BDMO 2017 National round Secondary 6

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ahmedittihad
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BDMO 2017 National round Secondary 6

Unread post by ahmedittihad » Fri Feb 10, 2017 9:30 pm

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.)
Frankly, my dear, I don't give a damn.

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Soumitra Das
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Re: BDMO 2017 National round Secondary 6

Unread post by Soumitra Das » 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.

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Tasnood
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Re: BDMO 2017 National round Secondary 6

Unread post by Tasnood » Thu Feb 01, 2018 7:55 pm

OK. A very big misunderstanding. I didn't see it was either a octagon or a octahedron! :oops:
Last edited by Tasnood on Fri Feb 02, 2018 10:49 pm, edited 2 times in total.

samiul_samin
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Re: BDMO 2017 National round Secondary 6

Unread post by samiul_samin » Fri Feb 02, 2018 2:08 am

But what will be happened if we rotate this octahedron like upside down?

Absur Khan Siam
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Re: BDMO 2017 National round Secondary 6

Unread post by Absur Khan Siam » Fri Feb 02, 2018 10:41 pm

It isn't Octahedron" at all! It is "Octagon".
This is octahedron :
Image
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

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Tasnood
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Re: BDMO 2017 National round Secondary 6

Unread post by Tasnood » Fri Feb 02, 2018 11:11 pm

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?

Absur Khan Siam
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Re: BDMO 2017 National round Secondary 6

Unread post by Absur Khan Siam » Sat Feb 03, 2018 12:40 pm

Tasnood wrote:
Fri Feb 02, 2018 11:11 pm
Right this time?
Yah, I think so...
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

samiul_samin
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Re: BDMO 2017 National round Secondary 6

Unread post by samiul_samin » Mon Feb 25, 2019 6:03 pm

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

soyeb pervez jim
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Re: BDMO 2017 National round Secondary 6

Unread post by soyeb pervez jim » Tue Feb 26, 2019 12:32 am

The question is same to 2000 AMC 12 problem 25,

samiul_samin
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Re: BDMO 2017 National round Secondary 6

Unread post by samiul_samin » Tue Feb 26, 2019 8:47 am

soyeb pervez jim wrote:
Tue Feb 26, 2019 12:32 am
The question is same to 2000 AMC 12 problem 25,
Duplicate question again! $4$ duplicate questions in $1$ National problemset!!!!

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