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

Revision as of 13:46, 25 January 2023

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, $a = b+2$ and $b = c + p_2$ . 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

Solution 2

As above, we must deduce that the sum of two primes must be equal to the third prime. Then, we can finish the solution using casework. If the primes are $2,3,5$ , then the smallest number can range between $1$ and $15$ . If the primes are $2,5,7$ , then the smallest number can range between $1$ and $13$ . If the primes are $2,11,13$ , then the smallest number can range between $1$ and $7$ . If the primes are $2,17,19$ , then the smallest prime can only be $1$ .

Adding all cases gets $15+13+7+1=36$ . However, due to the commutative property, we must multiply this by 2. For example, in the $2,17,19$ case the numbers can be $1,3,20$ or $1,18,20$ . Therefore the answer is $36\cdot2=\boxed{072}$ .

Note about solution 1: I don't think that works, because if for example there are 21 points on the circle, your solution would yield $19\cdot4=76$ , but there would be $8$ more solutions than if there are $20$ points. This is because the upper bound for each case increases by $1$ , but commutative property doubles it to be $4$ .

Video Solution by Power of Logic

https://youtu.be/m2Cm9r5_Jvs

@@ Line 26: / Line 26: @@
 Note about solution 1: I don't think that works, because if for example there are 21 points on the circle, your solution would yield <math>19\cdot4=76</math>, but there would be <math>8</math> more solutions than if there are <math>20</math> points. This is because the upper bound for each case increases by <math>1</math>, but commutative property doubles it to be <math>4</math>.
+==Video Solution by Power of Logic==
+https://youtu.be/m2Cm9r5_Jvs
 ==See Also==
 {{AIME box|year=2022|n=II|num-b=4|num-a=6}}
 {{MAA Notice}}

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

Page

Toolbox

Search