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

m (→‎Solution)
m (→‎Solution)
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==Solution==
 
==Solution==
  
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Let the centers of the circles containing arcs <math>\overarc{SR}</math> and <math>\overarc{TR}</math> be <math>S'</math> and <math>T'</math>, respectively. Extend <math>\overline{US}</math> and <math>\overline{UT}</math> to <math>S'</math> and <math>T'</math>, and connect point <math>S'</math> with point <math>T'</math>.  
+
Let the centers of the circles containing arcs <math>\overarc{SR}</math> and <math>\overarc{TR}</math> be <math>X</math> and <math>Y</math>, respectively. Extend <math>\overline{US}</math> and <math>\overline{UT}</math> to <math>X</math> and <math>Y</math>, and connect point <math>X</math> with point <math>Y</math>.  
−
<asy>draw((1,1.732)--(2,3.464)--(3,1.732));
+
<asy>
−
draw((1,1.732)--(0,0)--(4,0)--(3,1.732));
+
unitsize(1 cm);
−
draw(arc((0,0),(2,0),(1,1.732)));
+
pair U,S,T,R,X,Y;
−
draw(arc((4,0),(3,1.732),(2,0)));
+
U =(2,3.464);
−
label("$U$", (2,3.464), N);
+
S=(1,1.732);
−
label("$S$", (1,1.732), W);
+
T=(3,1.732);
−
label("$T$", (3,1.732), E);
+
R=(2,0);
−
label("$R$", (2,0), S);
+
X=(0,0);
−
label("$S'$", (0,0), W);
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Y=(4,0);
−
label("$T'$", (4,0), E);</asy>
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draw(U--S);
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We can clearly see that <math>\triangle US'T'</math> is an equilateral triangle, because two of its angles are <math>60^\circ</math>, which is the degree measure of <math>\frac{1}{6}</math> a circle. The area of the figure is equal to <math>[\triangle US'T']</math> minus the combined area of the <math>2</math> sectors of the circles. 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 US'T']</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>
+
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 <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>
  
 
==See Also==
 
==See Also==

Revision as of 11:21, 12 November 2018

Problem 25

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

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 $\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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