2021 Fall AMC 10B Problems/Problem 21

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.

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

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2021 Fall AMC 10B Problems/Problem 21

Problem 21

Solution 1