?

 

[soroban]

Hello, lookagain!

Given 5 distinct points in one plane, what is the maximum number of distinct pentagons
(convex and non-convex) possible that can be formed where the 5 points are vertices of a pentagon,
and you get to decide what the 5 distinct points are?

I'm not sure what that last phrase means . . .

Call the points $\displaystyle A,B,C,D,E.$

There are $\displaystyle 5!$ permutations of the points.
Each permutation determines a pentagon.
However, there is much duplication in the list.

The cyclic permutations $\displaystyle \begin{Bmatrix}ABCDE \\ BCDEA \\ CDEAB \\ DEABC \\ EABCD\end{Bmatrix}$ represent the same pentagon.

The reversals are also duplicates: $\displaystyle \begin{Bmatrix}ABCDE \\ EDCBA\end{Bmatrix}\;\; \begin{Bmatrix}CEADB \\ BDAEC\end{Bmatrix}\;\;\text{etc.}$

The answer is: .$\displaystyle \dfrac{120}{5\cdot2} \:=\:12$

 

[lookagain]

I'm not sure what that last phrase means . . .

It means that you aren't stuck with some random set of 5 distinct points to consider.

And you didn't consider that out of the 12 that you claim, that certain ones of those would
not make a pentagon? Certain ones that you mention could be (or are) self-crossing
pentagons. And they don't count.


Edit:

For example, suppose we decided to let the vertices of the pentagon be located at coordinates:

$\displaystyle A \ (3, 5), \ \ B \ (2, 2), \ \ C \ (4, 2), \ \ D \ (0, 0), \ \ and \ \ E \ (6, 0).$


The following orderings of points do not form/correspond to any convex nor any non-convex pentagons:


ABECD

ABEDC

ACDEB

ACDBE

 

[lookagain]

ADECB(A) and AEDBC(A) both qualify, right?

(I didn't mean to make you wait. I don't monitor the site as often as
certain others do.)

Yes, certainly, those both qualify. They are two of the ways out of a larger answer.

 

[lookagain]

ABDEC (of course)
.
ADECB ; AEDBC (per my previous post)
.
ABDCE ; ACEBD
.
ABCDE ; ACBED
.
So 7 if mirror image accepted, In all cases, let's count mirror images. (I could have given a set of
5 points having no symmetry.)

4 if not.
.
Should I take another trip to the corner?

Where is ADBCE (it's also the same as AECBD) in your list?

That would make eight.

 

[soroban]

Hello, lookagain!

I agree with your interpretation.

I had considered "non-convex" to include self-crossing pentagons,
. . which was still an interesting problem.


If we place the five points as you did:

                A
                ♥


             B♥   ♥C


     D♥                   ♥E

I found eight pentagons.

. . $\displaystyle \begin{array}{c}ABDEC(A) \\ ACBDE(A) \\ ABCDE(A) \\ ADBEC(A) \\ ADBCE(A) \\ ABDCE(A) \\ ADEBC(A) \\ ADECB(A) \end{array}$