Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2017 AMC 8 Problems/Problem 25

Revision as of 20:20, 30 January 2018 by G1zq (talk | contribs) (→‎Solution)

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 $S'$ and $T'$, respectively. Extend $\overline{US}$ and $\overline{UT}$ to $S'$ and $T'$, and connect point $S'$ with point $T'$. [asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw((1,1.732)--(0,0)--(4,0)--(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); label("$S'$", (0,0), W); label("$T'$", (4,0), E);[/asy] We can clearly see that $\triangle US'T'$ is an equilateral triangle, because two of its angles are $60^\circ$, which is the degree measure of $\frac{1}{6}$ a circle. The area of the figure is equal to the area of equilateral triangle $\triangle US'T'$ minus the combined area of the $2$ sectors of the circles. Using the area formula for an equilateral triangle, $\frac{\sqrt{3}}{4} \cdot a,$ where $a$ is the side length of the equilateral triangle, the area of $\triangle US'T'$ is $\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.$ The combined area of the $2$ sectors is $2$ times one sixth $\pi \cdot r^2$, which is $2 \cdot \frac 16 \cdot \pi \cdot 2^2 = \frac{4\pi}{3}.$ Our final answer is then $\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_25&oldid=90087"