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BDMO 2017 National round Secondary 6
Posted: Fri Feb 10, 2017 9:30 pm
by ahmedittihad
Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron?(Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)
Re: BDMO 2017 National round Secondary 6
Posted: Wed Apr 05, 2017 2:29 pm
by Soumitra Das
We can rotate the octahedron by 8\times3 or 24 ways(to prove it,just assign one side of the octahedron as A and one of it's neighbour as B and observe it's rotation). So,the total number of distinguishable octahedron is \frac{8!}{24} or 1680.
Re: BDMO 2017 National round Secondary 6
Posted: Thu Feb 01, 2018 7:55 pm
by Tasnood
OK. A very big misunderstanding. I didn't see it was either a octagon or a octahedron!
Re: BDMO 2017 National round Secondary 6
Posted: Fri Feb 02, 2018 2:08 am
by samiul_samin
But what will be happened if we rotate this octahedron like upside down?
Re: BDMO 2017 National round Secondary 6
Posted: Fri Feb 02, 2018 10:41 pm
by Absur Khan Siam
It isn't
Octahedron" at all! It is
"Octagon".
This is octahedron :
Re: BDMO 2017 National round Secondary 6
Posted: Fri Feb 02, 2018 11:11 pm
by Tasnood
Soumitra Das wrote: ↑Wed Apr 05, 2017 2:29 pm
We can rotate the octahedron by 8\times3 or 24 ways(to prove it,just assign one side of the octahedron as A and one of it's neighbour as B and observe it's rotation). So,the total number of distinguishable octahedron is \frac{8!}{24} or 1680.
He wants to mean:
The octahedron has $8$ surfaces. If we avoid the result of rotation, number of arrangement=$8!$
For a specific surface $A$, there are $3$ neighbor surface: $B^1,B^2,B^3$. $A$ can rotate and change its position in $8$ ways (At $8$ surfaces)
Whence, the serial $B^1B^2B^3,B^2B^3B^1,B^3B^1B^2$ are same, just the result of rotation. So, actually there are $8\times3$ arrangements that seem same due to rotation.
So, total amount of distingushable octahedron is $\frac{8!}{8\times3}=1680$
Right this time?
Re: BDMO 2017 National round Secondary 6
Posted: Sat Feb 03, 2018 12:40 pm
by Absur Khan Siam
Tasnood wrote: ↑Fri Feb 02, 2018 11:11 pm
Right this time?
Yah, I think so...
Re: BDMO 2017 National round Secondary 6
Posted: Mon Feb 25, 2019 6:03 pm
by samiul_samin
Soumitra Das wrote: ↑Wed Apr 05, 2017 2:29 pm
We can rotate the
octahedron by $8\times3=24$ ways(to prove it,just assign one side of the octahedron as $A$ and one of it's neighbour as $B$ and observe it's rotation). So,the total number of distinguishable octahedron is $\frac{8!}{24}=1680$.
.$LaTeX$ ed
Re: BDMO 2017 National round Secondary 6
Posted: Tue Feb 26, 2019 12:32 am
by soyeb pervez jim
The question is same to
2000 AMC 12 problem 25,
Re: BDMO 2017 National round Secondary 6
Posted: Tue Feb 26, 2019 8:47 am
by samiul_samin
Duplicate question again! $4$ duplicate questions in $1$ National problemset!!!!