Difference between revisions of "1981 AHSME Problems/Problem 22"

m (Solution 1(casework))
(Solution 1(casework))
 
Line 9: Line 9:
 
# 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>.
 
# 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>.
 
# 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>.
 
# 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>.
Note that options 2 and 3 will give us the same coordinates 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.  
+
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.  
 
<br>
 
<br>
 
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>.
 
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>.
 
<br>
 
<br>
 
(This was my first solution, apologies if it is bad).
 
(This was my first solution, apologies if it is bad).

Latest revision as of 20:46, 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).