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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==
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==Solution 1(casework)==
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No solutions yet!
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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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inputting...
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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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# 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 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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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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(This was my first solution, apologies if it is bad).

Latest revision as of 19:46, 9 September 2024

Problem

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(casework)

Restating the problem, we seek all the lines that will pass through ($i$, $j$, $k$), ($i + a$, $j + b$, $k + c$), ($i + 2a$, $j + 2b$, $k + 2c$), and ($i + 3a$, $j + 3b$, $k + 3c$), such that $i,j,k$ are positive integers, $a,b,c$ 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:

  1. Set $a/b/c$ to $0$. This then allows us to set the corresponding $i,j,k$ to any number from $1$ to $4$, inclusive.
  2. Set $a/b/c$ to $1$. This forces us to set the corresponding $i/j/k$ to $1$.
  3. Set $a/b/c$ to $-1$. This forces us to set the corresponding $i/j/k$ to $4$.

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.
All of this gives us $6$ ways to assign each coordinate, for a total of $216$. We then must subtract the ways to get a point ($4$ ways per coordinate, for a total of $64$). This leaves us with $152$. Finally, we divide by $2$ to account for mirror assignments giving us the same coordinate, for a final answer of $76$.
(This was my first solution, apologies if it is bad).

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