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Difference between revisions of "2017 AIME II Problems/Problem 14"

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A <math>10\times10\times10</math> grid of points consists of all points in space of the form <math>(i,j,k)</math>, where <math>i</math>, <math>j</math>, and <math>k</math> are integers between <math>1</math> and <math>10</math>, inclusive. Find the number of different lines that contain exactly <math>8</math> of these points.
 
A <math>10\times10\times10</math> grid of points consists of all points in space of the form <math>(i,j,k)</math>, where <math>i</math>, <math>j</math>, and <math>k</math> are integers between <math>1</math> and <math>10</math>, inclusive. Find the number of different lines that contain exactly <math>8</math> of these points.
  
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==Solution==
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==Solution 1==
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The easiest way to see the case where the lines are not parallel to the faces, is that a line through the point <math>(a,b,c)</math> must contain <math>(a \pm 1, b \pm 1, c \pm 1)</math> on it as well, as otherwise the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.
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<math>Case \textrm{ } 1:</math> The lines are not parallel to the faces
  
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We look at the one from <math>(1,1,1)</math> to <math>(10,10,10)</math>. The lower endpoint of the desired lines must contain both a 1 and a 3 (if min > 0 then the point <math>(a-1,b-1,c-1)</math> will also be on the line for example, 3 applies to the other end), so it can be <math>(1,1,3), (1,2,3), (1,3,3)</math>. Accounting for permutations, there are <math>12</math> ways, so there are <math>12 \cdot 4 = 48</math> different lines for this case. For the second case, we look at all the lines where the <math>x</math>, <math>y</math>, or <math>z</math> is the same for all the points in the line. WLOG, let the <math>x</math> value stay the same throughout, and let the line be parallel to the diagonal from <math>(1,1,1)</math> to <math>(1,10,10)</math>. For the line to have 8 points, the <math>y</math> and <math>z</math> must be 1 and 3 in either order, and the <math>x</math> value can be any value from 1 to 10. In addition, this line can be parallel to 6 face diagonals. So we get <math>2 \cdot 10 \cdot 6 = 120</math> possible lines for this case. The answer is, therefore, <math>120 + 48 = \boxed{168}</math>
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A line through the point <math>(a,b,c)</math> must contain <math>(a \pm 1, b \pm 1, c \pm 1)</math> on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.
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We look at the one from <math>(1,1,1)</math> to <math>(10,10,10)</math>. The lower endpoint of the desired lines must contain both a 1 and a 3, so it can be <math>(1,1,3), (1,2,3), (1,3,3)</math>. If <math>\textrm{min} > 0</math> then the point <math>(a-1,b-1,c-1)</math> will also be on the line for example, 3 applies to the other end.
 +
 
 +
Accounting for permutations, there are <math>12</math> ways, so there are <math>12 \cdot 4 = 48</math> different lines for this case.  
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<math>Case \textrm{ } 2:</math> The lines where the <math>x</math>, <math>y</math>, or <math>z</math> is the same for all the points on the line.  
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WLOG, let the <math>x</math> value stay the same throughout. Let the line be parallel to the diagonal from <math>(1,1,1)</math> to <math>(1,10,10)</math>. For the line to have 8 points, the <math>y</math> and <math>z</math> must be 1 and 3 in either order, and the <math>x</math> value can be any value from 1 to 10. In addition, this line can be parallel to 6 face diagonals. So we get <math>2 \cdot 10 \cdot 6 = 120</math> possible lines for this case.  
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 +
The answer is, therefore, <math>120 + 48 = \boxed{168}</math>
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==Solution 2==
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Look at one pair of opposite faces of the cube. There are <math>4</math> lines say <math>l_1, l_2, l_3, l_4</math> with exactly <math>8</math> collinear points on the top face. For each of these lines, draw a rectangular plane that consists of one of the <math>l_i</math> for <math>1 \leq i \leq 4</math> and perpendicular to the top face.
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There are <math>16</math> lines in total on this plane. <math>10</math> of which are parallel to one of the edges of the rectangular plane and <math>6</math> of which are diagonals. There are <math>3</math> pairs of opposite faces. So <math>3 \cdot 4 \cdot 16=192</math> lines.
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But we are overcounting the lines of the diagonals of those rectangular planes twice. There are <math>4</math> rectangular planes perpendicular to one pair of opposite faces. Thus <math>4 \cdot 6=24</math> lines are overcounted.
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So the answer is <math>192-24=\boxed{168}</math>.
  
 
=See Also=
 
=See Also=
 
{{AIME box|year=2017|n=II|num-b=13|num-a=15}}
 
{{AIME box|year=2017|n=II|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:27, 9 November 2020

Problem

A $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$, where $i$, $j$, and $k$ are integers between $1$ and $10$, inclusive. Find the number of different lines that contain exactly $8$ of these points.

Solution 1

$Case \textrm{ } 1:$ The lines are not parallel to the faces

A line through the point $(a,b,c)$ must contain $(a \pm 1, b \pm 1, c \pm 1)$ on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.

We look at the one from $(1,1,1)$ to $(10,10,10)$. The lower endpoint of the desired lines must contain both a 1 and a 3, so it can be $(1,1,3), (1,2,3), (1,3,3)$. If $\textrm{min} > 0$ then the point $(a-1,b-1,c-1)$ will also be on the line for example, 3 applies to the other end.

Accounting for permutations, there are $12$ ways, so there are $12 \cdot 4 = 48$ different lines for this case.


$Case \textrm{ } 2:$ The lines where the $x$, $y$, or $z$ is the same for all the points on the line.

WLOG, let the $x$ value stay the same throughout. Let the line be parallel to the diagonal from $(1,1,1)$ to $(1,10,10)$. For the line to have 8 points, the $y$ and $z$ must be 1 and 3 in either order, and the $x$ value can be any value from 1 to 10. In addition, this line can be parallel to 6 face diagonals. So we get $2 \cdot 10 \cdot 6 = 120$ possible lines for this case.

The answer is, therefore, $120 + 48 = \boxed{168}$

Solution 2

Look at one pair of opposite faces of the cube. There are $4$ lines say $l_1, l_2, l_3, l_4$ with exactly $8$ collinear points on the top face. For each of these lines, draw a rectangular plane that consists of one of the $l_i$ for $1 \leq i \leq 4$ and perpendicular to the top face.

There are $16$ lines in total on this plane. $10$ of which are parallel to one of the edges of the rectangular plane and $6$ of which are diagonals. There are $3$ pairs of opposite faces. So $3 \cdot 4 \cdot 16=192$ lines.

But we are overcounting the lines of the diagonals of those rectangular planes twice. There are $4$ rectangular planes perpendicular to one pair of opposite faces. Thus $4 \cdot 6=24$ lines are overcounted.

So the answer is $192-24=\boxed{168}$.

See Also

2017 AIME II (Problems • Answer Key • Resources)
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Problem 13 		Followed by
Problem 15
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