Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 AMC 12 Problems/Problem 25"

(→‎Solution 5 (Graph Theory))
m (→‎Solution 4)
(8 intermediate revisions by 2 users not shown)
Line 90: Line 90:
 
If we look at the base of an octahedron lying flat on a table, we can see there are 8 orbits since there are 8 colors to choose from as the base of the octahedron. We can also see that there are 3 stabilizers that keep the base the same color with the 0<math>^{\circ}</math> rotation, 120<math>^{\circ}</math> rotation, and 240<math>^{\circ}</math> rotation about the base. Using the orbit-stabilizer theorem, we then know that the number of rotational symmetries of an octahedron is <math>8 \cdot 3 = 24</math>. There are 8! ways to color the octahedron, and since rotations are indistinguishable, the answer comes out to be <math>8!/24 \Rightarrow \mathrm{(E)}</math>
 
If we look at the base of an octahedron lying flat on a table, we can see there are 8 orbits since there are 8 colors to choose from as the base of the octahedron. We can also see that there are 3 stabilizers that keep the base the same color with the 0<math>^{\circ}</math> rotation, 120<math>^{\circ}</math> rotation, and 240<math>^{\circ}</math> rotation about the base. Using the orbit-stabilizer theorem, we then know that the number of rotational symmetries of an octahedron is <math>8 \cdot 3 = 24</math>. There are 8! ways to color the octahedron, and since rotations are indistinguishable, the answer comes out to be <math>8!/24 \Rightarrow \mathrm{(E)}</math>
  
−
~Hithere22702
+
~kempwood
  
 
==Solution 5 (Graph Theory)==
 
==Solution 5 (Graph Theory)==
Line 96: Line 96:
 
[[File:2000AMC12P25.png|500px|center]]
 
[[File:2000AMC12P25.png|500px|center]]
  
−
This problem can be approached by [https://en.wikipedia.org/wiki/Graph_theory Graph Theory]. Note that each face of the octahedron is connected to 3 other faces. We use the following graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge.
+
This problem can be approached by [https://en.wikipedia.org/wiki/Graph_theory Graph Theory]. Note that each face of the octahedron is connected to 3 other faces. We use the above graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge.
  
−
Now the problem becomes how many distinguishable ways are there to color the <math>8</math> vertices of the graph above where two colored graphs are distinguishable if neither can be rotated to look just like the other.
+
Now the problem becomes how many distinguishable ways to color the <math>8</math> vertices such that two colored graphs are distinguishable if neither can be rotated and reflected to become the other.
  
−
Notice that once the outer 4 vertices with the hallow circles are colored, no matter how the inner 4 vertices with the black circles are colored, the resulting graphs are distinguishable graphs.
+
Notice that once the outer 4 vertices are colored, no matter how the inner 4 vertices are colored, the resulting graphs are distinguishable graphs.
  
−
There are <math>\binom{8}{4}</math> ways to color outer 4 vertices with the hallow circles.
+
There are <math>8</math> colors and <math>4</math> outer vertices, therefore there are <math>\binom{8}{4}</math> ways to color outer 4 vertices. Combination is used because the coloring has to be distinguishable when rotated and reflected. There are <math>4</math> colors left, therefore there are <math>4!</math> ways to color inner 4 vertices. Permutation is used because the coloring of the inner vertices have no restrictions. In total that is <math>\binom{8}{4} \cdot 4! = \boxed{\textbf{(E) } 1680 }</math>
  
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
−
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
+
==Solution 6==
 +
Let the colors be <math>1</math> to <math>8</math> inclusive, then rotate the octahedron such that color <math>1</math> is on top. You have <math>7</math> choices of what color is on the bottom, WLOG <math>2</math>. Then, there's two rings of each <math>3</math> colors on the top and bottom. For the top ring, you can choose any <math>3</math> out of the <math>6</math> remaining colors, and there's two ways to orient them. The octahedron is now fixed in place, so you can have <math>3!</math> ways to put the three remaining colors in three spaces.
 +
In total this is <math>7 \cdot \binom{6}{3} \cdot 2 \cdot 3!=\boxed{\textbf{(E) } 1680 }</math>
 +
 
 +
-mathfan2020
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 20:04, 23 December 2023

Problem

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.)

[asy] import three; import math; unitsize(1.5cm); currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7)); [/asy]

$\textbf {(A)}\ 210 \qquad \textbf {(B)}\ 560 \qquad \textbf {(C)}\ 840 \qquad \textbf {(D)}\ 1260 \qquad \textbf {(E)}\ 1680$

Solution 1

Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.

[asy] size(8cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0.2,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); draw(A--B--E--cycle); draw(A--C--D--cycle); draw(F--C--B--cycle); draw(F--D--E--cycle,dotted+linewidth(0.7)); draw(surface(A--B--C--cycle),rgb(1,.6,.6),nolight);[/asy]

There are $7!$ ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is $7!/3 = 1680$.

