2021 Fall AMC 10B Problems/Problem 18
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 
![[asy] size(10cm,0); path p = box((0,0), (1,1)); draw(p, black + linewidth(2.0pt)); draw(rotate(30,(1/2,1/2))*p,black + linewidth(2.0pt)); draw(rotate(60,(1/2,1/2))*p,black + linewidth(2.0pt)); draw((0.2113248,1)--(0.7886752,1),white + linewidth(2.0pt)); draw((0.2113248,0)--(0.7886752,0),white + linewidth(2.0pt)); draw((1,0.2113248)--(1,0.7886752),white + linewidth(2.0pt)); draw((0,0.2113248)--(0,0.7886752),white + linewidth(2.0pt)); draw((0.5,1.07735026918)--(0,0.7886752),white + linewidth(2.0pt)); draw((0.5,1.07735026918)--(0,0.7886752),black+dashed+linewidth(2.0pt)); [/asy]](http://latex.artofproblemsolving.com/2/7/d/27dfd4852f123f02482ceb2f59ccad2c9543c7da.png)
Someone pls finish this
Solution 1
First note the useful fact that if 
is the circumradius of a dodecagon (
-gon) the area of the figure is 
If we connect the vertices of the 
squares we get a dodecagon. The radius of circumcircle of the dodecagon is simply half the diagonal of the square, which is 
Thus the area of the dodecagon is 
But, the problem asks for the area of figure of rotated squares. This area is the area of the dodecagon, which was found, subtracting the isosceles triangles, which are formed when connecting the vertices of the squares to created the dodecagon. To find this area, we need to know the base of the isosceles triangle, call this 
Then, we can use Law of Cosines, on the triangle that is formed from the two vertices of the square and the center of the square. After computing, we get that 
Realize that the isosceles are congruent with an angle measure of 
this means that we can create 
congruent equilateral triangles with side length of 
The area of the equilateral triangle is 
Thus, the area of all the twelve small equilateral traingles are 
. Thus, the requested area is 
Thus, 
Thus, the answer is 
~NH14
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 (30-60-90 Triangles)
To make things simpler, let's take only the original sheet and the 30 degree rotated sheet. Then the diagram is this;
![[asy] size(10cm,0); path p = box((0,0), (1,1)); draw(p, black + linewidth(2.0pt)); draw(rotate(30,(1/2,1/2))*p,black + linewidth(2.0pt)); /*Rotate 60 degrees*/ [/asy]](http://latex.artofproblemsolving.com/1/d/1/1d18fa448a9d67e0133a366e5ace7d169451f836.png)
The area of this diagram is the original square plus the area of the four triangles that 'jut' out of the square. Because the square is rotated 
, each triangle is a 30-60-90 triangle. Similarly, the triangles that are bounded on the inside of the original square outside of the rotated square are also congruent 30-60-90 triangles. Noting this, we can do some labelling:
![[asy] size(10cm,0); path p = box((0,0), (1,1)); draw(p, black + linewidth(2.0pt)); draw(rotate(30,(1/2,1/2))*p,black + linewidth(2.0pt)); /*Rotate 60 degrees*/ label("$y$",(0.1,-0.05)); label("$x$",(0.4,0.05)); label("$y\sqrt{3}$",(0.8,-0.05)); label("$\frac{x}{2}$",(0.22,-0.12)); label("$\frac{x\sqrt{3}}{2}$",(0.5,-0.15)); label("$2y$",(0.8,0.15)); label("$y$",(1.05,0.1)); label("$\frac{x}{2}$",(1.12,0.22)); [/asy]](http://latex.artofproblemsolving.com/9/c/d/9cd2cf7a32b282cd81ea1059fb9218a0b425d1df.png)
Since the side lengths of the squares must be the same, and they are both 6, we have a system of equations;
![\[y+x+y\sqrt{3} = 6\]](http://latex.artofproblemsolving.com/7/a/2/7a22ed6b2792c6e6060f9e191dd39c0bca789e17.png)
![\[\frac{x\sqrt{3}}{2} + 2y + \frac{x}{2} = 6\]](http://latex.artofproblemsolving.com/8/2/4/824bd2dad1bac1009e31fa02d89ebc4ee2bdb625.png)
We solve this to get 
and 
.
The area of each triangle is 
by plugging in 
.
The rotated 60 degree square is the same thing as rotating it 30 degrees counterclockwise, so it's triangles that jut out of the square will be congruent to the triangles we have found, and therefore they will have the same area.
Unfortunately, when drawing all three squares, we see the two triangles overlap; take the very top for example.
![[asy] import olympiad; size(10cm); pair A,B,C,D,E,F,G; A = (0.211,0); B=(0.1547,0); C = (0.63397,0); D = (0.789,0); E = (0.31666,0.1823); F=(0.5,0.077); G=(0.68334,0.1823); draw((0,0)--(1,0),black+linewidth(2pt)); draw((0.211,0)--(0.31666,0.1823)--(0.63397,0)--cycle); draw((0.1547,0)--(0.789,0)--(0.68334,0.1823)--cycle); [/asy]](http://latex.artofproblemsolving.com/9/9/d/99d24a9517f1406829b0b8b6524e609bb93f098c.png)
The area of this shape is twice the area of each of the triangles that we have already found minus the area of the small triangle that is overlapped by the two by PIE.
~KingRavi
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
_Problem_15,_sol.png){kind=link}