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Difference between revisions of "2017 AMC 8 Problems/Problem 25"

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==Problem 25==
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==Problem==
  
 
In the figure shown, <math>\overline{US}</math> and <math>\overline{UT}</math> are line segments each of length 2, and <math>m\angle TUS = 60^\circ</math>. Arcs <math>\overarc{TR}</math> and <math>\overarc{SR}</math> are each one-sixth of a circle with radius 2. What is the area of the region shown?  
 
In the figure shown, <math>\overline{US}</math> and <math>\overline{UT}</math> are line segments each of length 2, and <math>m\angle TUS = 60^\circ</math>. Arcs <math>\overarc{TR}</math> and <math>\overarc{SR}</math> are each one-sixth of a circle with radius 2. What is the area of the region shown?  
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<math>\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}</math>
 
<math>\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}</math>
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==Solution==
 
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We can extend <math>\overline{US}</math> and <math>\overline{UT}</math> into an equilateral triangle of side length <math>4</math> with area <math>\dfrac{\sqrt{3}}{4}\cdot4^2=4\sqrt{3}</math>. We then must subtract the two 1/6-circles, which each have radius <math>2</math>, so their total area is <math>2\cdot\dfrac{\pi\cdot2^2}{6}=\dfrac{4\pi}{3}</math>. Our answer is therefore <math>\boxed{\textbf{(B)}\ 4\sqrt{3}-\dfrac{4\pi}{3}}</math>.
 
  
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==See Also==
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==Solution 1==
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{{AMC8 box|year=2017|num-b=24|after=Last Problem}}
 
  
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{{MAA Notice}}
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<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>
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In addition to the given diagram, we can draw lines <math>\overline{SR}</math> and <math>\overline{RT}.</math> The area of rhombus <math>SRTU</math> is half the product of its diagonals, which is <math>\frac{2\sqrt3 \cdot 2}{2}=2\sqrt3</math>. 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 <math>\frac{1}{6}</math>, 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 <math>2(\frac{4 \pi}{6}-\sqrt3).</math> The area of rhombus <math>SRTU</math> minus the circular segments is <math>2\sqrt3-\frac{4 \pi}{3}+2\sqrt3= \boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.</math>
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~PEKKA
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==Video Solutions==
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https://youtu.be/sVclz6EmpEU
 +
 
 +
~savannahsolver

Latest revision as of 22:44, 18 October 2024

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

[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 $2\sqrt3-\frac{4 \pi}{3}+2\sqrt3= \boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$

~PEKKA

Video Solutions

https://youtu.be/sVclz6EmpEU

~savannahsolver

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