Difference between revisions of "2021 Fall AMC 12B Problems/Problem 19"

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Revision as of 10:40, 24 November 2021

The following problem is from both the 2021 Fall AMC 10B #21 and 2021 Fall AMC 12B #19, so both problems redirect to this page.

Problem 19

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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
2021 Fall AMC 12B (ProblemsAnswer KeyResources)
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 12 Problems and Solutions

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