Difference between revisions of "2017 AMC 10A Problems/Problem 23"
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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 linesthat 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>. | 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 linesthat 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>. | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2017|ab=A|num-b=22|num-a=24}} | ||
| + | {{MAA Notice}} | ||
Revision as of 20:06, 8 February 2017
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
There are a total of 
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 
degenerate triangles, 4 linesthat go through exactly 4 points, which contributes 
degenerate triangles, and 16 lines that go through exactly three points, which contributes 
degenerate triangles. Subtracting these degenerate triangles, we get an answer of 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
