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

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== Solution ==
 
== Solution ==
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Let <math>x</math> denote the common side length of the rhombi.
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Let <math>y</math> denote one of the smaller interior [[angle]]s of rhombus <math> \mathcal{P} </math>. Then <math>x^2\sin(y)=\sqrt{2006}</math>.  We also see that <math>\displaystyle K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y</math>.  Thus <math>K</math> can be any positive integer in the [[interval]] <math>(0, 2\sqrt{2006})</math>.  
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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.
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<math>2\sqrt{2006} = \sqrt{8024}</math> and <math>89^2 = 7921 < 8024 < 8100 = 90^2</math>, so <math>K</math> can be any [[integer]] between 1 and 89, inclusive. Thus the number of positive values for <math>K</math> is 089.
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There are 6 possible colors for the center triangle.
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*There are <math>{6\choose3} = 20</math> possible choices for the three outer triangles, if all three have different colors.
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*There are <math>6\cdot 5 = 30</math> (or <math>2 {6\choose2}</math>) possible choices for the three outer triangles, if two are one color and the third is a different color.
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*There are <math>{6\choose1} = 6</math> possible choices for the three outer triangles, if all three are the same color.
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Thus, in total we have <math>6\cdot(20 + 30 + 6) = 336</math> total possibilities.
  
 
== See also ==
 
== See also ==

Revision as of 15:35, 25 September 2007

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?

Solution

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.

See also

2006 AIME I (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
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