Difference between revisions of "2017 AMC 10A Problems/Problem 23"
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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>. | 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>. | ||
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>. | 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>. | ||
| − | 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> | + | 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>. |
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{(\mathbf{B})\text{ }2148}</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{B})\text{ }2148}</math> | ||
Revision as of 11:56, 9 December 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
We can solve this by finding all the combinations, then subtraction the ones that are on the same line. There are 
points in all, from 
to 
, so 
is 
, which simplifies to 
.
Now we count the ones that are on the same line. We see that any three points chosen from and 
would be on the same line, so 
is 
, and there are rows, columns, and 
long diagonals, so that results in 
.
We can also count the ones with 
on a diagonal. That is 
, which is 4, and there are of those diagonals, so that results in 
.
We can count the ones with only 
on a diagonal, and there are diagonals like that, so that results in .
We can also count the ones with a slope of 
, , 
, or 
, with points in each. There are 
of them, so that results in .
Finally, we subtract all the ones in a line from , so we have 
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. 
