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

(Undo revision 29756 by Doonie (Talk))
(→‎Solution 2 rigorous)
 
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<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>
  
−
__TOC__
+
== Solution 1 ==
−
== Solution ==
+
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.
−
=== 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 <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.  
+
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.  
−
<center><asy>
+
<center>
−
unitsize(0.3inch);
+
<asy>
−
import three;  
+
import three;
−
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)
+
draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle);
−
  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));
−
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);
+
draw((2,0,0)--(2,0,2));
−
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);
−
/* 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);
 +
</asy>
 +
</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: 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 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>.
 +
<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);
 +
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>
 +
</center>
  
−
<center><asy>
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== Solution 2  ==
−
unitsize(0.3inch);
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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
−
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 */
 
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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); */
 
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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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=== Solution 2 (rigorous) ===
 
−
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>
−
 
 
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.
  
−
See [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=48 Math Jam] solution.
+
== See Also ==
−
 
 
−
{{solution}}
 
−
 
 
−
== 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 00:36, 26 May 2024

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 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 $\sqrt{2}$; 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 $9 \cdot 8 = 72$ 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: 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 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 $\frac{8 \cdot 2}{2} = 8$ of these equilateral triangles, for a total of $\boxed{\textbf{(C) }80}$.

[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 lengths of the equilateral triangle must be $\sqrt{d_x^2 + d_y^2 + d_z^2}, 0 \le d_x, d_y, d_z \le 2$, which yields the possible edge lengths of

$\sqrt{0^2+0^2+1^2}=\sqrt{1},\ \sqrt{0^2+1^2+1^2}=\sqrt{2},\ \sqrt{1^2+1^2+1^2}=\sqrt{3},$ $\ \sqrt{0^2+0^2+2^2}=\sqrt{4},\ \sqrt{0^2+1^2+2^2}=\sqrt{5},\ \sqrt{1^2+1^2+2^2}=\sqrt{6},$ $\ \sqrt{0^2+2^2+2^2}=\sqrt{8},\ \sqrt{1^2+2^2+2^2}=\sqrt{9},\ \sqrt{2^2+2^2+2^2}=\sqrt{12}$

Some casework shows that $\sqrt{2},\ \sqrt{6},\ \sqrt{8}$ are the only lengths that work, from which we can use the same counting argument as above.

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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