2014 AMC 10A Problem 23

WLOG, let the width of the rectangle be $1$ and the length be $\sqrt3.$ Draw the line perpendicular to the midpoint of the fold, as shown below. Note that $QR = \dfrac{2\sqrt3}{3}$ and $$QT = \sqrt{1^2 + \left(\dfrac{\sqrt3}{3}\right)^2}$$$$= \sqrt{1 + \dfrac{1}{3}} = \sqrt{\dfrac{4}{3}}.$$ This tells us $$QT = \dfrac{2\sqrt3}{3} = QR.$$ This means that $\triangle QRT$ is equilateral. Similarly, $\triangle RTS$ is equilateral. This makes the two triangles congruent. This means that after the rectangle gets folded, this area will be overlapped. The area of the rectangle is $1 \cdot \sqrt3 = \sqrt3.$ The side length of this triangle is $QT = \dfrac{2\sqrt3}{3}.$ The area of it is then $$\left(\dfrac{2\sqrt3}{3}\right)^2 \cdot \dfrac{\sqrt3}{4} = \dfrac{\sqrt3}{3}.$$ The area of the folded figure is then $$\sqrt3 - \dfrac{\sqrt3}{3} = \dfrac{2\sqrt3}{3}.$$ The desired ratio is then $$\dfrac{2\sqrt3}{3} \div \sqrt3 = \dfrac{2}{3}.$$ Therefore $B:A=2:3$. Thus, C is the correct answer.