[asy] size(8cm); defaultpen(0.5); import three; import math; currentprojection=orthographic(2,0,1); triple A=(0,0,1); triple B=(sqrt(2)/2,sqrt(2)/2,0); triple C=(sqrt(2)/2,-sqrt(2)/2,0); triple D=(-sqrt(2)/2,-sqrt(2)/2,0); triple E=(-sqrt(2)/2,sqrt(2)/2,0); triple F=(0,0,-1); triple right=(0,1,0); picture p = new picture, r = new picture, s = new picture; draw(p,A--B--E--cycle); draw(p,A--C--D--cycle); draw(p,F--C--B--cycle); draw(p,F--D--E--cycle,dotted+linewidth(0.7)); draw(p,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(p,surface(A--B--E--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*p); draw(r,A--B--E--cycle); draw(r,A--C--D--cycle); draw(r,F--C--B--cycle); draw(r,F--D--E--cycle,dotted+linewidth(0.7)); draw(r,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(r,surface(A--C--D--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(2*right)*r); draw(s,A--B--E--cycle); draw(s,A--C--D--cycle); draw(s,F--C--B--cycle); draw(s,F--D--E--cycle,dotted+linewidth(0.7)); draw(s,surface(A--B--C--cycle),rgb(1,.6,.6),nolight); draw(s,surface(B--C--F--cycle),rgb(1,1,.6),nolight); add(scale3(2.2)*shift(4*right)*s); [/asy]

Solution 2

We consider the dual of the octahedron, the cube; a cube can be inscribed in an octahedron with each of its vertices at a face of the octahedron. So the problem is equivalent to finding the number of ways to color the vertices of a cube.

Select any vertex and call it $A$; there are $8$ color choices for this vertex, but this vertex can be rotated to any of $8$ locations. After fixing $A$, we pick another vertex $B$ adjacent to $A$. There are seven color choices for $B$, but there are only three locations to which $B$ can be rotated to (since there are three edges from $A$). The remaining six vertices can be colored in any way and their locations are now fixed. Thus the total number of ways is $\frac{8}{8} \cdot \frac{7}{3} \cdot 6! = 1680 \Rightarrow \mathrm{(E)}$.

Solution 3

There are 8! ways to place eight colors on a fixed octahedron. An octahedron has six vertices, of which one can face the top, and for any vertex that faces the top, there are four different triangles around that vertex that can be facing you. Thus there are 6*4 = 24 ways to orient an octahedron, and $8!/24 = 1680 \Rightarrow \mathrm{(E)}$

Solution 4

If we look at the base of an octahedron lying flat on a table, we can see there are 8 orbits since there are 8 colors to choose from as the base of the octahedron. We can also see that there are 3 stabilizers that keep the base the same color with the 0$^{\circ}$ rotation, 120$^{\circ}$ rotation, and 240$^{\circ}$ rotation about the base. Using the orbit-stabilizer theorem, we then know that the number of rotational symmetries of an octahedron is $8 \cdot 3 = 24$. There are 8! ways to color the octahedron, and since rotations are indistinguishable, the answer comes out to be $8!/24 \Rightarrow \mathrm{(E)}$

~kempwood

Solution 5 (Graph Theory)

2000AMC12P25.png

This problem can be approached by Graph Theory. Note that each face of the octahedron is connected to 3 other faces. We use the above graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge.

Now the problem becomes how many distinguishable ways to color the $8$ vertices such that two colored graphs are distinguishable if neither can be rotated and reflected to become the other.

Notice that once the outer 4 vertices are colored, no matter how the inner 4 vertices are colored, the resulting graphs are distinguishable graphs.

There are $8$ colors and $4$ outer vertices, therefore there are $\binom{8}{4}$ ways to color outer 4 vertices. Combination is used because the coloring has to be distinguishable when rotated and reflected. There are $4$ colors left, therefore there are $4!$ ways to color inner 4 vertices. Permutation is used because the coloring of the inner vertices have no restrictions. In total that is $\binom{8}{4} \cdot 4! = \boxed{\textbf{(E) } 1680 }$

~isabelchen

Solution 6

Let the colors be $1$ to $8$ inclusive, then rotate the octahedron such that color $1$ is on top. You have $7$ choices of what color is on the bottom, WLOG $2$. Then, there's two rings of each $3$ colors on the top and bottom. For the top ring, you can choose any $3$ out of the $6$ remaining colors, and there's two ways to orient them. The octahedron is now fixed in place, so you can have $3!$ ways to put the three remaining colors in three spaces. In total this is $7 \cdot \binom{6}{3} \cdot 2 \cdot 3!=\boxed{\textbf{(E) } 1680 }$

-mathfan2020

Video Solution

https://youtu.be/8WRpCVwQNBo

See Also

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_25&oldid=208259"