Difference between revisions of "2017 AMC 10A Problems/Problem 23"
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− | ==Solution== | + | ==Solution 1== |
We can solve this by finding all the combinations, then subtracting the ones that are on the same line. There are <math>25</math> points in all, from <math>(1,1)</math> to <math>(5,5)</math>, so <math>\dbinom{25}3</math> is <math>\frac{25\cdot 24\cdot 23}{3\cdot 2 \cdot 1}</math>, which simplifies to <math>2300</math>. | We can solve this by finding all the combinations, then subtracting the ones that are on the same line. There are <math>25</math> points in all, from <math>(1,1)</math> to <math>(5,5)</math>, so <math>\dbinom{25}3</math> is <math>\frac{25\cdot 24\cdot 23}{3\cdot 2 \cdot 1}</math>, which simplifies to <math>2300</math>. | ||
Now we count the ones that are on the same line. We see that any three points chosen from <math>(1,1)</math> and <math>(1,5)</math> would be on the same line, so <math>\dbinom53</math> is <math>10</math>, and there are <math>5</math> rows, <math>5</math> columns, and <math>2</math> long diagonals, so that results in <math>120</math>. | Now we count the ones that are on the same line. We see that any three points chosen from <math>(1,1)</math> and <math>(1,5)</math> would be on the same line, so <math>\dbinom53</math> is <math>10</math>, and there are <math>5</math> rows, <math>5</math> columns, and <math>2</math> long diagonals, so that results in <math>120</math>. | ||
We can also count the ones with <math>4</math> on a diagonal. That is <math>\dbinom43</math>, which is 4, and there are <math>4</math> of those diagonals, so that results in <math>16</math>. | We can also count the ones with <math>4</math> on a diagonal. That is <math>\dbinom43</math>, which is 4, and there are <math>4</math> of those diagonals, so that results in <math>16</math>. | ||
We can count the ones with only <math>3</math> on a diagonal, and there are <math>4</math> diagonals like that, so that results in <math>4</math>. | We can count the ones with only <math>3</math> on a diagonal, and there are <math>4</math> diagonals like that, so that results in <math>4</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. | + | 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{(\ | + | 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 == |
− | Video Solution: | + | |
− | https://www.youtube.com/watch?v= | + | 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>10 \times 10 = 100</math> cases where the 3 points chosen make up a vertical or horizontal line. | ||
+ | |||
+ | There are <math>2\left(1+\dbinom{4}3+\dbinom{5}3+\dbinom{4}3+1\right)=40</math> cases where the 3 points all land on the diagonals of the square. | ||
+ | |||
+ | 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. The answer would be <math>2300-152=\boxed{(\textbf{B})\ 2148}</math> | ||
+ | |||
+ | ~MrThinker | ||
+ | |||
+ | ==Video Solution by Pi Academy== | ||
+ | |||
+ | https://youtu.be/7XAw78CZ3fY?si=HxmOK65niihjrpmq | ||
+ | |||
+ | ~ Pi Academy | ||
+ | |||
+ | ==Video Solutions== | ||
+ | Video Solutions: | ||
+ | |||
+ | https://www.youtube.com/watch?v=wfWsolGGfNY | ||
+ | https://www.youtube.com/watch?v=LCvDL-SMknI | ||
+ | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2017|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] |
Latest revision as of 19:54, 11 October 2024
Contents
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
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
Video Solution by Pi Academy
https://youtu.be/7XAw78CZ3fY?si=HxmOK65niihjrpmq
~ Pi Academy
Video Solutions
Video Solutions:
https://www.youtube.com/watch?v=wfWsolGGfNY
https://www.youtube.com/watch?v=LCvDL-SMknI
See Also
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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