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Difference between revisions of "2017 AMC 10A Problems/Problem 23"

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==Solution==
 
==Solution==
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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>
 
  
 
==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}}

Revision as of 17:19, 3 February 2018

Problem

How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$, inclusive?

$\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300$


Solution

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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