2024 AMC 12B Problem 19

Question

Equilateral

\triangle ABC

with side length

14

is rotated about its center by angle

\theta,

where

0 \lt \theta \lt 60^\circ,

to form

\triangle DEF.

See the figure. The area of hexagon

ADBECF

is

91\sqrt3.

What is

\tan\theta?

\dfrac{3}{4}

\dfrac{5\sqrt3}{11}

\dfrac{4}{5}

\dfrac{11}{13}

\dfrac{7\sqrt3}{13}

All six vertices lie on the circle of radius

R = \dfrac{14}{\sqrt3};

the hexagon’s central angles alternate

\theta

and

120^\circ - \theta

The area is

\tfrac12 R^2\bigl(3\sin\theta + 3\sin(120^\circ - \theta)\bigr)

= 98\bigl(\sin\theta + \sin(120^\circ - \theta)\bigr);

simplify with sum-to-product to solve for

\theta

The six vertices lie on the circumcircle of radius

R = \dfrac{14}{\sqrt3},

so

R^2 = \dfrac{196}{3}.

Going around, the central angles alternate between

\theta

(three times) and

120^\circ - \theta

(three times). The cyclic-hexagon area is

\begin{aligned} &\tfrac12 R^2\bigl(3\sin\theta + 3\sin(120^\circ - \theta)\bigr) \\ &= 98\bigl(\sin\theta + \sin(120^\circ - \theta)\bigr). \end{aligned}

By sum-to-product,

\sin\theta + \sin(120^\circ - \theta)

= 2\sin 60^\circ\cos(\theta - 60^\circ)

= \sqrt3\cos(60^\circ - \theta).

Setting the area to

91\sqrt3

gives

\sqrt3\cos(60^\circ - \theta)

= \dfrac{91\sqrt3}{98} = \dfrac{13\sqrt3}{14},

so

\cos(60^\circ - \theta) = \dfrac{13}{14}

and

\sin(60^\circ - \theta) = \dfrac{3\sqrt3}{14}.

Then

\tan(60^\circ - \theta) = \dfrac{3\sqrt3}{13},

and

\begin{aligned} &\tan\theta = \tan\bigl(60^\circ - (60^\circ - \theta)\bigr) \\ &= \frac{\sqrt3 - \frac{3\sqrt3}{13}}{1 + \sqrt3\cdot\frac{3\sqrt3}{13}} \\ &= \frac{\frac{10\sqrt3}{13}}{\frac{22}{13}} \\ &= \frac{5\sqrt3}{11}. \end{aligned}

Thus, the correct answer is B.

Problem 19 in other years