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Difference between revisions of "1981 AHSME Problems/Problem 22"
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<math>\textbf{(A)}\ 60\qquad\textbf{(B)}\ 64\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 100</math>
 
<math>\textbf{(A)}\ 60\qquad\textbf{(B)}\ 64\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 100</math>
  
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==Solution 1(casework)==
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==Solution 1==
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Restating the problem, we seek all the lines that will pass through (<math>i</math>, <math>j</math>, <math>k</math>), (<math>i + a</math>, <math>j + b</math>, <math>k + c</math>), (<math>i + 2a</math>, <math>j + 2b</math>, <math>k + 2c</math>), and (<math>i + 3a</math>, <math>j + 3b</math>, <math>k + 3c</math>), such that <math>i,j,k</math> are positive integers, <math>a,b,c</math> are integers, and all of our points are between 1 and 4, inclusive. With this constraint in mind, we realize that for each coordinate, we have three choices:
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# Set <math>a/b/c</math> to <math>0</math>. This then allows us to set the corresponding <math>i,j,k</math> to any number from <math>1</math> to <math>4</math>, inclusive.
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Let <math>(i, j, k)</math> be the first point whose coordinates are positive integers at most <math>4</math> that a line passes through when being traced in a certain direction.  Then the next three lattice points the line passes through must be in the form <math>(i + a, j + b, k + c)</math>, <math>(i + 2a, j + 2b, k + 2c)</math>, and <math>(i + 3a, j + 3b, k + 3c)</math>, where <math>a, b, c</math> are integers. 
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# Set <math>a/b/c</math> to <math>1</math>. This forces us to set the corresponding <math>i/j/k</math> to <math>1</math>.
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# Set <math>a/b/c</math> to <math>-1</math>. This forces us to set the corresponding <math>i/j/k</math> to <math>4</math>.
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Note that if <math>a \geq 2</math>, <math>i + 3a \geq 1 + 3 \cdot 2 = 7</math>, which is too large.  Therefore <math>a \leq 1</math>, and by similar logic <math>b \leq 1</math> and <math>c \leq 1</math>.  Also, if <math>a \leq -2</math>, <math>i + 3a \leq 4 + 3(-2) = -2</math>, which is too small.  Therefore, <math>a \geq -1</math>, and by similar logic <math>b \geq -1</math> and <math>c \geq -1</math>. So <math>a, b, c \in \{-1, 0, 1\}</math>.
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Note that options 2 and 3 will give us the same points if we mirror the assignments of each coordinate. Also note that we cannot set all three coordinates to not change, as that would be a point.
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<br>
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If <math>a = 1</math>, then <math>1 \leq i, i+1, i+2, i+3 \leq 4</math>.  In this case, only <math>i = 1</math> is valid.
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All of this gives us <math>6</math> ways to assign each coordinate, for a total of <math>216</math>. We then must subtract the ways to get a point (<math>4</math> ways per coordinate, for a total of <math>64</math>). This leaves us with <math>152</math>. Finally, we divide by <math>2</math> to account for mirror assignments giving us the same coordinate, for a final answer of <math>76</math>.
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<br>
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If <math>a = 0</math>, then <math>1 \leq i \leq 4</math>.  In this case, <math>i = 1, 2, 3, 4</math> are all valid.
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(This was my first solution, apologies if it is bad).
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If <math>a = -1</math>, then <math>1 \leq i, i-1, i-2, i-3 \leq 4</math>. In this case, only <math>i = 4</math> is valid.
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Therefore, <math>(a, i) \in {(1, 1), (0, 1), (0, 2), (0, 3), (0, 4), (-1, 4)}</math>. By similar logic, <math>(b, j)</math> and <math>(c, k)</math> must also be in this set.  
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If <math>a = b = c = 0</math>, then all four points are <math>(i, j, k)</math>, so at least one of <math>a, b, c</math> must be nonzero. Therefore, there are <math>6^3 - 4^3 = 216 - 64 = 152</math> choices for <math>(i, j, k, a, b, c)</math>. However, each line can be determined by two different values of <math>(i, j, k, a, b, c)</math>, as the line can be traced in two different directions. For instance, <math>(i, j, k, a, b, c) = (4, 1, 3, -1, 1, 0)</math> and <math>(i, j, k, a, b, c) = (1, 4, 3, 1, -1, 0)</math> determine the line containing <math>(4, 1, 3), (3, 2, 3), (2, 3, 3), (1, 4, 3)</math>. Therefore, there are <math>\frac{152}{2} = \boxed{(\textbf{D})\ 76}</math> such lines.
  
 
==See also==
 
==See also==

Revision as of 13:30, 20 August 2025

Problem 22

How many lines in a three dimensional rectangular coordinate system pass through four distinct points of the form $(i, j, k)$, where $i$, $j$, and $k$ are positive integers not exceeding four?

$\textbf{(A)}\ 60\qquad\textbf{(B)}\ 64\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 100$

Solution 1

Let $(i, j, k)$ be the first point whose coordinates are positive integers at most $4$ that a line passes through when being traced in a certain direction. Then the next three lattice points the line passes through must be in the form $(i + a, j + b, k + c)$, $(i + 2a, j + 2b, k + 2c)$, and $(i + 3a, j + 3b, k + 3c)$, where $a, b, c$ are integers.

Note that if $a \geq 2$, $i + 3a \geq 1 + 3 \cdot 2 = 7$, which is too large. Therefore $a \leq 1$, and by similar logic $b \leq 1$ and $c \leq 1$. Also, if $a \leq -2$, $i + 3a \leq 4 + 3(-2) = -2$, which is too small. Therefore, $a \geq -1$, and by similar logic $b \geq -1$ and $c \geq -1$. So $a, b, c \in \{-1, 0, 1\}$.

If $a = 1$, then $1 \leq i, i+1, i+2, i+3 \leq 4$. In this case, only $i = 1$ is valid.

If $a = 0$, then $1 \leq i \leq 4$. In this case, $i = 1, 2, 3, 4$ are all valid.

If $a = -1$, then $1 \leq i, i-1, i-2, i-3 \leq 4$. In this case, only $i = 4$ is valid.

Therefore, $(a, i) \in {(1, 1), (0, 1), (0, 2), (0, 3), (0, 4), (-1, 4)}$. By similar logic, $(b, j)$ and $(c, k)$ must also be in this set.

If $a = b = c = 0$, then all four points are $(i, j, k)$, so at least one of $a, b, c$ must be nonzero. Therefore, there are $6^3 - 4^3 = 216 - 64 = 152$ choices for $(i, j, k, a, b, c)$. However, each line can be determined by two different values of $(i, j, k, a, b, c)$, as the line can be traced in two different directions. For instance, $(i, j, k, a, b, c) = (4, 1, 3, -1, 1, 0)$ and $(i, j, k, a, b, c) = (1, 4, 3, 1, -1, 0)$ determine the line containing $(4, 1, 3), (3, 2, 3), (2, 3, 3), (1, 4, 3)$. Therefore, there are $\frac{152}{2} = \boxed{(\textbf{D})\ 76}$ such lines.

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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