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

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<math>\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300</math>
 
<math>\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300</math>
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==Solution==
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There are a total of <math>\binom{25}{3}=2300</math> triangles. However, some of them are degenerate if <math>3</math> or more points lie on the same line. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), 4 lines with slope <math>\pm 1</math> that go through exactly 4 points, and 16 lines (<math>3*4+4</math>) with slope <math>\pm 0.5, \pm 1, \pm 2</math>. Subtracting these degenerate triangles, we get an answer of <math>2300-12*10-4*4-16=2180-32=\boxed{\textbf{(B)}2148}</math>.

Revision as of 17:38, 8 February 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

There are a total of $\binom{25}{3}=2300$ triangles. However, some of them are degenerate if $3$ or more points lie on the same line. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), 4 lines with slope $\pm 1$ that go through exactly 4 points, and 16 lines ($3*4+4$) with slope $\pm 0.5, \pm 1, \pm 2$. Subtracting these degenerate triangles, we get an answer of $2300-12*10-4*4-16=2180-32=\boxed{\textbf{(B)}2148}$.

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