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

Revision as of 15:41, 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

@@ Line 5: / Line 5: @@
 ==Solution 1 ==
-Let <math>a</math>, <math>b</math>, and <math>c</math> be the vertex of a triangle that satisfies this problem, where <math>a < b < c</math>.
+Let <math>a</math>, <math>b</math>, and <math>c</math> be the vertex of a triangle that satisfies this problem, where <math>a > b > c</math>.
 <cmath>a - b = p_1</cmath>
 <cmath>b - c = p_2</cmath>

2022 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2022 AIME II Problems/Problem 5"

Revision as of 15:41, 20 February 2022

Problem

Solution 1

See Also