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Difference between revisions of "2022 AIME II Problems/Problem 5"

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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
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Note: I believe that you also need <math>20 \geq a</math>, so you cannot just simply do <math>18 \cdot 4</math>. In addition, it is possible for <cmath>b - c = 2</cmath> <cmath>a - b = p_1</cmath> to occur. You would need to do casework to make sure <math>a</math> does not go above 20.
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Note: I believe that you also need <math>20 \geq a</math>, so you cannot just simply do <math>18 \cdot 4</math>. In addition, it is possible for <cmath>b - c = 2</cmath> <cmath>a - b = p_1</cmath> to occur. You would need to do casework to make sure <math>a</math> does not go above 20.
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If you did the casework, it should come out to be <math>(15+13+7+1) \cdot 2 = \boxed{\textbf{072}}</math>. It just so happens that <math>72 = 18 \cdot 4</math>.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=II|num-b=4|num-a=6}}
 
{{AIME box|year=2022|n=II|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:46, 30 August 2022

Problem

Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points.

Solution

Let $a$, $b$, and $c$ be the vertex of a triangle that satisfies this problem, where $a > b > c$. \[a - b = p_1\] \[b - c = p_2\] \[a - c = p_3\]

$p_3 = a - c = a - b + b - c = p_1 + p_2$. Because $p_3$ is the sum of two primes, $p_1$ and $p_2$, $p_1$ or $p_2$ must be $2$. Let $p_1 = 2$, then $p_3 = p_2 + 2$. There are only $8$ primes less than $20$: $2, 3, 5, 7, 11, 13, 17, 19$. Only $3, 5, 11, 17$ plus $2$ equals another prime. $p_2 \in \{ 3, 5, 11, 17 \}$.

Once $a$ is determined, $b = a+2$ and $c = b + p_2$. There are $18$ values of $a$ where $a+2 \le 20$, and $4$ values of $p_2$. Therefore the answer is $18 \cdot 4 = \boxed{\textbf{072}}$

~isabelchen

Note: I believe that you also need $20 \geq a$, so you cannot just simply do $18 \cdot 4$. In addition, it is possible for \[b - c = 2\] \[a - b = p_1\] to occur. You would need to do casework to make sure $a$ does not go above 20.

If you did the casework, it should come out to be $(15+13+7+1) \cdot 2 = \boxed{\textbf{072}}$. It just so happens that $72 = 18 \cdot 4$.

See Also

2022 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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