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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>?
+
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>
  
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__TOC__
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== Solution 1 ==
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== Solution ==
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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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=== Solution 1 (non-rigorous) ===
 
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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.
 
  
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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.  
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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.  
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<center><asy>
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<center>
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unitsize(0.3inch);
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<asy>
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import three;  
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import three;
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currentprojection=perspective(5,4,3); pointpen = black; pathpen=black; pen l = linewidth(0.5); pen d = linewidth(0.8);
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unitsize(1cm);
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int i,j,k;
+
size(200);
−
for(i=0;i<=2;++i)
+
currentprojection=perspective(1/3,-1,1/2);
−
for(j=0;j<=2;++j)
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draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle);
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  for(k=0;k<=2;++k)
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draw((0,0,0)--(0,0,2));
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  dot((i,j,k));
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draw((0,2,0)--(0,2,2));
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/* draw in cube edges */
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draw((2,2,0)--(2,2,2));
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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);
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draw((2,0,0)--(2,0,2));
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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);  
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draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle);
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/* draw in equilateral triangle */
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draw((2,0,0)--(0,2,0)--(0,0,2)--cycle,green);
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D((2,0,2)--(0,0,0)--(2,2,0)--(2,0,2),d); D((0,0,1)--(1,0,2)--(0,1,2)--(0,0,1),d);  
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draw((1,0,0)--(0,1,0)--(0,0,1)--cycle,red);
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/* darken points: D((2,0,2)^^(0,0,0)^^(2,2,0),d); D((0,0,1)^^(1,0,2)^^(0,1,2),d); */
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label("$x=2$",(1,0,0),S);
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</asy></center>
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label("$z=2$",(2,2,1),E);
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label("$y=2$",(2,1,0),SE);
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</asy>
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</center>
  
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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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(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. 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.   
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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>
 +
<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);
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draw((1,0,0)--(2,2,1)--(0,1,2)--cycle,blue);
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label("$x=2$",(1,0,0),S);
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label("$z=2$",(2,2,1),E);
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label("$y=2$",(2,1,0),SE);
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label("$A$",(0,2,0), NW);
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label("$B$",(2,0,2), NW);
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</asy>
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</center>
  
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<center><asy>
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== Solution 2  ==
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unitsize(0.3inch);
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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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import three;
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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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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);
 
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/* draw in equilateral triangle */
 
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D((1,0,0)--(2,2,1)--(0,1,2)--(1,0,0),d);
 
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/* darken points: D((1,0,0)^^(2,2,1)^^(0,1,2),d); */
 
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</asy></center>
 
  
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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>.  
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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.
  
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=== Solution 2 (rigorous) ===
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== See Also ==
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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
 
−
<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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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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See [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=48 Math Jam] solution.
 
−
 
 
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== 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]]
 +
{{MAA Notice}}

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)
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

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