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, $120 + 48 = \boxed{168}$

2017 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2017 AIME II Problems/Problem 14

Problem

Solution

See Also