2017 AMC 8 Problems/Problem 25

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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

Let the centers of the circles containing arcs $\overarc{SR}$ and $\overarc{TR}$ be $X$ and $Y$, respectively. Extend $\overline{US}$ and $\overline{UT}$ to $X$ and $Y$, and connect point $X$ with point $Y$. [asy] unitsize(1 cm); pair U,S,T,R,X,Y; U =(2,3.464); S=(1,1.732); T=(3,1.732); R=(2,0); X=(0,0); Y=(4,0); draw(U--S); draw(S--U--T); draw(S--X--Y--T,red); draw(arc(X,R,S),red); draw(arc(Y,T,R),red); label("$U$",U, N); label("$S$", S, W); label("$T$", T, E); label("$R$", R, S); label("$X$",X, W); 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 $2\sqrt3-\frac{4 \pi}{3}+2\sqrt3= \boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$


Video Solutions



- Happytwin





Solution 3 & Video

First, we must make it like an equilateral triangle. If you add two of the same arcs, you will get an equilateral triangle. The goal is to find ONE of the arcs and multiply it by two. Then subtract it from the equilateral triangle. If you look at the triangle, T and S are midpoints of the equilateral triangle. So now we can find the arc.These arcs are $\frac{2}{6}\pi$ so now you multiply by 2 so $\frac{4}{3}\pi$. And subtract. Equilateral's area is $4\sqrt{3}$. So now you do $4 \sqrt 3 - \frac{4}{3} \pi$. That is the answer and thank you.

~ math.is.hard.to.understand

Here's a video to understand it (credits to ~ pi_is_3.14)


See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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