Difference between revisions of "2021 AMC 12B Problems/Problem 15"
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− | ==Problem | + | {{duplicate|[[2021 AMC 10B Problems#Problem 20|2021 AMC 10B #20]] and [[2021 AMC 12B Problems#Problem 15|2021 AMC 12B #15]]}} |
− | 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 | + | |
+ | ==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 */ | ||
Line 16: | Line 18: | ||
label("A",A,N); | label("A",A,N); | ||
label("B",B,W); | label("B",B,W); | ||
− | label("C",C, | + | label("C",C,S); |
− | label("D",D, | + | 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== |
+ | |||
+ | Let <math>M</math> be the midpoint of <math>CD</math>. Noting that <math>AED</math> and <math>ABC</math> are <math>120</math> - <math>30</math> - <math>30</math> triangles because of the equilateral triangles, <cmath>AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}.</cmath> Also, <math>[AED]=2\cdot2\cdot\frac{1}{2}\cdot\sin{120^\circ}=\sqrt{3}</math> (or split <math>\triangle AED</math> into two <math>30-60-90</math> triangles) and so <cmath>[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{\textbf{(D)} ~23}.</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | <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 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 | |
− | + | ==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}} |
Revision as of 17:00, 3 October 2022
- 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 Solution 2
- 4 Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area)
- 5 Video Solution by Hawk Math
- 6 Video Solution by Mathematical Dexterity (Basic Area Formulas)
- 7 Video Solution by TheBeautyofMath
- 8 Video Solution by Interstigation (Ignore Useless Segments)
- 9 Video Solution by The Power of Logic
- 10 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
Let be the midpoint of . Noting that and are - - triangles because of the equilateral triangles, Also, (or split into two triangles) and so
Solution 2
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 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
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.