Difference between revisions of "2022 AIME II Problems/Problem 5"
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Twenty distinct points are marked on a circle and labeled <math>1</math> through <math>20</math> 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 <math>20</math> points. | Twenty distinct points are marked on a circle and labeled <math>1</math> through <math>20</math> 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 <math>20</math> points. | ||
− | ==Solution== | + | ==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>. | ||
+ | <cmath>a - b = p_1</cmath> | ||
+ | <cmath>b - c = p_2</cmath> | ||
+ | <cmath>a - c = p_3</cmath> | ||
+ | |||
+ | <math>p_3 = a - c = a - b + b - c = p_1 + p_2</math>. Because <math>p_3</math> is the sum of two primes, <math>p_1</math> and <math>p_2</math>, <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_3 = p_2 + 2</math>. There are only <math>8</math> primes less than <math>20</math>: <math>2, 3, 5, 7, 11, 13, 17, 19</math>. Only <math>3, 5, 11, 17</math> plus <math>2</math> equals another prime. <math>p_2 \in \{ 3, 5, 11, 17 \}</math>. | ||
+ | |||
+ | Once <math>a</math> is determined, <math>a = b+2</math> and <math>b = c + p_2</math>. There are <math>18</math> values of <math>a</math> where <math>a+2 \le 20</math>, and <math>4</math> values of <math>p_2</math>. Therefore the answer is <math>18 \cdot 4 = \boxed{\textbf{072}}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ===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 <math>a + 2 \leq 20</math>. | ||
+ | |||
+ | ==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 <math>2,3,5</math>, then the smallest number can range between <math>1</math> and <math>15</math>. | ||
+ | If the primes are <math>2,5,7</math>, then the smallest number can range between <math>1</math> and <math>13</math>. | ||
+ | If the primes are <math>2,11,13</math>, then the smallest number can range between <math>1</math> and <math>7</math>. | ||
+ | If the primes are <math>2,17,19</math>, then the smallest number can only be <math>1</math>. | ||
+ | |||
+ | Adding all cases gets <math>15+13+7+1=36</math>. However, due to the commutative property, we must multiply this by 2. For example, in the <math>2,17,19</math> case the numbers can be <math>1,3,20</math> or <math>1,18,20</math>. Therefore the answer is <math>36\cdot2=\boxed{072}</math>. | ||
+ | |||
+ | 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/iI2ZpdpGNyc | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=II|num-b=4|num-a=6}} | {{AIME box|year=2022|n=II|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
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.