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

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<math>p_3 = a - c = a - b + b - c = p_1 + p_2</math>. Because <math>p_3</math> is the sum of two primes, <math>p_1</math> and <math>p_2</math>, <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_3 = p_2 + 2</math>. There are only <math>8</math> primes less than <math>20</math>: <math>2, 3, 5, 7, 11, 13, 17, 19</math>. Only <math>3, 5, 11, 17</math> plus <math>2</math> equals another prime. <math>p_2 \in \{ 3, 5, 11, 17 \}</math>.
 
<math>p_3 = a - c = a - b + b - c = p_1 + p_2</math>. Because <math>p_3</math> is the sum of two primes, <math>p_1</math> and <math>p_2</math>, <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_3 = p_2 + 2</math>. There are only <math>8</math> primes less than <math>20</math>: <math>2, 3, 5, 7, 11, 13, 17, 19</math>. Only <math>3, 5, 11, 17</math> plus <math>2</math> equals another prime. <math>p_2 \in \{ 3, 5, 11, 17 \}</math>.
  
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Once <math>a</math> is determined, <math>b = a+2</math> and <math>c = b + p_2</math>. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>, and <math>4</math> values of <math>p_2</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{72}}</math>
+
Once <math>a</math> is determined, <math>b = a+2</math> and <math>c = b + p_2</math>. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>, and <math>4</math> values of <math>p_2</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{072}}</math>
  
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]

Revision as of 06:42, 20 February 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 1

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

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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