Difference between revisions of "2022 AIME II Problems/Problem 5"
(→Solution 1) |
(→Note: This solution seems incorrect.) |
||
Line 17: | Line 17: | ||
===Note: This solution seems incorrect.=== | ===Note: This solution seems incorrect.=== | ||
− | Although the answer is correct, solution 2 below is a more accurate way to approach this problem. | + | Although the answer is correct, solution 2 below is a more accurate way to approach this problem. I agree, I don't get how <math>a + 2 \leq 20</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 11:08, 28 January 2024
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.