Difference between revisions of "2021 AMC 12B Problems/Problem 15"
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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]]}} | ||
+ | |||
==Problem== | ==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 | + | 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 */ | ||
Line 25: | Line 27: | ||
==Solution 1== | ==Solution 1== | ||
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− | |||
− | |||
− | |||
<asy> | <asy> | ||
− | /* Made by samrocksnature, adapted by Tucker */ | + | /* Made by samrocksnature, adapted by Tucker, then adjusted by samrocksnature again, then adjusted by erics118 xD*/ |
pair A=(-2.4638,4.10658); | pair A=(-2.4638,4.10658); | ||
pair B=(-4,2.6567453480756127); | pair B=(-4,2.6567453480756127); | ||
Line 37: | Line 35: | ||
pair D=(-1.464483379039766,0.6335248637894945); | pair D=(-1.464483379039766,0.6335248637894945); | ||
pair E=(-0.956630463955801,2.6567453480756127); | pair E=(-0.956630463955801,2.6567453480756127); | ||
− | pair F=(-1. | + | pair F=(-1.85,2); |
pair G=(-3.1,2); | 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--B--C--D--E--A); | ||
draw(A--C--A--D); | draw(A--C--A--D); | ||
Line 47: | Line 49: | ||
label("E",E,dir(0)); | label("E",E,dir(0)); | ||
dot(A^^B^^C^^D^^E); | dot(A^^B^^C^^D^^E); | ||
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− | |||
− | |||
− | |||
</asy> | </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>. | 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) == | == Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area) == | ||
Line 62: | Line 69: | ||
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== | ||
https://www.youtube.com/watch?v=p4iCAZRUESs | 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== | ==Video Solution by TheBeautyofMath== | ||
Line 67: | Line 77: | ||
~IceMatrix | ~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 | ||
==See Also== | ==See Also== |
Latest revision as of 22:16, 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.
Contents
- 1 Problem
- 2 Solution 1
- 3 Video Solution (🚀Under 3 min!🚀)
- 4 Video Solution(always chill)
- 5 Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area)
- 6 Video Solution by Hawk Math
- 7 Video Solution by Mathematical Dexterity (Basic Area Formulas)
- 8 Video Solution by TheBeautyofMath
- 9 Video Solution by Interstigation (Ignore Useless Segments)
- 10 Video Solution by The Power of Logic
- 11 Remark
- 12 See Also
Problem
The figure is constructed from line segments, each of which has length
. The area of pentagon
can be written as
, where
and
are positive integers. What is
Solution 1
Draw diagonals and
to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles
and
, they each have area
. For triangle
, we can see that
and
. Using Pythagorean Theorem, the altitude for this triangle is
, so the area is
. Adding each part up, we get
.
Video Solution (🚀Under 3 min!🚀)
~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)
~ 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)
~Interstigation
Video Solution by The Power of Logic
https://www.youtube.com/watch?v=f8L5K2yIXUc
~The Power of Logic
Remark
This configuration of congruent line segments is known as the Moser Spindle https://en.wikipedia.org/wiki/Moser_spindle , and can be used to demonstrate that
colors are not sufficient to color all of the points in the plane such that points that are
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.