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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> sets of three points. However, some of them form degenerate triangles (i.e., they have area of 0) if the three points are collinear. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), which contributes <math>\binom{5}{3} \cdot 12 = 120</math> degenerate triangles, 4 lines that go through exactly 4 points, which contributes <math>\binom{4}{3} \cdot 4 = 16</math> degenerate triangles, and 16 lines that go through exactly three points, which contributes <math>\binom{3}{3} \cdot 16 = 16</math> degenerate triangles. Subtracting these degenerate triangles, we get an answer of <math>2300-120-16-16=2300-152=\boxed{\textbf{(B) }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 23:12, 8 October 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?

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

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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