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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 21"

(Created page with "==Problem 21== Regular polygons with <math>5,</math> <math>6,</math> <math>7,</math> and <math>8{ }</math> sides are inscribed in the same circle. No two of the polygons share...")
 
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==Solution 1==
 
==Solution 1==
 
Imagine we have <math>2</math> regular polygons with <math>m</math> and <math>n</math> sides and <math>m>n</math> inscribed in a circle without sharing a vertex. We see that each side of the polygon with <math>n</math> sides (the polygon with fewer sides) will be intersected twice.
 
Imagine we have <math>2</math> regular polygons with <math>m</math> and <math>n</math> sides and <math>m>n</math> inscribed in a circle without sharing a vertex. We see that each side of the polygon with <math>n</math> sides (the polygon with fewer sides) will be intersected twice.
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(We can see this because to have a vertex of the m-gon on an arc subtended by a side of the n-gon, there will be one intersection to “enter” the arc and one to “exit” the arc. ~KingRavi)
  
  
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~kingofpineapplz
 
~kingofpineapplz
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==See Also==
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{{AMC10 box|year=2021 Fall|ab=B|num-a=22|num-b=20}}
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{{MAA Notice}}

Revision as of 00:57, 23 November 2021

Problem 21

Regular polygons with $5,$ $6,$ $7,$ and $8{ }$ sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?

$(\textbf{A})\: 52\qquad(\textbf{B}) \: 56\qquad(\textbf{C}) \: 60\qquad(\textbf{D}) \: 64\qquad(\textbf{E}) \: 68$

Solution 1

Imagine we have $2$ regular polygons with $m$ and $n$ sides and $m>n$ inscribed in a circle without sharing a vertex. We see that each side of the polygon with $n$ sides (the polygon with fewer sides) will be intersected twice. (We can see this because to have a vertex of the m-gon on an arc subtended by a side of the n-gon, there will be one intersection to “enter” the arc and one to “exit” the arc. ~KingRavi)


This means that we will end up with $2$ times the number of sides in the polygon with fewer sides.


If we have polygons with $5,$ $6,$ $7,$ and $8{ }$ sides, we need to consider each possible pair of polygons and count their intersections.

Throughout 6 of these pairs, the $5$-sided polygon has the least number of sides $3$ times, the $6$-sided polygon has the least number of sides $2$ times, and the $7$-sided polygon has the least number of sides $1$ time.


Therefore the number of intersections is $2\cdot(3\cdot5+2\cdot6+1\cdot7)=\boxed{(\textbf{E}) \:68}$.

~kingofpineapplz


See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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