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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 19"
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{{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]]}}
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==Problem==
 
==Problem==
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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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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>
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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)
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(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.
  
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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.
 +
 
 +
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.
 +
 
 +
Therefore the number of intersections is <math>2\cdot(3\cdot5+2\cdot6+1\cdot7)=\boxed{(\textbf{E}) \:68}</math>.
  
 +
~kingofpineapplz
  
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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.
+
==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:
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<asy>
 +
/* Made by MRENTHUSIASM */
  
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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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size(350);
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path p5 = polygon(5);
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path p6 = polygon(6);
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path p7 = rotate(180)*polygon(7);
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path p8 = polygon(8);
  
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draw(p5,red);
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draw(p6,green);
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draw(p7,blue);
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draw(p8,olive);
  
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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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draw(Circle(origin,1));
  
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~kingofpineapplz
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dot(intersectionpoints(p5,p6),linewidth(2.5));
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dot(intersectionpoints(p5,p7),linewidth(2.5));
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dot(intersectionpoints(p5,p8),linewidth(2.5));
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dot(intersectionpoints(p6,p7),linewidth(2.5));
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dot(intersectionpoints(p6,p8),linewidth(2.5));
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dot(intersectionpoints(p7,p8),linewidth(2.5));
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</asy>
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~MRENTHUSIASM
  
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
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~savannahsolver
 
~savannahsolver
 +
==Video Solution by TheBeautyofMath==
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https://youtu.be/yTQSKinIo8g
  
 
==See Also==
 
==See Also==

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

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