Difference between revisions of "2005 AMC 12A Problems/Problem 25"
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| − | The three dimensional distance formula shows that the lengths of the equilateral triangle must be <math>\sqrt{d_x^2 + d_y^2 + d_z^2}, 0 \le d_x, d_y, d_z \le 2</math>, which yields the possible edge lengths of | + | The three-dimensional [[distance formula]] shows that the lengths of the equilateral triangle must be <math>\sqrt{d_x^2 + d_y^2 + d_z^2}, 0 \le d_x, d_y, d_z \le 2</math>, which yields the possible edge lengths of |
<center><math>\sqrt{0^2+0^2+1^2}=\sqrt{1},\ \sqrt{0^2+1^2+1^2}=\sqrt{2},\ \sqrt{1^2+1^2+1^2}=\sqrt{3},</math> <math>\ \sqrt{0^2+0^2+2^2}=\sqrt{4},\ \sqrt{0^2+1^2+2^2}=\sqrt{5},\ \sqrt{1^2+1^2+2^2}=\sqrt{6},</math> <math>\ \sqrt{0^2+2^2+2^2}=\sqrt{8},\ \sqrt{1^2+2^2+2^2}=\sqrt{9},\ \sqrt{2^2+2^2+2^2}=\sqrt{12}</math></center> | <center><math>\sqrt{0^2+0^2+1^2}=\sqrt{1},\ \sqrt{0^2+1^2+1^2}=\sqrt{2},\ \sqrt{1^2+1^2+1^2}=\sqrt{3},</math> <math>\ \sqrt{0^2+0^2+2^2}=\sqrt{4},\ \sqrt{0^2+1^2+2^2}=\sqrt{5},\ \sqrt{1^2+1^2+2^2}=\sqrt{6},</math> <math>\ \sqrt{0^2+2^2+2^2}=\sqrt{8},\ \sqrt{1^2+2^2+2^2}=\sqrt{9},\ \sqrt{2^2+2^2+2^2}=\sqrt{12}</math></center> | ||
| − | If we manage to be awesome, we can solve any IMO problem using Jason's theorem. | + | If we manage to be awesome, we can solve any IMO problem using [[Jason's theorem]]. |
Some casework shows that <math>\sqrt{2},\ \sqrt{6},\ \sqrt{8}</math> are the only lengths that work, from which we can use the same counting argument as above. | Some casework shows that <math>\sqrt{2},\ \sqrt{6},\ \sqrt{8}</math> are the only lengths that work, from which we can use the same counting argument as above. | ||
Revision as of 20:15, 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 
.
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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. 
