Difference between revisions of "2017 AMC 8 Problems/Problem 25"
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==Solution== | ==Solution== | ||
− | Let the centers of the circles containing arcs <math>\overarc{SR}</math> and <math>\overarc{TR}</math> be <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> | + | <asy> |
− | + | unitsize(1 cm); | |
− | draw(arc( | + | pair U,S,T,R,X,Y; |
− | draw(arc( | + | U =(2,3.464); |
− | label("$U$", | + | S=(1,1.732); |
− | label("$S$", | + | T=(3,1.732); |
− | label("$T$", | + | R=(2,0); |
− | label("$R$", | + | X=(0,0); |
− | label("$ | + | Y=(4,0); |
− | label("$ | + | draw(U--S); |
− | We can clearly see that <math>\triangle | + | 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, and are line segments each of length 2, and . Arcs and are each one-sixth of a circle with radius 2. What is the area of the region shown?
Solution
Let the centers of the circles containing arcs and be and , respectively. Extend and to and , and connect point with point . We can clearly see that is an equilateral triangle, because the problem states that . We can figure out that and because they are of a circle. The area of the figure is equal to minus the combined area of the sectors of the circles(in red). Using the area formula for an equilateral triangle, where is the side length of the equilateral triangle, is The combined area of the sectors is , which is Thus, our final answer is
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.