Difference between revisions of "2017 AIME II Problems/Problem 14"
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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. | 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 | + | 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> |
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=13|num-a=15}} | {{AIME box|year=2017|n=II|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:45, 23 March 2017
Problem
A grid of points consists of all points in space of the form , where , , and are integers between and , inclusive. Find the number of different lines that contain exactly 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 must contain 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 to . The lower endpoint of the desired lines must contain both a 0 and a 2 (if min > 0 then the point will also be on the line for example, 2 applies to the other end), so it can be . Accounting for permutations, there are ways, so there are different lines for this case. The answer is, therefore,
See Also
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 |
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