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

(Solution 1)
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==Solution 1 ==
 
==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</math> < <math>b</math> < <math>c</math>.
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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>.
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<cmath>a - b = p_1</cmath>
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<cmath>b - c = p_2</cmath>
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<cmath>a - b = p_3 = a - b + b - c = p_1 + p_2</cmath>
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Because <math>p_3 = p_1 + p_2</math>, so <math>p_1</math> or <math>p_2</math> must be <math>2</math>. Let <math>p_1 = 2</math>, then <math>p_2 \in \{ 3, 5, 11, 17 \}</math>.
  
 
To be continued......
 
To be continued......

Revision as of 03:53, 19 February 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 1

Let $a$, $b$, and $c$ be the vertex of a triangle that satisfies this problem, where $a < b < c$. \[a - b = p_1\] \[b - c = p_2\] \[a - b = p_3 = a - b + b - c = p_1 + p_2\]

Because $p_3 = p_1 + p_2$, so $p_1$ or $p_2$ must be $2$. Let $p_1 = 2$, then $p_2 \in \{ 3, 5, 11, 17 \}$.

To be continued......

~isabelchen

See Also

2022 AIME II (ProblemsAnswer KeyResources)
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

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