2024 AMC 12A Problems/Problem 18

Solution 1

Let the midpoint of $AC$ be $P$ We see that no matter how many moves we do, $P$ stays where it is $\[\]$ Now we can find the angle of rotation ( $\angle APB$ ) per move with the following steps: $AP^2=(\frac{1}{2})^2+(1+\frac{\sqrt{3}}{2})^2=2+\sqrt{3}$ $1^2=AP^2+AP^2-2(AP)(AP)cos\angle APB$ $1=2(2+\sqrt{3})(1-cos\angle APB)$ $cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}$ $cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\cdot\frac{4-2\sqrt{3}}{4-2\sqrt{3}}$ $cos\angle APB=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$ $\angle APB=30^\circ$ Since Vertex $C$ is the closest one and $\angle BPC=360-180-30=150$

Vertex C will land on Vertex B when $\frac{150}{30}+1=\fbox{(A) 6}$ cards are placed

Page

Toolbox

Search

2024 AMC 12A Problems/Problem 18

Solution 1