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

Revision as of 03:56, 19 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 $a < b < c$ . $a - b = p_1$ $b - c = p_2$ $a - b = p_3 = a - b + b - c = p_1 + p_2$

Because $p_3 = p_1 + p_2$ , so $p_1$ or $p_2$ must be $2$ . Let $p_1 = 2$ , then $p_2 \in \{ 3, 5, 11, 17 \}$ .

Once $a$ is determined, $b$ and $c$ are determined. There are $18$ values of $a$ where $a+2 \le 20$ . Therefore the answer is $18 \cdot 4 = \boxed{\textbf{72}}$

~isabelchen

@@ Line 12: / Line 12: @@
 Because <math>p_3 = p_1 + p_2</math>, so <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_2 \in \{ 3, 5, 11, 17 \}</math>.
-To be continued......
+Once <math>a</math> is determined, <math>b</math> and <math>c</math> are determined. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{72}}</math>
 ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]

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

Revision as of 03:56, 19 February 2022

Problem

Solution 1

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