Difference between revisions of "2005 AMC 12A Problems/Problem 25"
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== Solution == | == Solution == | ||
=== Solution 1 (non-rigorous) === | === 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. | + | 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 <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. | |
− | + | NOTE: Connect the centers of the faces will actually give an [[octahedron]], not a cube, because it only has <math>6</math> vertices. | |
− | Now, we look for any additional equilateral triangles. | + | Now, we look for any additional equilateral triangles. Connecting the midpoints of three non-adjacent, non-parallel edges also gives us more equilateral triangles. 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> of these equilateral triangles, for a total of <math>80 \Longrightarrow \mathrm{(C)}</math>. |
− | + | {{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> | ||
=== Solution 2 (rigorous) === | === Solution 2 (rigorous) === |
Revision as of 20:12, 7 November 2013
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. 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.
NOTE: Connect 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. Notice that picking these three edges leaves two vertices alone, and that picking any two opposite vertices determine two equilateral triangles. Hence there are of these equilateral triangles, for a total of .
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
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
If we manage to be awesome, we can solve any IMO problem using Jason's theorem. Some casework shows that are the only lengths that work, from which we can use the same counting argument as above.
See Math Jam solution.
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.