2022 AIME II Problems/Problem 5

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 $(15+13+7+1) \cdot 2 = \boxed{{072}}$ . It just so happens that $72 = 18 \cdot 4$ .

2022 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2022 AIME II Problems/Problem 5

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