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?
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5 distinct points in a plane ---> max. number of pentagons that can be formed
- Thread starter lookagain
- Start date
Hello, lookagain!
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: .\(\displaystyle \dfrac{120}{5\cdot2} \:=\:12\)
I'm not sure what that last phrase means . . .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?
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: .\(\displaystyle \dfrac{120}{5\cdot2} \:=\:12\)
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
Last edited:
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.
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.
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:
Code:
A
♥
B♥ ♥C
D♥ ♥E. . \(\displaystyle \begin{array}{c}ABDEC(A) \\ ACBDE(A) \\ ABCDE(A) \\ ADBEC(A) \\ ADBCE(A) \\ ABDCE(A) \\ ADEBC(A) \\ ADECB(A) \end{array}\)