2022 AIME II Problems/Problem 5
Contents
Problem
Twenty distinct points are marked on a circle and labeled through 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 points.
Solution 1
Let , , and be the vertex of a triangle that satisfies this problem, where .
. Because is the sum of two primes, and , or must be . Let , then . There are only primes less than : . Only plus equals another prime. .
Once is determined, and . There are values of where , and values of . Therefore the answer is
Note: This solution seems incorrect.
Although the answer is correct, solution 2 below is a more accurate way to approach this problem. I agree, I don't get how .
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 , then the smallest number can range between and . If the primes are , then the smallest number can range between and . If the primes are , then the smallest number can range between and . If the primes are , then the smallest number can only be .
Adding all cases gets . However, due to the commutative property, we must multiply this by 2. For example, in the case the numbers can be or . Therefore the answer is .
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 , but there would be more solutions than if there are points. This is because the upper bound for each case increases by , but commutative property doubles it to be .
Video Solution by Power of Logic
See Also
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 | ||
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