Difference between revisions of "2021 Fall AMC 12B Problems/Problem 15"
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C1=dir(x-2z); C2=dir(x+y-2z); C3=dir(x+2y-2z); C4=dir(x+3y-2z); | C1=dir(x-2z); C2=dir(x+y-2z); C3=dir(x+2y-2z); C4=dir(x+3y-2z); | ||
draw(A1--A2--A3--A4--A1, gray+0.25+dashed); | draw(A1--A2--A3--A4--A1, gray+0.25+dashed); | ||
− | + | filldraw(B1--B2--B3--B4--cycle, white, gray+dashed+linewidth(0.25)); | |
− | + | filldraw(C1--C2--C3--C4--cycle, white, gray+dashed+linewidth(0.25)); | |
dot(O); | dot(O); | ||
pair P1,P2,P3,P4,Q1,Q2,Q3,Q4,R1,R2,R3,R4; | pair P1,P2,P3,P4,Q1,Q2,Q3,Q4,R1,R2,R3,R4; | ||
Line 33: | Line 33: | ||
pair C=(3,3); | pair C=(3,3); | ||
pair D=(3,1.732); | pair D=(3,1.732); | ||
− | draw(A--(0,3)--C--(3,0)--A, lightgray); | + | draw(A--(0,3)--C--(3,0)--A, lightgray+dashed); |
− | draw(A--B--C--D-- | + | draw(A--B--C--A); |
+ | draw(A--D--C, gray); | ||
label("$A$",A,W); | label("$A$",A,W); | ||
label("$B$",B,N); | label("$B$",B,N); | ||
Line 42: | Line 43: | ||
label("$F$",(3,0),E); | label("$F$",(3,0),E); | ||
</asy> | </asy> | ||
− | + | The <math>24</math>-sided polygon is made out of <math>24</math> shapes like <math>\triangle ABC</math>. Then <math>\angle BAC=360^\circ/24=15^\circ</math>, and <math>\angle EAC = 45^\circ</math>, so <math>\angle{EAB} = 30^{\circ}</math>. Then <math>EB=AE\tan 30^\circ = \sqrt{3}</math>; therefore <math>BC=EC-EB=3-\sqrt{3}</math>. Thus | |
+ | <cmath>[ABC] = \frac{BC}{EC}\cdot [ACE] = \frac{3-\sqrt{3}}{3}\cdot \frac 92 </cmath>and the required area is <math>24\cdot[ABC] =108-36\sqrt{3}</math>. Finally <math>108+36+3=\boxed{(\textbf{E})\ 147}</math>. | ||
+ | ~lopkiloinm | ||
+ | |||
+ | Note: Drop an altitude from <math>A</math> to <math>\overline{BC}</math> to construct point <math>E</math>. This creates right triangles. ~erringbubble | ||
== Solution 2 == | == Solution 2 == | ||
Line 72: | Line 77: | ||
~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | == Solution 3 (complex number & coordinate geometry)== | ||
+ | |||
+ | [[Image:2021_amc_12b_fall_P15_S3.png|thumb|center|800px]] | ||
+ | |||
+ | set A = 3+3i , A' , B' rotate 30 degree from A, B | ||
+ | |||
+ | A'= A <math> \cdot e^i30^\circ = (3+3i)*(\sqrt{3}/2 + 1/2 i) =( \sqrt{3}/2 - 1/2) + (1/2 + \sqrt{3}/2) i </math> | ||
+ | |||
+ | line A'B' <math> \frac{y-(1/2 + \sqrt{3}/2)}{x - ( \sqrt{3}/2 - 1/2)} = Tan(90\circ+30\circ) = -\sqrt{3} </math> | ||
+ | |||
+ | intersect with line y=3 at point <math>E_{x} = \sqrt{3}</math> , then length <math> AE = A_{x} - E_{x} = 3- \sqrt{3}</math> , | ||
+ | |||
+ | use shoelace or <math>\triangle OAE </math> = 1/2 * AE * AB/2 = 1/2 * <math>(3- \sqrt{3})</math> * 3 | ||
+ | |||
+ | total area = 24 * <math>\triangle OAE </math> = = 108 - 36 <math>\sqrt{3}</math> the answer is <math>\boxed{\textbf{(E) }147}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | |||
+ | ==Video Solution (Just 4 min!)== | ||
+ | https://youtu.be/u_8EWGBErs8 | ||
+ | |||
+ | <i> ~Education, the Study of Everything </i> | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/YD9J394zeig | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Latest revision as of 20:41, 10 October 2024
Contents
[hide]Problem
Three identical square sheets of paper each with side length are stacked on top of each other. The middle sheet is rotated clockwise about its center and the top sheet is rotated clockwise about its center, resulting in the -sided polygon shown in the figure below. The area of this polygon can be expressed in the form , where , , and are positive integers, and is not divisible by the square of any prime. What is ?
Solution 1
The -sided polygon is made out of shapes like . Then , and , so . Then ; therefore . Thus and the required area is . Finally . ~lopkiloinm
Note: Drop an altitude from to to construct point . This creates right triangles. ~erringbubble
Solution 2
As shown in Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png, all 12 vertices of three squares form a regular dodecagon (12-gon). Denote by the center of this dodecagon.
Hence, .
Because the length of a side of a square is 6, .
Hence, .
We notice that . Hence, .
Therefore, the area of the region that three squares cover is
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 3 (complex number & coordinate geometry)
set A = 3+3i , A' , B' rotate 30 degree from A, B
A'= A
line A'B'
intersect with line y=3 at point , then length ,
use shoelace or = 1/2 * AE * AB/2 = 1/2 * * 3
total area = 24 * = = 108 - 36 the answer is .
Video Solution (Just 4 min!)
~Education, the Study of Everything
Video Solution by TheBeautyofMath
~IceMatrix
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.