Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2 (extreme last-minute desperate guess))
Line 12: Line 12:
 
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. Note that there are <math>3</math> such lines, for each slope, present in the grid. In total, this 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. Note that there are <math>3</math> such lines, for each slope, present in the grid. In total, this 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>
−
 
−
==Solution 2 (extreme last-minute desperate guess) ==
 
−
We know there are <math>25</math> points, and we need to pick <math>3</math>. <math>\dbinom{25}3=2300</math>. But we notice that this includes colinear points. Answer choices <math>\textbf{(C)}</math> and <math>\textbf{(D)}</math> seem a bit too whole (perfect multiple of 10). <math>\textbf{(A)}</math>
 
−
seems unlikely, so it must be <math>\boxed{\textbf{(B) }2148}</math>
 
  
 
==See Also==
 
==See Also==

Revision as of 21:11, 28 August 2022

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

We can solve this by finding all the combinations, then subtracting the ones that are on the same line. There are $25$ points in all, from $(1,1)$ to $(5,5)$, so $\dbinom{25}3$ is $\frac{25\cdot 24\cdot 23}{3\cdot 2 \cdot 1}$, which simplifies to $2300$. Now we count the ones that are on the same line. We see that any three points chosen from $(1,1)$ and $(1,5)$ would be on the same line, so $\dbinom53$ is $10$, and there are $5$ rows, $5$ columns, and $2$ long diagonals, so that results in $120$. We can also count the ones with $4$ on a diagonal. That is $\dbinom43$, which is 4, and there are $4$ of those diagonals, so that results in $16$. We can count the ones with only $3$ on a diagonal, and there are $4$ diagonals like that, so that results in $4$. We can also count the ones with a slope of $\frac12$, $2$, $-\frac12$, or $-2$, with $3$ points in each. Note that there are $3$ such lines, for each slope, present in the grid. In total, this results in $12$. Finally, we subtract all the ones in a line from $2300$, so we have $2300-120-16-4-12=\boxed{(\mathbf{B})\text{ }2148}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_23&oldid=177636"