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| ==Solution== | | ==Solution== |
| We can solve this by finding all the combinations, then subtraction 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>.
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| 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>(1,4)</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>.
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| 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>.
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| 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>.
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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. There are <math>12</math> of them, so that results in <math>12</math>.
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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{(\mathbf{B})\text{ }2148}</math>
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| ==See Also== | | ==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}} |
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
See Also
These problems are copyrighted © by the Mathematical Association of America. 
in the coordinate plane, where 
and 
are integers between 
and 
, inclusive?
