2017 AMC 8 Problems/Problem 25

Problem

In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$ . Arcs $\overarc{TR}$ and $\overarc{SR}$ are each one-sixth of a circle with radius 2. What is the area of the region shown?

$[asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("$U$", (2,3.464), N); label("$S$", (1,1.732), W); label("$T$", (3,1.732), E); label("$R$", (2,0), S);[/asy]$

$\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$

Solution 1

label(" $Y$ ", Y, E); </asy> We can clearly see that $\triangle UXY$ is an equilateral triangle, because the problem states that $m\angle TUS = 60^\circ$ . We can figure out that $m\angle SXR= 60^\circ$ and $m\angle TYR = 60^\circ$ because they are $\frac{1}{6}$ of a circle. The area of the figure is equal to $[\triangle UXY]$ minus the combined area of the $2$ sectors of the circles (in red). Using the area formula for an equilateral triangle, $\frac{a^2\sqrt{3}}{4},$ where $a$ is the side length of the equilateral triangle, $[\triangle UXY]$ is $\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.$ The combined area of the $2$ sectors is $2\cdot\frac16\cdot\pi r^2$ , which is $\frac 13\pi \cdot 2^2 = \frac{4\pi}{3}.$ Thus, our final answer is $\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$

Solution 2

$[asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("$U$", (2,3.464), N); label("$S$", (1,1.732), W); label("$T$", (3,1.732), E); label("$R$", (2,0), S);[/asy]$

In addition to the given diagram, we can draw lines $\overline{SR}$ and $\overline{RT}.$ The area of rhombus $SRTU$ is half the product of its diagonals, which is $\frac{2\sqrt3 \cdot 2}{2}=2\sqrt3$ . However, we have to subtract off the circular segments. The area of those can be found by computing the area of the circle with radius 2, multiplying it by $\frac{1}{6}$ , then finally subtracting the area of an equilateral triangle with a side length 2 from the sector. The sum of the areas of the circular segments is $2(\frac{4 \pi}{6}-\sqrt3).$ The area of rhombus $SRTU$ minus the circular segments is $2\sqrt3-\frac{4 \pi}{3}+2\sqrt3= \boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$