Difference between revisions of "2017 AMC 10A Problems/Problem 23"
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Finally, we subtract all the ones in a line from <math>2300</math>, so we have <math>2300-120-16-4-12=\boxed{(\textbf{B})\ 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{(\textbf{B})\ 2148}</math> | ||
− | ==Solution 2 ( | + | ==Solution 2 (Concise)== |
There are <math>5 \times 5 = 25</math> total points in all. So, there are <math>\dbinom{25}3 = 2300</math> ways to choose the three vertices for the triangle. However, there are some cases where they 3 points chosen results in a straight line. | There are <math>5 \times 5 = 25</math> total points in all. So, there are <math>\dbinom{25}3 = 2300</math> ways to choose the three vertices for the triangle. However, there are some cases where they 3 points chosen results in a straight line. | ||
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There are <math>3 \times 4=12</math> ways where the 3 points make the a slope of <math>\frac{1}{2}</math>, <math>-\frac{1}{2}</math>, <math>2</math>, and <math>-2</math>. | There are <math>3 \times 4=12</math> ways where the 3 points make the a slope of <math>\frac{1}{2}</math>, <math>-\frac{1}{2}</math>, <math>2</math>, and <math>-2</math>. | ||
− | Hence, there are <math>100+40+12=152</math> cases where the chosen 3 points make a line. | + | Hence, there are <math>100+40+12=152</math> cases where the chosen 3 points make a line. The answer would be <math>2300-152=\boxed{(\textbf{B})\ 2148}</math> |
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~MrThinker | ~MrThinker |
Revision as of 21:20, 13 October 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 1
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 (Concise)
There are total points in all. So, there are ways to choose the three vertices for the triangle. However, there are some cases where they 3 points chosen results in a straight line.
There are cases where the 3 points chosen make up a vertical or horizontal line.
There are cases where the 3 points all land on the diagonals of the square.
There are ways where the 3 points make the a slope of , , , and .
Hence, there are cases where the chosen 3 points make a line. The answer would be
~MrThinker
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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