Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "{{duplicate|2021 Fall AMC 10B #21 and 2021 Fall AMC 12B #19}} ==Problem 19== Regular polygo...")
 
(→‎Remark)
 
(12 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 
{{duplicate|[[2021 Fall AMC 10B Problems#Problem 21|2021 Fall AMC 10B #21]] and [[2021 Fall AMC 12B Problems#Problem 19|2021 Fall AMC 12B #19]]}}
 
{{duplicate|[[2021 Fall AMC 10B Problems#Problem 21|2021 Fall AMC 10B #21]] and [[2021 Fall AMC 12B Problems#Problem 19|2021 Fall AMC 12B #19]]}}
−
==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?
+
==Problem==
 +
Regular polygons with <math>5,6,7,</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?
  
 
<math>(\textbf{A})\: 52\qquad(\textbf{B}) \: 56\qquad(\textbf{C}) \: 60\qquad(\textbf{D}) \: 64\qquad(\textbf{E}) \: 68</math>
 
<math>(\textbf{A})\: 52\qquad(\textbf{B}) \: 56\qquad(\textbf{C}) \: 60\qquad(\textbf{D}) \: 64\qquad(\textbf{E}) \: 68</math>
  
−
==Solution 1==
+
==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)
 +
 
 +
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.
  
−
This means that we will end up with <math>2</math> times the number of sides in the polygon with fewer sides.
+
Therefore the number of intersections is <math>2\cdot(3\cdot5+2\cdot6+1\cdot7)=\boxed{(\textbf{E}) \:68}</math>.
 +
 
 +
~kingofpineapplz
 +
 
 +
==Remark==
 +
For regular polygons with <math>5,6,7,</math> and <math>8</math> sides, the <math>68</math> points of intersection inside the circle are shown below:
 +
<asy>
 +
/* Made by MRENTHUSIASM */
  
 +
size(350);
 +
path p5 = polygon(5);
 +
path p6 = polygon(6);
 +
path p7 = rotate(180)*polygon(7);
 +
path p8 = polygon(8);
  
−
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.
+
draw(p5,red);
 +
draw(p6,green);
 +
draw(p7,blue);
 +
draw(p8,olive);
  
−
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.
+
draw(Circle(origin,1));
  
 +
dot(intersectionpoints(p5,p6),linewidth(2.5));
 +
dot(intersectionpoints(p5,p7),linewidth(2.5));
 +
dot(intersectionpoints(p5,p8),linewidth(2.5));
 +
dot(intersectionpoints(p6,p7),linewidth(2.5));
 +
dot(intersectionpoints(p6,p8),linewidth(2.5));
 +
dot(intersectionpoints(p7,p8),linewidth(2.5));
 +
</asy>
 +
~MRENTHUSIASM
  
−
Therefore the number of intersections is <math>2\cdot(3\cdot5+2\cdot6+1\cdot7)=\boxed{(\textbf{E}) \:68}</math>.
+
==Video Solution by Interstigation==
 +
https://youtu.be/7cfZwwYSttQ
  
−
~kingofpineapplz
+
~Interstigation
  
 +
==Video Solution 2 by WhyMath==
 +
https://youtu.be/5nHMBfDyvps
  
 +
~savannahsolver
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/yTQSKinIo8g
  
 
==See Also==
 
==See Also==
−
{{AMC10 box|year=2021 Fall|ab=B|num-a=20|num-b=18}}
 
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=22|num-b=20}}
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=22|num-b=20}}
 +
{{AMC12 box|year=2021 Fall|ab=B|num-a=20|num-b=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:11, 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.

Problem

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

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

Remark

For regular polygons with $5,6,7,$ and $8$ sides, the $68$ points of intersection inside the circle are shown below: [asy] /* Made by MRENTHUSIASM */ size(350); path p5 = polygon(5); path p6 = polygon(6); path p7 = rotate(180)*polygon(7); path p8 = polygon(8); draw(p5,red); draw(p6,green); draw(p7,blue); draw(p8,olive); draw(Circle(origin,1)); dot(intersectionpoints(p5,p6),linewidth(2.5)); dot(intersectionpoints(p5,p7),linewidth(2.5)); dot(intersectionpoints(p5,p8),linewidth(2.5)); dot(intersectionpoints(p6,p7),linewidth(2.5)); dot(intersectionpoints(p6,p8),linewidth(2.5)); dot(intersectionpoints(p7,p8),linewidth(2.5)); [/asy] ~MRENTHUSIASM

Video Solution by Interstigation

https://youtu.be/7cfZwwYSttQ

~Interstigation

Video Solution 2 by WhyMath

https://youtu.be/5nHMBfDyvps

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/yTQSKinIo8g

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_19&oldid=186463"