Difference between revisions of "2005 AMC 12A Problems/Problem 25"
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Let <math>S</math> be the [[set]] of all [[point]]s with [[coordinate]]s <math>(x,y,z)</math>, where x, y, and z are each chosen from the set <math>\{0,1,2\}</math>. How many [[equilateral]] [[triangle]]s all have their [[vertices]] in <math>S</math>? | Let <math>S</math> be the [[set]] of all [[point]]s with [[coordinate]]s <math>(x,y,z)</math>, where x, y, and z are each chosen from the set <math>\{0,1,2\}</math>. How many [[equilateral]] [[triangle]]s all have their [[vertices]] in <math>S</math>? | ||
| − | <math> | + | <math>(\mathrm {A}) \ 72\qquad (\mathrm {B}) \ 76 \qquad (\mathrm {C})\ 80 \qquad (\mathrm {D}) \ 84 \qquad (\mathrm {E})\ 88</math> |
| − | (\mathrm {A}) \ 72\qquad (\mathrm {B}) \ 76 \qquad (\mathrm {C})\ 80 \qquad (\mathrm {D}) \ 84 \qquad (\mathrm {E})\ 88 | ||
| − | </math> | ||
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Take an unit cube. We 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. 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, giving us 9 cubes and <math>9 \cdot 8 = 72</math> equilateral triangles. | Take an unit cube. We 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. 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, giving us 9 cubes and <math>9 \cdot 8 = 72</math> equilateral triangles. | ||
| − | + | {{image}} | |
It may be tempting to connect the centers of the faces and to call that a cube, but a quick look at this tells us that that figure is actually an [[octahedron]]. | It may be tempting to connect the centers of the faces and to call that a cube, but a quick look at this tells us that that figure is actually an [[octahedron]]. | ||
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Now, we look for any additional equilateral triangles. Since the space diagonals of the cube cannot make an equilateral triangle, we will assume symmetry in the cube. A bit more searching shows us that connecting three non-adjacent, non-parallel edges, and connecting their midpoints gives us another equilateral triangle. | Now, we look for any additional equilateral triangles. Since the space diagonals of the cube cannot make an equilateral triangle, we will assume symmetry in the cube. A bit more searching shows us that connecting three non-adjacent, non-parallel edges, and connecting their midpoints gives us another equilateral triangle. | ||
| − | + | {{image}} | |
Notice that picking these three edges leaves two vertices alone, and that picking any two opposite vertices determine two equilateral triangles. Hence there are <math>\frac{8 \cdot 2}{2} = 8</math> additional equilateral triangles, for a total of <math>80 \Longrightarrow \mathrm{(C)}</math>. | Notice that picking these three edges leaves two vertices alone, and that picking any two opposite vertices determine two equilateral triangles. Hence there are <math>\frac{8 \cdot 2}{2} = 8</math> additional equilateral triangles, for a total of <math>80 \Longrightarrow \mathrm{(C)}</math>. | ||
=== Solution 2 (rigorous) === | === Solution 2 (rigorous) === | ||
| − | See [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=48 Math Jam] solution | + | See [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=48 Math Jam] solution. |
| − | + | {{solution}} | |
| + | {{image}} | ||
== See also == | == See also == | ||
{{AMC12 box|year=2005|ab=A|num-b=24|after=Last question}} | {{AMC12 box|year=2005|ab=A|num-b=24|after=Last question}} | ||
| − | |||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 02:44, 23 April 2008
Problem
Let 
be the set of all points with coordinates 
, where x, y, and z 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 triangles left.
Take an unit cube. We 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. 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, giving us 9 cubes and 
equilateral triangles.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
It may be tempting to connect the centers of the faces and to call that a cube, but a quick look at this tells us that that figure is actually an octahedron.
Now, we look for any additional equilateral triangles. Since the space diagonals of the cube cannot make an equilateral triangle, we will assume symmetry in the cube. A bit more searching shows us that connecting three non-adjacent, non-parallel edges, and connecting their midpoints gives us another equilateral triangle.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Notice that picking these three edges leaves two vertices alone, and that picking any two opposite vertices determine two equilateral triangles. Hence there are 
additional equilateral triangles, for a total of 
.
Solution 2 (rigorous)
See Math Jam solution.
This problem needs a solution. If you have a solution for it, please help us out by adding it.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
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 | |
