Difference between revisions of "2024 AMC 12A Problems/Problem 18"
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Now we can find the angle of rotation (<math>\angle APB</math>) per move with the following steps: | Now we can find the angle of rotation (<math>\angle APB</math>) per move with the following steps: | ||
<cmath>AP^2=(\frac{1}{2})^2+(1+\frac{\sqrt{3}}{2})^2=2+\sqrt{3}</cmath> | <cmath>AP^2=(\frac{1}{2})^2+(1+\frac{\sqrt{3}}{2})^2=2+\sqrt{3}</cmath> | ||
| − | <cmath>1^2=AP^2+AP^2-2(AP)(AP)cos\angle APB</cmath> | + | <cmath>1^2=AP^2+AP^2-2(AP)(AP)\cos\angle APB</cmath> |
| − | <cmath>1=2(2+\sqrt{3})(1-cos\angle APB)</cmath> | + | <cmath>1=2(2+\sqrt{3})(1-\cos\angle APB)</cmath> |
| − | <cmath>cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}</cmath> | + | <cmath>\cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}</cmath> |
| − | < | + | <math>\cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\cdot\frac{4-2\sqrt{3}}{4-2\sqrt{3}}<cmath> |
| − | <cmath>cos\angle APB=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}< | + | </cmath>\cos\angle APB=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}<cmath> |
| − | <cmath>\angle APB=30^\circ</ | + | </cmath>\angle APB=30^\circ</math><math> |
| − | Since Vertex <math>C< | + | Since Vertex </math>C<math> is the closest one and <cmath>\angle BPC=360-180-30=150</cmath> |
| − | Vertex C will land on Vertex B when <math>\frac{150}{30}+1=\fbox{(A) 6} | + | Vertex C will land on Vertex B when </math>\frac{150}{30}+1=\fbox{(A) 6}$ cards are placed. |
~lptoggled, minor Latex edits by eevee9406 | ~lptoggled, minor Latex edits by eevee9406 | ||
Revision as of 20:06, 8 November 2024
Problem
On top of a rectangular card with sides of length 
and 
, an identical card is placed so that two of their diagonals line up, as shown (
, in this case).
![[asy] defaultpen(fontsize(12)+0.85); size(150); real h=2.25; pair C=origin,B=(0,h),A=(1,h),D=(1,0),Dp=reflect(A,C)*D,Bp=reflect(A,C)*B; pair L=extension(A,Dp,B,C),R=extension(Bp,C,A,D); draw(L--B--A--Dp--C--Bp--A); draw(C--D--R); draw(L--C^^R--A,dashed+0.6); draw(A--C,black+0.6); dot("$C$",C,2*dir(C-R)); dot("$A$",A,1.5*dir(A-L)); dot("$B$",B,dir(B-R)); [/asy]](http://latex.artofproblemsolving.com/5/5/7/55798e965d22d0047d5702a95bff6b78fb909bf7.png)
Continue the process, adding a third card to the second, and so on, lining up successive diagonals after rotating clockwise. In total, how many cards must be used until a vertex of a new card lands exactly on the vertex labeled 
in the figure?
Solution 1
Let the midpoint of 
be 
.
We see that no matter how many moves we do, stays where it is.
Now we can find the angle of rotation (
) per move with the following steps:
![\[AP^2=(\frac{1}{2})^2+(1+\frac{\sqrt{3}}{2})^2=2+\sqrt{3}\]](http://latex.artofproblemsolving.com/d/f/f/dff05d07a9bbbf2b47f7c7f52d5bce32d8184d8c.png)
![\[1^2=AP^2+AP^2-2(AP)(AP)\cos\angle APB\]](http://latex.artofproblemsolving.com/b/d/c/bdc2e055b50b7014a7ee1ae103769759b254e7d5.png)
![\[1=2(2+\sqrt{3})(1-\cos\angle APB)\]](http://latex.artofproblemsolving.com/e/e/6/ee68163c696a7a21c0eb024105a8aad1b84c048d.png)
![\[\cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\]](http://latex.artofproblemsolving.com/7/6/a/76aaa00a93a1d508ed7891bd6ab5a94940d42e38.png)
C$is the closest one and <cmath>\angle BPC=360-180-30=150</cmath>
Vertex C will land on Vertex B when$ (Error compiling LaTeX. Unknown error_msg)\frac{150}{30}+1=\fbox{(A) 6}$ cards are placed.
~lptoggled, minor Latex edits by eevee9406
See also
| 2024 AMC 12A (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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
