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

(Problem 23)
(Solution)
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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>3</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>3</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. There are <math>12</math> of them, so that 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. There are <math>12</math> of them, so that 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{(A) 2128}</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{A}) 2128}</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 17:18, 22 November 2017

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

We can solve this by finding all the combinations, then subtraction the ones that are on the same line. There are $25$ points in all, from $(1,1)$ to $(5,5)$, so $\dbinom{25}3$ is $\frac{25\cdot 24\cdot 23}{3\cdot 2 \cdot 1}$, which simplifies to $2300$. Now we count the ones that are on the same line. We see that any three points chosen from $(1,1)$ and $(1,4)$ would be on the same line, so $\dbinom53$ is $10$, and there are $5$ rows, $5$ columns, and $2$ long diagonals, so that results in $120$. We can also count the ones with $4$ on a diagonal. That is $\dbinom43$, which is 4, and there are $4$ of those diagonals, so that results in $16$. We can count the ones with only $3$ on a diagonal, and there are $4$ diagonals like that, so that results in $3$. We can also count the ones with a slope of $\frac12$, $2$, $-\frac12$, or $-2$, with $3$ points in each. There are $12$ of them, so that results in $12$. Finally, we subtract all the ones in a line from $2300$, so we have $2300-120-16-4-12=\boxed{(\mathbf{A}) 2128}$

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 10 Problems and Solutions

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