BDMO 2017 National round Secondary 6
- ahmedittihad
- Posts:181
- Joined:Mon Mar 28, 2016 6:21 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.
- Soumitra Das
- Posts:5
- Joined:Mon Apr 03, 2017 1:59 pm
Re: BDMO 2017 National round Secondary 6
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
OK. A very big misunderstanding. I didn't see it was either a octagon or a octahedron!
Last edited by Tasnood on Fri Feb 02, 2018 10:49 pm, edited 2 times in total.
-
- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Re: BDMO 2017 National round Secondary 6
But what will be happened if we rotate this octahedron like upside down?
-
- Posts:65
- Joined:Tue Dec 08, 2015 4:25 pm
- Location:Bashaboo , Dhaka
Re: BDMO 2017 National round Secondary 6
It isn't Octahedron" at all! It is "Octagon".
This is octahedron :
This is octahedron :
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
Re: BDMO 2017 National round Secondary 6
He wants to mean:Soumitra Das wrote: ↑Wed Apr 05, 2017 2:29 pmWe 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.
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?
-
- Posts:65
- Joined:Tue Dec 08, 2015 4:25 pm
- Location:Bashaboo , Dhaka
Re: BDMO 2017 National round Secondary 6
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid
-
- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Re: BDMO 2017 National round Secondary 6
.$LaTeX$ edSoumitra Das wrote: ↑Wed Apr 05, 2017 2:29 pmWe 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 $\frac{8!}{24}=1680$.
-
- Posts:21
- Joined:Sat Jan 28, 2017 11:06 pm
Re: BDMO 2017 National round Secondary 6
The question is same to 2000 AMC 12 problem 25,
-
- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Re: BDMO 2017 National round Secondary 6
Duplicate question again! $4$ duplicate questions in $1$ National problemset!!!!