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

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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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==Solution 2==
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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.
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If the primes are <math>2,3,5</math>, then the smallest number can range between <math>1</math> and <math>15</math>.
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If the primes are <math>2,5,7</math>, then the smallest number can range between <math>1</math> and <math>13</math>.
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If the primes are <math>2,11,13</math>, then the smallest number can range between <math>1</math> and <math>7</math>.
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If the primes are <math>2,17,19</math>, then the smallest prime can only be <math>1</math>.
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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>.
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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>.
  
 
==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 13:42, 26 December 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

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 $b$ where $b+2 \le 20$, 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 $4$.

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. AMC logo.png

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