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

Latest revision as of 08:59, 14 December 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

Solution 2 (tiny bit intuitional)

We can extend $\overline{US}$ , $\overline{UT}$ to $X$ and $Y$ , respectively, such that $X$ and $Y$ are collinear to point $R$ . Connect $\overline{XY}$ . We can see points $X$ , $Y$ are probably circle centers of arc $SR$ , $TR$ , respectively. So, $\overline{XS} = 2 = \overline{TY}$ . Thus, $\triangle{UXY}$ is equilateral. The area of $\triangle{UXY}$ is $\frac{\sqrt{3}}{4} \cdot 4^2$ , or $4\sqrt{3}$ , and both one sixth circles total up to $\frac{4\pi}{3}$ . Finally, the answer is $\boxed{\textbf{(B)} 4\sqrt{3}-\frac{4\pi}{3}}$ .

~ lovelearning999

Video Solutions

https://youtu.be/sVclz6EmpEU

~savannahsolver

@@ Line 8: / Line 8: @@
 ==Solution 1==
-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>
-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 <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>
-==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>
@@ Line 41: / Line 15: @@
 ~PEKKA
-==Video Solutions==
+== Solution 2 (tiny bit intuitional) ==
-https://youtu.be/LT4gyH--328
-https://youtu.be/wc5rGulTTR8 - Happytwin
+We can extend <math>\overline{US}</math>, <math>\overline{UT}</math> to <math>X</math> and <math>Y</math>, respectively, such that <math>X</math> and <math>Y</math> are collinear to point <math>R</math>. Connect <math>\overline{XY}</math>. We can see points <math>X</math>, <math>Y</math> are probably circle centers of arc <math>SR</math>, <math>TR</math>, respectively. So, <math>\overline{XS} = 2 = \overline{TY}</math>. Thus, <math>\triangle{UXY}</math> is equilateral. The area of <math>\triangle{UXY}</math> is <math>\frac{\sqrt{3}}{4} \cdot 4^2</math>, or <math>4\sqrt{3}</math>, and both one sixth circles total up to <math>\frac{4\pi}{3}</math>. Finally, the answer is <math>\boxed{\textbf{(B)} 4\sqrt{3}-\frac{4\pi}{3}}</math>.
-https://youtu.be/uvwLT5xBNdU ~DSA_Catachu
+~ lovelearning999
-https://youtu.be/aE0oAq4Q_Ks
+==Video Solutions==
-https://euclideanmathcircle.wixsite.com/emc1/videos?wix-vod-video-id=3a7970c3cd01453aa4263a8be7998588&wix-vod-comp-id=comp-kn8844mv
-==See Also==
+https://youtu.be/sVclz6EmpEU
-{{AMC8 box|year=2017|num-b=24|after=Last Problem}}
-{{MAA Notice}}
+~savannahsolver

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