Revision as of 16:17, 15 April 2017

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

The easiest way to see the case where the lines are not parallel to the faces, is that 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 (if 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), so it can be $(1,1,3), (1,2,3), (1,3,3)$ . Accounting for permutations, there are $12$ ways, so there are $12 \cdot 4 = 48$ different lines for this case. For the second case, we look at all the lines where the $x$ , $y$ , or $z$ is the same for all the points in the line. WLOG, let the $x$ value stay the same throughout, and 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, $120 + 48 = \boxed{168}$

@@ Line 5: / Line 5: @@
 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.
-We look at the one from <math>(0,0,0)</math> to <math>(10,10,10)</math>. The lower endpoint of the desired lines must contain both a 0 and a 2 (if min > 0 then the point <math>(a-1,b-1,c-1)</math> will also be on the line for example, 2 applies to the other end), so it can be <math>(0,0,2), (0,1,2), (0,2,2)</math>. Accounting for permutations, there are <math>12</math> ways, so there are <math>12 \cdot 4 = 48</math> different lines for this case. The answer is, therefore, <math>120 + 48 = \boxed{168}</math>
+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>
 =See Also=
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Difference between revisions of "2017 AIME II Problems/Problem 14"

Revision as of 16:17, 15 April 2017

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