Difference between revisions of "2021 Fall AMC 12B Problems/Problem 19"
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==Solution== | ==Solution== | ||
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. | ||
| − | (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) | + | (We can see this because to have a vertex of the <math>m</math>-gon on an arc subtended by a side of the <math>n</math>-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. | This means that we will end up with <math>2</math> times the number of sides in the polygon with fewer sides. | ||
| + | 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. | ||
| − | + | Throughout <math>6</math> 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>. | Therefore the number of intersections is <math>2\cdot(3\cdot5+2\cdot6+1\cdot7)=\boxed{(\textbf{E}) \:68}</math>. | ||
~kingofpineapplz | ~kingofpineapplz | ||
| + | |||
| + | ==Remark== | ||
==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
Revision as of 01:25, 12 January 2023
- 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.
Contents
Problem
Regular polygons with 
and 
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?
Solution
Imagine we have 
regular polygons with 
and 
sides and 
inscribed in a circle without sharing a vertex. We see that each side of the polygon with sides (the polygon with fewer sides) will be intersected twice.
(We can see this because to have a vertex of the -gon on an arc subtended by a side of the -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 times the number of sides in the polygon with fewer sides.
If we have polygons with and 
sides, we need to consider each possible pair of polygons and count their intersections.
Throughout 
of these pairs, the 
-sided polygon has the least number of sides 
times, the -sided polygon has the least number of sides times, and the 
-sided polygon has the least number of sides 
time.
Therefore the number of intersections is 
.
~kingofpineapplz
Remark
Video Solution by Interstigation
~Interstigation
Video Solution 2 by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
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 | ||
| 2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
