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| − | ==Problem 21==
| + | #REDIRECT [[2021_Fall_AMC_12B_Problems/Problem_19]] |
| − | 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 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?
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| − | <math>(\textbf{A})\: 52\qquad(\textbf{B}) \: 56\qquad(\textbf{C}) \: 60\qquad(\textbf{D}) \: 64\qquad(\textbf{E}) \: 68</math>
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| − | ==Solution 1==
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| − | 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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| − | This means that we will end up with <math>2</math> times the number of sides in the polygon with fewer sides.
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| − | If we have polygons with <math>5,</math> <math>6,</math> <math>7,</math> and <math>8{ }</math> sides, we need to consider each possible pair of polygons and count their intersections.
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| − | Throughout 6 of these pairs, the <math>5</math>-sided polygon has the least number of sides <math>3</math> times, the <math>6</math>-sided polygon has the least number of sides <math>2</math> times, and the <math>7</math>-sided polygon has the least number of sides <math>1</math> time.
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| − | Therefore the number of intersections is <math>2\cdot(3\cdot5+2\cdot6+1\cdot7)=\boxed{(\textbf{E}) \:68}</math>.
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| − | ~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}}
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