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

Revision as of 12:49, 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

@@ Line 15: / Line 15: @@
 ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
-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.
-If you did the casework, it should come out to be <math>(15+13+7+1) \cdot 2 = \boxed{{072}}</math>.
 ==See Also==
 {{AIME box|year=2022|n=II|num-b=4|num-a=6}}
 {{MAA Notice}}

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

Revision as of 12:49, 30 August 2022

Problem

Solution

See Also