Difference between revisions of "2005 AMC 12A Problems/Problem 25"
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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 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 <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 = | + | 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. |
<center><asy> | <center><asy> | ||
unitsize(0.3inch); | unitsize(0.3inch); |
Revision as of 20:05, 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 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.
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 the midpoints of three non-adjacent, non-parallel edges gives us more equilateral triangles.
unitsize(0.3inch); import three; currentprojection=perspective(5,4,3); pointpen = black; pathpen=black; pen l = linewidth(0.5); pen d = linewidth(0.8); int i,j,k; for(i=0;i<=2;++i) for(j=0;j<=2;++j) for(k=0;k<=2;++k) dot((i,j,k)); /* draw in cube edges */ D((0,0,0)--(0,0,2)--(0,2,2)--(0,2,0)--(0,0,0)--(0,2,0)--(2,2,0)--(2,0,0)--(0,0,0)--(2,0,0)--(2,0,2)--(0,0,2)--(0,0,0),l); D((2,2,2)--(0,2,2)--(0,2,0)--(2,2,0)--(2,2,2)--(2,2,0)--(2,0,0)--(2,0,2)--(2,2,2)--(2,0,2)--(0,0,2)--(0,2,2)--(2,2,2),l); /* draw in equilateral triangle */ D((1,0,0)--(2,2,1)--(0,1,2)--(1,0,0),d); /* darken points: D((1,0,0)^^(2,2,1)^^(0,1,2),d); */ (Error making remote request. Unknown error_msg)
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)
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.