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Difference between revisions of "2005 AMC 12A Problems/Problem 25"

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== Problem ==
 
== Problem ==
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Let <math>S</math> be the [[set]] of all [[point]]s with [[coordinate]]s <math>(x,y,z)</math>, where <math>x</math>, <math>y</math>, and <math>z</math> are each chosen from the set <math>\{0,1,2\}</math>. How many [[equilateral]] [[triangle]]s all have their [[vertices]] in <math>S</math>?
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Let <math>S</math> be the set of all points with coordinates <math>(x,y,z)</math>, where <math>x</math>, <math>y</math>, and <math>z</math> are each chosen from the set <math>\{0,1,2\}</math>. How many equilateral triangles all have their vertices in <math>S</math>?
  
 
<math>(\mathrm {A}) \ 72\qquad (\mathrm {B}) \ 76 \qquad (\mathrm {C})\ 80 \qquad (\mathrm {D}) \ 84 \qquad (\mathrm {E})\ 88</math>
 
<math>(\mathrm {A}) \ 72\qquad (\mathrm {B}) \ 76 \qquad (\mathrm {C})\ 80 \qquad (\mathrm {D}) \ 84 \qquad (\mathrm {E})\ 88</math>
  
 
== Solution 1 ==
 
== Solution 1 ==
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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 size of triangles left.
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For this solution, we will just find as many equilateral triangles as possible, until it becomes intuitive that there are no more size of triangles left.
  
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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.  
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First, we observe that we can form an equilateral triangle with vertices in <math>S</math> by taking any point in <math>S</math> and connecting it to the <math>2</math> adjacent points. 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 (the red triangle in the diagram below). Each of these triangles is determined by one vertex of the cube, so in one cube we have <math>8</math> equilateral triangles. We have <math>8</math> unit cubes, as well as the entire <math>2 \times 2 \times 2</math> cube (giving the green triangle in the diagram), for a total of <math>8+1 = 9</math> cubes, and thus <math>9 \cdot 8 = 72</math> equilateral triangles.  
 
<center>
 
<center>
 
<asy>
 
<asy>
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</center>
 
</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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(Note that 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>.
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Now we look for any further equilateral triangles. Connecting the midpoints of <math>3</math> non-adjacent, non-parallel edges indeed gives us more equilateral triangles (e.g. the blue triangle in the diagram below).  Notice that picking these <math>3</math> edges leaves <math>2</math> vertices alone (labelled A and B in the diagram), and that picking any <math>2</math> opposite vertices determines <math>2</math> equilateral triangles. Hence there are <math>\frac{8 \cdot 2}{2} = 8</math> of these equilateral triangles, so adding these to the triangles already found above gives a total of <math>72+8 = \boxed{\textbf{(C) }80}</math>.
 
<center>
 
<center>
 
<asy>
 
<asy>
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</center>
 
</center>
  
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== Solution 2 rigorous ==
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== Solution 2 ==
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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 side length of the equilateral triangle must be <math>\sqrt{d_x^2 + d_y^2 + d_z^2}</math> with <math>\left\lvert d_x\right\rvert,\left\lvert d_y\right\rvert,\left\lvert d_z\right\rvert \in \{0-0,1-0,1-1,2-0,2-1,2-2\} = \{0,1,2\}</math>, so the possible side lengths are
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<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>
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<cmath>\begin{align*}&\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}, \\ &\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}, \\ &\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}.\end{align*}</cmath>
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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.
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 +
Some casework shows that <math>\sqrt{2}</math>, <math>\sqrt{6}</math>, and <math>\sqrt{8}</math> are the only lengths that work, after which we can complete the problem using the same counting argument as in Solution 1.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 15:35, 1 July 2025

Problem

Let $S$ be the set of all points with coordinates $(x,y,z)$, where $x$, $y$, and $z$ are each chosen from the set $\{0,1,2\}$. How many equilateral triangles all have their vertices in $S$?

$(\mathrm {A}) \ 72\qquad (\mathrm {B}) \ 76 \qquad (\mathrm {C})\ 80 \qquad (\mathrm {D}) \ 84 \qquad (\mathrm {E})\ 88$

Solution 1

For this solution, we will just find as many equilateral triangles as possible, until it becomes intuitive that there are no more size of triangles left.

First, we observe that we can form an equilateral triangle with vertices in $S$ by taking any point in $S$ and connecting it to the $2$ adjacent points. This triangle will have a side length of $\sqrt{2}$; a quick further examination of this cube will show us that this is the only possible side length (the red triangle in the diagram below). 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, as well as the entire $2 \times 2 \times 2$ cube (giving the green triangle in the diagram), for a total of $8+1 = 9$ cubes, and thus $9 \cdot 8 = 72$ equilateral triangles.

[asy] import three; unitsize(1cm); size(200); currentprojection=perspective(1/3,-1,1/2); draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle); draw((0,0,0)--(0,0,2)); draw((0,2,0)--(0,2,2)); draw((2,2,0)--(2,2,2)); draw((2,0,0)--(2,0,2)); draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle); draw((2,0,0)--(0,2,0)--(0,0,2)--cycle,green); draw((1,0,0)--(0,1,0)--(0,0,1)--cycle,red); label("$x=2$",(1,0,0),S); label("$z=2$",(2,2,1),E); label("$y=2$",(2,1,0),SE); [/asy]

(Note that connecting the centers of the faces will actually give an octahedron, not a cube, because it only has $6$ vertices.)

Now we look for any further equilateral triangles. Connecting the midpoints of $3$ non-adjacent, non-parallel edges indeed gives us more equilateral triangles (e.g. the blue triangle in the diagram below). Notice that picking these $3$ edges leaves $2$ vertices alone (labelled A and B in the diagram), and that picking any $2$ opposite vertices determines $2$ equilateral triangles. Hence there are $\frac{8 \cdot 2}{2} = 8$ of these equilateral triangles, so adding these to the triangles already found above gives a total of $72+8 = \boxed{\textbf{(C) }80}$.

[asy] import three; unitsize(1cm); size(200); currentprojection=perspective(1/3,-1,1/2); draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle); draw((0,0,0)--(0,0,2)); draw((0,2,0)--(0,2,2)); draw((2,2,0)--(2,2,2)); draw((2,0,0)--(2,0,2)); draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle); draw((1,0,0)--(2,2,1)--(0,1,2)--cycle,blue); label("$x=2$",(1,0,0),S); label("$z=2$",(2,2,1),E); label("$y=2$",(2,1,0),SE); label("$A$",(0,2,0), NW); label("$B$",(2,0,2), NW); [/asy]

Solution 2

The three-dimensional distance formula shows that the side length of the equilateral triangle must be $\sqrt{d_x^2 + d_y^2 + d_z^2}$ with $\left\lvert d_x\right\rvert,\left\lvert d_y\right\rvert,\left\lvert d_z\right\rvert \in \{0-0,1-0,1-1,2-0,2-1,2-2\} = \{0,1,2\}$, so the possible side lengths are \begin{align*}&\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}, \\ &\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}, \\ &\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}.\end{align*}

Some casework shows that $\sqrt{2}$, $\sqrt{6}$, and $\sqrt{8}$ are the only lengths that work, after which we can complete the problem using the same counting argument as in Solution 1.

See Also

2005 AMC 12A (Problems • Answer Key • Resources)
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