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Difference between revisions of "2021 AMC 12B Problems/Problem 15"

(→‎Problem 20)
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==Problem 20==
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{{duplicate|[[2021 AMC 10B Problems#Problem 20|2021 AMC 10B #20]] and [[2021 AMC 12B Problems#Problem 15|2021 AMC 12B #15]]}}
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The figure is constructed from <math>11</math> line segments, each of which has length <math>2</math>. The area of pentagon <math>ABCDE</math> can be written is <math>\sqrt{m} + \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers. What is <math>m + n ?</math>
+
 
 +
==Problem==
 +
The figure is constructed from <math>11</math> line segments, each of which has length <math>2</math>. The area of pentagon <math>ABCDE</math> can be written as <math>\sqrt{m} + \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers. What is <math>m + n ?</math>
 
<asy>
 
<asy>
 
/* Made by samrocksnature */
 
/* Made by samrocksnature */
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label("A",A,N);
 
label("A",A,N);
 
label("B",B,W);
 
label("B",B,W);
−
label("C",C,W);
+
label("C",C,S);
−
label("D",D,E);
+
label("D",D,S);
 
label("E",E,dir(0));
 
label("E",E,dir(0));
 
dot(A^^B^^C^^D^^E^^F^^G);
 
dot(A^^B^^C^^D^^E^^F^^G);
Line 24: Line 26:
 
<math>\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24</math>
 
<math>\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24</math>
  
−
==Solution==
+
==Solution 1==
 +
 
 +
<asy>
 +
/* Made by samrocksnature, adapted by Tucker, then adjusted by samrocksnature again, then adjusted by erics118 xD*/
 +
pair A=(-2.4638,4.10658);
 +
pair B=(-4,2.6567453480756127);
 +
pair C=(-3.47132,0.6335248637894945);
 +
pair D=(-1.464483379039766,0.6335248637894945);
 +
pair E=(-0.956630463955801,2.6567453480756127);
 +
pair F=(-1.85,2);
 +
pair G=(-3.1,2);
 +
draw(A--G--A--F, lightgray);
 +
draw(B--F--C, lightgray);
 +
draw(E--G--D, lightgray);
 +
dot(F^^G, lightgray);
 +
draw(A--B--C--D--E--A);
 +
draw(A--C--A--D);
 +
label("A",A,N);
 +
label("B",B,W);
 +
label("C",C,S);
 +
label("D",D,S);
 +
label("E",E,dir(0));
 +
dot(A^^B^^C^^D^^E);
 +
</asy>
 +
 
 +
Draw diagonals <math>AC</math> and <math>AD</math> to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles <math>ABC</math> and <math>ADE</math>, they each have area <math>2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}</math>. For triangle <math>ACD</math>, we can see that <math>AC=AD=2\sqrt{3}</math> and <math>CD=2</math>. Using Pythagorean Theorem, the altitude for this triangle is <math>\sqrt{11}</math>, so the area is <math>\sqrt{11}</math>. Adding each part up, we get <math>2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}</math>.
 +
 
 +
==Video Solution (🚀Under 3 min!🚀)==
 +
https://youtu.be/1CAbbfArA6w
 +
 
 +
<i>~Education, the Study of Everything </i>
 +
 
 +
 
 +
==Video Solution(always chill)==
 +
https://www.youtube.com/watch?v=CSR-rJ-3hys&ab_channel=Chillin
 +
 
 +
== Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area) ==
 +
https://youtu.be/QtSbAKUb1VE
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution by Hawk Math==
 +
https://www.youtube.com/watch?v=p4iCAZRUESs
 +
 
 +
==Video Solution by Mathematical Dexterity (Basic Area Formulas)==
 +
https://www.youtube.com/watch?v=7kDTlVixsD0
 +
 
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/FV9AnyERgJQ?t=1226
 +
 
 +
~IceMatrix
 +
==Video Solution by Interstigation (Ignore Useless Segments)==
 +
https://youtu.be/9eChInz03UQ
 +
 
 +
~Interstigation
 +
==Video Solution by The Power of Logic==
 +
https://www.youtube.com/watch?v=f8L5K2yIXUc
 +
 
 +
~The Power of Logic
 +
 
 +
==Remark==
 +
This configuration of <math>11</math> congruent line segments is known as the Moser Spindle https://en.wikipedia.org/wiki/Moser_spindle , and can be used to demonstrate that <math>3</math> colors are not sufficient to color all of the points in the plane such that points that are <math>1</math> unit apart have different colors. Finding the minimum such number of colors is a famous unsolved problem: the Nelson-Hadwiger problem. See: https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem
 +
 
 +
~hailstone
  
−
Let <math>M</math> be the midpoint of <math>CD</math>.
+
==See Also==
 +
{{AMC12 box|year=2021|ab=B|num-b=14|num-a=16}}
 +
{{AMC10 box|year=2021|ab=B|num-b=19|num-a=21}}
 +
{{MAA Notice}}

Latest revision as of 21:16, 5 June 2024

The following problem is from both the 2021 AMC 10B #20 and 2021 AMC 12B #15, so both problems redirect to this page.

Problem

The figure is constructed from $11$ line segments, each of which has length $2$. The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$, where $m$ and $n$ are positive integers. What is $m + n ?$ [asy] /* Made by samrocksnature */ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-2,2); pair G=(-3,2); draw(A--B--C--D--E--A); draw(A--F--A--G); draw(B--F--C); draw(E--G--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E^^F^^G); [/asy]

$\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24$

Solution 1

[asy] /* Made by samrocksnature, adapted by Tucker, then adjusted by samrocksnature again, then adjusted by erics118 xD*/ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-1.85,2); pair G=(-3.1,2); draw(A--G--A--F, lightgray); draw(B--F--C, lightgray); draw(E--G--D, lightgray); dot(F^^G, lightgray); draw(A--B--C--D--E--A); draw(A--C--A--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E); [/asy]

Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$, they each have area $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$. For triangle $ACD$, we can see that $AC=AD=2\sqrt{3}$ and $CD=2$. Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$, so the area is $\sqrt{11}$. Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}$.

Video Solution (🚀Under 3 min!🚀)

https://youtu.be/1CAbbfArA6w

~Education, the Study of Everything


Video Solution(always chill)

https://www.youtube.com/watch?v=CSR-rJ-3hys&ab_channel=Chillin

Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area)

https://youtu.be/QtSbAKUb1VE

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=p4iCAZRUESs

Video Solution by Mathematical Dexterity (Basic Area Formulas)

https://www.youtube.com/watch?v=7kDTlVixsD0

Video Solution by TheBeautyofMath

https://youtu.be/FV9AnyERgJQ?t=1226

~IceMatrix

Video Solution by Interstigation (Ignore Useless Segments)

https://youtu.be/9eChInz03UQ

~Interstigation

Video Solution by The Power of Logic

https://www.youtube.com/watch?v=f8L5K2yIXUc

~The Power of Logic

Remark

This configuration of $11$ congruent line segments is known as the Moser Spindle https://en.wikipedia.org/wiki/Moser_spindle , and can be used to demonstrate that $3$ colors are not sufficient to color all of the points in the plane such that points that are $1$ unit apart have different colors. Finding the minimum such number of colors is a famous unsolved problem: the Nelson-Hadwiger problem. See: https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem

~hailstone

See Also

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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