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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 1==
 
==Solution 1==
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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>.
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We can solve this by finding all the combinations, then subtracting the ones that are on the same line. There are <math>25</math> points in all, from <math>(1,1)</math> to <math>(5,5)</math>, so <math>\dbinom{25}3</math> is <math>\frac{25\cdot 24\cdot 23}{3\cdot 2 \cdot 1}</math>, which simplifies to <math>2300</math>.
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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>(1,5)</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>.
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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>.
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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>.
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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>.
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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>
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==Solution 2 ==
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 +
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.
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There are <math>10 \times 10 = 100</math> cases where the 3 points chosen make up a vertical or horizontal line.
  
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==Solution 2==
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There are <math>2\left(1+\dbinom{4}3+\dbinom{5}3+\dbinom{4}3+1\right)=40</math> cases where the 3 points all land on the diagonals of the square.
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We can find out that the least number of digits the number <math>N</math> is <math>142</math>, with <math>141</math> <math>9</math>'s and one <math>5</math>.
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By randomly mixing the digits up, we are likely to get: <math>9999</math>...<math>9995999</math>...<math>9999</math>.
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There are <math>3 \times 4=12</math> ways where the 3 points make the a slope of <math>\frac{1}{2}</math>, <math>-\frac{1}{2}</math>, <math>2</math>, and <math>-2</math>.
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By adding 1 to this number, we get: <math>9999</math>...<math>9996000</math>...<math>0000</math>.
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−
We can subtract 6 from every available choice, and see if the number is divisible by 9 afterwards.
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Hence, there are <math>100+40+12=152</math> cases where the chosen 3 points make a line. The answer would be <math>2300-152=\boxed{(\textbf{B})\ 2148}</math>
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After subtracting 6 from every number, we can conclude that <math>1233</math> (originally <math>1239</math>) is the only number divisible by 9.  
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So our answer is <math>1239</math>
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~MrThinker
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==Video Solutions==
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Video Solutions:
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 +
https://www.youtube.com/watch?v=wfWsolGGfNY
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 +
https://www.youtube.com/watch?v=LCvDL-SMknI
  
 
==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}}
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 +
[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 23:08, 7 October 2023

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 1

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 $2300-120-16-4-12=\boxed{(\textbf{B})\ 2148}$

Solution 2

There are $5 \times 5 = 25$ total points in all. So, there are $\dbinom{25}3 = 2300$ 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 $10 \times 10 = 100$ cases where the 3 points chosen make up a vertical or horizontal line.

There are $2\left(1+\dbinom{4}3+\dbinom{5}3+\dbinom{4}3+1\right)=40$ cases where the 3 points all land on the diagonals of the square.

There are $3 \times 4=12$ ways where the 3 points make the a slope of $\frac{1}{2}$, $-\frac{1}{2}$, $2$, and $-2$.

Hence, there are $100+40+12=152$ cases where the chosen 3 points make a line. The answer would be $2300-152=\boxed{(\textbf{B})\ 2148}$

~MrThinker

Video Solutions

Video Solutions:

https://www.youtube.com/watch?v=wfWsolGGfNY

https://www.youtube.com/watch?v=LCvDL-SMknI

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
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