Difference between revisions of "2017 AMC 10A Problems/Problem 23"
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Finally, we subtract all the ones in a line from <math>2300</math>, so we have <math>2300-120-16-4-12=\boxed{(\textbf{B})\ 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{(\textbf{B})\ 2148}</math> | ||
| − | ==Solution 2 | + | ==Solution 2 == |
There are <math>5 \times 5 = 25</math> total points in all. So, there are <math>\dbinom{25}3 = 2300</math> ways to choose the three vertices for the triangle. However, there are some cases where they 3 points chosen results in a straight line. | There are <math>5 \times 5 = 25</math> total points in all. So, there are <math>\dbinom{25}3 = 2300</math> ways to choose the three vertices for the triangle. However, there are some cases where they 3 points chosen results in a straight line. | ||
Revision as of 11:49, 8 November 2022
Contents
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 1
We can solve this by finding all the combinations, then subtracting 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. Note that there are such lines, for each slope, present in the grid. In total, this results in 
.
Finally, we subtract all the ones in a line from , so we have 
Solution 2
There are 
total points in all. So, there are 
ways to choose the three vertices for the triangle. However, there are some cases where they 3 points chosen results in a straight line.
There are 
cases where the 3 points chosen make up a vertical or horizontal line.
There are 
cases where the 3 points all land on the diagonals of the square.
There are 
ways where the 3 points make the a slope of 
, 
, , and .
Hence, there are 
cases where the chosen 3 points make a line. The answer would be 
~MrThinker
See Also
Video Solution:
https://www.youtube.com/watch?v=wfWsolGGfNY
https://www.youtube.com/watch?v=LCvDL-SMknI
| 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 | ||
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