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

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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 1==
 
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label("<math>Y</math>", Y, E);
 
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</asy>
 
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We can clearly see that <math>\triangle UXY</math> is an equilateral triangle, because the problem states that <math>m\angle TUS = 60^\circ</math>. We can figure out that <math>m\angle SXR= 60^\circ</math> and <math>m\angle TYR = 60^\circ</math> because they are <math>\frac{1}{6}</math> of a circle. The area of the figure is equal to <math>[\triangle UXY]</math> minus the combined area of the <math>2</math> sectors of the circles (in red). Using the area formula for an equilateral triangle, <math>\frac{a^2\sqrt{3}}{4},</math> where <math>a</math> is the side length of the equilateral triangle, <math>[\triangle UXY]</math> is <math>\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.</math> The combined area of the <math>2</math> sectors is <math>2\cdot\frac16\cdot\pi r^2</math>, which is <math>\frac 13\pi \cdot 2^2 = \frac{4\pi}{3}.</math> Thus, our final answer is <math>\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.</math>
 
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==Solution 2==
 
==Solution 2==
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==Video Solutions==
 
==Video Solutions==
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https://youtu.be/LT4gyH--328
 
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https://youtu.be/wc5rGulTTR8
 
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- Happytwin
 
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https://youtu.be/aE0oAq4Q_Ks
 
  
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https://euclideanmathcircle.wixsite.com/emc1/videos?wix-vod-video-id=3a7970c3cd01453aa4263a8be7998588&wix-vod-comp-id=comp-kn8844mv
 
  
 
https://youtu.be/sVclz6EmpEU
 
https://youtu.be/sVclz6EmpEU
  
 
~savannahsolver
 
~savannahsolver

Revision as of 10:06, 24 July 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 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 $2\sqrt3-\frac{4 \pi}{3}+2\sqrt3= \boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$

~PEKKA

Video Solutions

https://youtu.be/sVclz6EmpEU

~savannahsolver

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