Difference between revisions of "2006 AIME II Problems/Problem 8"

Revision as of 23:47, 12 January 2018

Problem

There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.

Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?

$[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy]$

Solution 1

If two of our big equilateral triangles have the same color for their center triangle and the same multiset of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection. Thus, to make two triangles distinct, they must differ either in their center triangle or in the collection of colors which make up their outer three triangles.

There are 6 possible colors for the center triangle.

There are ${6\choose3} = 20$ possible choices for the three outer triangles, if all three have different colors.
There are $6\cdot 5 = 30$ (or $2 {6\choose2}$ ) possible choices for the three outer triangles, if two are one color and the third is a different color.
There are ${6\choose1} = 6$ possible choices for the three outer triangles, if all three are the same color.

Thus, in total we have $6\cdot(20 + 30 + 6) = 336$ total possibilities.

Solution 2 (Easy burnsides)

We apply burnsides lemma and consider 3 rotations of 120 degrees, 240 degrees, and 0 degrees. We also consider three reflections from the three lines of symmetry in the triangle. Thus, we have to divide by $3+3=6$ for our final count.

Case 1: 0 degree rotation. This is known as the identity rotation, and there are $6^4=1296$ choices because we don't have any restrictions.

Case 2: 120 degree rotation. Note that the three "outer" sides of the triangle have to be the same during this, and the middle one can be anything. We have $6*6=36$ choices from this.

Case 3: 240 degree rotation. Similar to the 120 degree rotation, each must be the same except for the middle. We have $6*6=36$ choices from this.

Case 4: symmetry about lines. We multiply by 3 for these because the amount of colorings fixed under symmetry are the exact same each time. Two triangles do not change under this, and they must be the same. The other two triangles (1 middle and 1 outer) can be anything because they stay the same during the reflection. We have $6*6*6=216$ ways for one symmetry. There are 3 symmetries, so there are $216*3=648$ combinations in all.

Now, we add our cases up: $1296+36+36+648=2016$ . We have to divide by 6, so $2016/6=\boxed{336}$ distinguishable ways to color the triangle.

@@ Line 15: / Line 15: @@
 </asy>
-== Solution ==
+== Solution 1==
 If two of our big equilateral triangles have the same color for their center triangle and the same [[multiset]] of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection.  Thus, to make two triangles distinct, they must differ either in their center triangle or in the collection of colors which make up their outer three triangles.
@@ Line 25: / Line 25: @@
 Thus, in total we have <math>6\cdot(20 + 30 + 6) = 336</math> total possibilities.
+== Solution 2 (Easy burnsides) ==
+We apply burnsides lemma and consider 3 rotations of 120 degrees, 240 degrees, and 0 degrees. We also consider three reflections from the three lines of symmetry in the triangle. Thus, we have to divide by <math>3+3=6</math> for our final count.
+Case 1: 0 degree rotation. This is known as the identity rotation, and there are <math>6^4=1296</math> choices because we don't have any restrictions.
+Case 2: 120 degree rotation. Note that the three "outer" sides of the triangle have to be the same during this, and the middle one can be anything. We have <math>6*6=36</math> choices from this.
+Case 3: 240 degree rotation. Similar to the 120 degree rotation, each must be the same except for the middle. We have <math>6*6=36</math> choices from this.
+Case 4: symmetry about lines. We multiply by 3 for these because the amount of colorings fixed under symmetry are the exact same each time. Two triangles do not change under this, and they must be the same. The other two triangles (1 middle and 1 outer) can be anything because they stay the same during the reflection. We have <math>6*6*6=216</math> ways for one symmetry. There are 3 symmetries, so there are <math>216*3=648</math> combinations in all.
+Now, we add our cases up: <math>1296+36+36+648=2016</math>. We have to divide by 6, so <math>2016/6=\boxed{336}</math> distinguishable ways to color the triangle.
 == See also ==

2006 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

Difference between revisions of "2006 AIME II Problems/Problem 8"

Revision as of 23:47, 12 January 2018

Contents

Problem

Solution 1

Solution 2 (Easy burnsides)

See also