Difference between revisions of "2024 AMC 12A Problems/Problem 18"
(Created page with "==Solution 1== Let the midpoint of <math>AC</math> be <math>P</math> We see that no matter how many moves we do, <math>P</math> stays where it is Now we can find the angle of...") |
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Let the midpoint of <math>AC</math> be <math>P</math> | Let the midpoint of <math>AC</math> be <math>P</math> | ||
We see that no matter how many moves we do, <math>P</math> stays where it is | We see that no matter how many moves we do, <math>P</math> stays where it is | ||
| + | <cmath></cmath> | ||
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> | ||
| Line 8: | Line 9: | ||
<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> | ||
<cmath>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{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> |
| − | < | + | <cmath>\angle APB=30^\circ</cmath> |
| − | Since Vertex <math>C</math> is the closest one and < | + | 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}</math> cards are placed | Vertex C will land on Vertex B when <math>\frac{150}{30}+1=\fbox{(A) 6}</math> cards are placed | ||
Revision as of 17:57, 8 November 2024
Solution 1
Let the midpoint of 
be 
We see that no matter how many moves we do, stays where it is
![\[\]](http://latex.artofproblemsolving.com/5/7/f/57f1406b65f9f2e6a365b7356da731780b42cceb.png)
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/1/c/a/1cad7b86f2632ba08b46c0a65589d255d6f215f0.png)
![\[1=2(2+\sqrt{3})(1-cos\angle APB)\]](http://latex.artofproblemsolving.com/4/d/c/4dc8eb60c3639ed1cddee3b1ed8b8f80221e44ed.png)
![\[cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\]](http://latex.artofproblemsolving.com/6/d/8/6d8d083cb983cb1ee515b3cdc50c18cee2cb4fd6.png)
![\[cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\cdot\frac{4-2\sqrt{3}}{4-2\sqrt{3}}\]](http://latex.artofproblemsolving.com/5/2/a/52a1a8e329e1e30ca87b405af8b0ec5595e69029.png)
![\[cos\angle APB=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\]](http://latex.artofproblemsolving.com/2/8/1/281c0e470f0008cc3cdd14e1125ea245409c2f95.png)
![\[\angle APB=30^\circ\]](http://latex.artofproblemsolving.com/d/c/f/dcf08b2f3f240bc1d7a58bcb4adcb770c6197932.png)
Since Vertex 
is the closest one and ![\[\angle BPC=360-180-30=150\]](http://latex.artofproblemsolving.com/7/8/4/784e6bc311b766e34cf6b61553c50bd62a2b81ce.png)
Vertex C will land on Vertex B when 
cards are placed
