Difference between revisions of "2005 AMC 12A Problems/Problem 25"
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First, try to make three of its vertices form an equilateral triangle. This we find is possible by taking any [[vertex]], and connecting the three adjacent vertices into a triangle. This triangle will have a side length of <math>\sqrt{2}</math>; a quick further examination of this cube will show us that this is the only possible side length (red triangle in diagram). Each of these triangles is determined by one vertex of the cube, so in one cube we have 8 equilateral triangles. We have 8 unit cubes, and then the entire cube (green triangle), giving us 9 cubes and <math>9 \cdot 8 = 72</math> equilateral triangles. | First, try to make three of its vertices form an equilateral triangle. This we find is possible by taking any [[vertex]], and connecting the three adjacent vertices into a triangle. This triangle will have a side length of <math>\sqrt{2}</math>; a quick further examination of this cube will show us that this is the only possible side length (red triangle in diagram). Each of these triangles is determined by one vertex of the cube, so in one cube we have 8 equilateral triangles. We have 8 unit cubes, and then the entire cube (green triangle), giving us 9 cubes and <math>9 \cdot 8 = 72</math> equilateral triangles. | ||
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label("$y=2$",(2,1,0),SE); | label("$y=2$",(2,1,0),SE); | ||
</asy> | </asy> | ||
+ | </center> | ||
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+ | NOTE: Connecting the centers of the faces will actually give an [[octahedron]], not a cube, because it only has <math>6</math> vertices. | ||
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+ | Now, we look for any additional equilateral triangles. Connecting the midpoints of three non-adjacent, non-parallel edges also gives us more equilateral triangles (blue triangle). Notice that picking these three edges leaves two vertices alone (labelled A and B), and that picking any two opposite vertices determine two equilateral triangles. Hence there are <math>\frac{8 \cdot 2}{2} = 8</math> of these equilateral triangles, for a total of <math>\boxed{\textbf{(C) }80}</math>. | ||
+ | <center> | ||
<asy> | <asy> | ||
import three; | import three; |
Revision as of 21:32, 30 January 2015
Problem
Let be the set of all points with coordinates , where , , and are each chosen from the set . How many equilateral triangles all have their vertices in ?
Solution
Solution 1 (non-rigorous)
For this solution, we will just find as many solutions as possible, until it becomes intuitive that there are no more size of triangles left.
First, try to make three of its vertices form an equilateral triangle. This we find is possible by taking any vertex, and connecting the three adjacent vertices into a triangle. This triangle will have a side length of ; a quick further examination of this cube will show us that this is the only possible side length (red triangle in diagram). Each of these triangles is determined by one vertex of the cube, so in one cube we have 8 equilateral triangles. We have 8 unit cubes, and then the entire cube (green triangle), giving us 9 cubes and equilateral triangles.
NOTE: Connecting the centers of the faces will actually give an octahedron, not a cube, because it only has vertices.
Now, we look for any additional equilateral triangles. Connecting the midpoints of three non-adjacent, non-parallel edges also gives us more equilateral triangles (blue triangle). Notice that picking these three edges leaves two vertices alone (labelled A and B), and that picking any two opposite vertices determine two equilateral triangles. Hence there are of these equilateral triangles, for a total of .
Solution 2 (rigorous)
The three-dimensional distance formula shows that the lengths of the equilateral triangle must be , which yields the possible edge lengths of
Some casework shows that are the only lengths that work, from which we can use the same counting argument as above.
See also
2005 AMC 12A (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.