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; |
Revision as of 08:55, 7 June 2022
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
We can see that this shape is made out of shapes like . Note that and because . Then ; therefore . Thus and the required area is . Finally . ~lopkiloinm
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)
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.