Difference between revisions of "2017 AMC 10A Problems/Problem 23"
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We can also count the ones with a slope of <math>\frac12</math>, <math>2</math>, <math>-\frac12</math>, or <math>-2</math>, with <math>3</math> points in each. Note that there are <math>3</math> such lines, for each slope, present in the grid. In total, this results in <math>12</math>. | We can also count the ones with a slope of <math>\frac12</math>, <math>2</math>, <math>-\frac12</math>, or <math>-2</math>, with <math>3</math> points in each. Note that there are <math>3</math> such lines, for each slope, present in the grid. In total, this results in <math>12</math>. | ||
Finally, we subtract all the ones in a line from <math>2300</math>, so we have <math>2300-120-16-4-12=\boxed{(\mathbf{B})\text{ }2148}</math> | Finally, we subtract all the ones in a line from <math>2300</math>, so we have <math>2300-120-16-4-12=\boxed{(\mathbf{B})\text{ }2148}</math> | ||
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+ | ==Solution 2 (extreme last minute desperate guess) (if you can call it a solution)== | ||
+ | We know that there are 25 points, and we need to pick 3. <math>\dbinom{25}3=2300</math>. But we notice that this includes colinear points. Answer choices C and D seem a bit too whole (perfect multiple of 10). We do an "eenie meenie miney mo" between A and B and arrive at B. | ||
==See Also== | ==See Also== |
Revision as of 23:03, 13 June 2022
Contents
[hide]Problem
How many triangles with positive area have all their vertices at points in the coordinate plane, where and are integers between and , inclusive?
Solution
We can solve this by finding all the combinations, then subtracting the ones that are on the same line. There are points in all, from to , so is , which simplifies to . Now we count the ones that are on the same line. We see that any three points chosen from and would be on the same line, so is , and there are rows, columns, and long diagonals, so that results in . We can also count the ones with on a diagonal. That is , which is 4, and there are of those diagonals, so that results in . We can count the ones with only on a diagonal, and there are diagonals like that, so that results in . We can also count the ones with a slope of , , , or , with points in each. Note that there are such lines, for each slope, present in the grid. In total, this results in . Finally, we subtract all the ones in a line from , so we have
Solution 2 (extreme last minute desperate guess) (if you can call it a solution)
We know that there are 25 points, and we need to pick 3. . But we notice that this includes colinear points. Answer choices C and D seem a bit too whole (perfect multiple of 10). We do an "eenie meenie miney mo" between A and B and arrive at B.
See Also
Video Solution:
https://www.youtube.com/watch?v=wfWsolGGfNY
https://www.youtube.com/watch?v=LCvDL-SMknI
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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All AMC 10 Problems and Solutions |
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